Chapter 8: Electrostatics
In the 1830s, the English scientist Michael Faraday built a large wooden frame in his laboratory and covered it completely with metal foil and wire mesh. He then stepped inside it, carrying his most sensitive charge-detecting instruments, and asked that the outer surface of the enclosure be charged using a powerful electrostatic machine.
Outside, the enclosure crackled. Sparks jumped from its surface, and anyone who touched it would have received a severe shock. Inside, Faraday’s instruments recorded nothing at all — not the faintest trace of an electric field. He confirmed the same idea with a much simpler arrangement, lowering a charged metal ball into a metal pail and showing that the charge always moved to the outer surface of the pail, leaving the inner region untouched.
Faraday had demonstrated something that still protects people and equipment today: a closed conductor shields whatever lies within it. The charge settles on the outside; the interior stays electrically quiet. The same principle is used in aircraft bodies, MRI scanning rooms, screened cables, and — as you will soon see — in the metal body of a car during a thunderstorm.
The interesting question is why. Why must the field inside a conductor be exactly zero? Why does charge refuse to remain in the interior? Answering this means understanding how charge, electric field, potential and energy are linked to one another.
Diagram
Fig. 8.0 – A person seated inside a metal-mesh enclosure with sparks striking its outer surface, and a charge detector inside reading zero.
As you work through this chapter, you will be able to answer each of these questions for yourself.
8.1 Introduction
Why We Return to Gauss’ Law
In Std XI you studied Gauss’ law, which gives the relationship between electric charge and the electric field it produces. In this chapter we put that law to work.
Gauss’ law does more than state a relationship. It gives us an equivalent, and often far easier, method of finding electric field intensity. Instead of adding up the contributions of every small piece of charge one by one, we relate the value of the field on a chosen closed surface to the total charge enclosed by that surface.
This makes Gauss’ law a powerful tool. It is especially useful for calculating the electric field when the field originates from a charge distribution that has sufficient symmetry — for example, a charged sphere, a long charged wire, or a large charged sheet.
The Statement of the Law
Before writing the law, recall two ideas it depends on. Electric flux is a measure of how much of the electric field passes through a given surface. For a small area element \(d\vec{s}\) on the surface, the flux contributed by that element is \(\vec{E}\cdot d\vec{s}\), where the direction of \(d\vec{s}\) is taken along the outward normal to the surface at that point. A Gaussian surface is an imaginary closed surface that we ourselves choose, drawn through the region of interest so that the calculation becomes simple.
With these ideas in place, Gauss’ law is written as
\[\phi = \oint \vec{E} \cdot \overrightarrow{ds} = \frac{q}{\varepsilon_0} \qquad \text{--- (8.1)}\]
Here \(\phi\) is the total flux coming out of a closed surface, measured in N m² C⁻¹; \(\vec{E}\) is the electric field intensity at a point on that surface, in N C⁻¹; \(\overrightarrow{ds}\) is a small area element of the surface, in m²; \(q\) is the total charge inside the closed surface, in coulomb (C); and \(\varepsilon_0\) is the permittivity of free space, whose value is \(8.85 \times 10^{-12}\) C² N⁻¹ m⁻². The circle on the integral sign is a reminder that the integration is carried out over a closed surface.
What the Law Is Telling Us Physically
Read Eq. (8.1) carefully and one point stands out: the right-hand side contains only the charge inside the surface. The size of the surface, its shape, and where exactly the charge sits inside it make no difference to the total flux.
Charges lying outside the closed surface are not absent from the picture — they do contribute to the value of \(\vec{E}\) at points on the surface. But their field lines enter the surface on one side and leave on the other, so the flux they bring in is exactly cancelled by the flux they take out. Their net contribution to the total flux is zero. This is easier to see in a sketch such as Fig. 8.1a.
Diagram
Fig. 8.1a – An irregular closed surface with two charges inside it and one charge outside, showing field lines of the outside charge entering and leaving the surface.
Why Symmetry Is the Deciding Factor
Gauss’ law is true for every closed surface, without exception. But it is only useful for finding \(\vec{E}\) when the charge distribution is symmetric enough.
The reason is simple. In Eq. (8.1), \(\vec{E}\) sits inside an integral. To pull it out and solve for it, we must be able to say that \(E\) has the same magnitude at every point of the chosen surface, and that the angle between \(\vec{E}\) and \(d\vec{s}\) is the same everywhere. Only a symmetric charge distribution, paired with a wisely chosen Gaussian surface, lets us say this.
That is why the three cases we take up next — a uniformly charged spherical shell, an infinitely long straight charged wire, and a large charged plane sheet — are exactly the three basic symmetries: spherical, cylindrical and planar.
Keep these five steps in mind as you read the next section — every derivation there follows exactly this pattern, and questions in examinations are usually built around one of these steps.
8.2 Application of Gauss’ Law
In this section we shall see how to obtain the electric field intensity for some symmetric charge configurations, with the help of some examples.
Three configurations are taken up, one after the other: a uniformly charged spherical shell, an infinitely long straight charged wire, and a large charged plane sheet. In each case the charge is spread out over a region rather than sitting at a single point, so the amount of charge is described by a charge density — charge per unit length, per unit area, or per unit volume, whichever suits the shape.
Each derivation follows the same five steps listed in the previous section, so watch for that pattern rather than trying to memorise three separate results.
8.2.1 Electric Field Intensity due to Uniformly Charged Spherical Shell or Hollow Sphere
Setting Up the Problem
Consider a sphere of radius \(R\) with its centre at O, charged to a uniform surface charge density \(\sigma\) (C/m²), placed in a dielectric medium of permittivity \(\varepsilon\), where \(\varepsilon = \varepsilon_0 k\).
Surface charge density simply means the charge carried by each square metre of the surface. Since the total surface area of the sphere is \(4\pi R^2\), the total charge on the sphere is
\[q = \sigma \times 4\pi R^2\]
The quantity \(k\) is the dielectric constant of the surrounding medium; for air or vacuum \(k = 1\), so that \(\varepsilon = \varepsilon_0\). We shall work in air throughout this derivation.
By Gauss’ theorem, the net flux through a closed Gaussian surface is
\[\phi = \frac{q}{\varepsilon_0} \quad \text{(for air/vacuum, } k = 1\text{)}\]
where \(q\) is the total charge inside the closed surface.
Choosing the Gaussian Surface
To find the electric field intensity at a point P, at a distance \(r\) from the centre of the charged sphere, imagine a concentric Gaussian sphere of radius \(r\) passing through P, as shown in Fig. 8.1. Let \(ds\) be a small area around the point P on this Gaussian surface.
Diagram
Fig. 8.1 – A uniformly charged spherical shell of radius R with a dotted concentric Gaussian sphere of radius r passing through point P, and radially outward field arrows.
Two facts follow purely from the symmetry of the arrangement, and both are needed before the integral can be solved. First, due to symmetry and the spheres being concentric, the electric field at each point on the Gaussian surface has the same magnitude \(E\) and is directed radially outward. Second, the angle between the direction of \(E\) and the normal to the surface of the sphere (\(ds\)) is zero, i.e. \(\cos\theta = 1\).
\[\therefore \; \vec{E}\cdot \overrightarrow{ds} = E\,ds\cos\theta = E\,ds\]
\[\therefore \; \text{flux } d\phi \text{ through the area } ds = E\,ds\]
Carrying Out the Integration
The total electric flux through the Gaussian surface is obtained by adding the contributions of all such small areas:
\[\phi = \oint \vec{E}\cdot \overrightarrow{ds} = \oint E\,ds = E \oint ds\]
Now \(\oint ds\) is nothing but the total surface area of the Gaussian sphere of radius \(r\), which is \(4\pi r^2\).
\[\therefore \; \phi = E\,4\pi r^2 \qquad \text{--- (8.2)}\]
From equations (8.1) and (8.2),
\[\frac{q}{\varepsilon_0} = E\,4\pi r^2\]
\[\therefore \; E = \frac{q}{4\pi \varepsilon_0 r^2} \qquad \text{--- (8.3)}\]
Since \(q = \sigma \times 4\pi R^2\),
\[E = \frac{\sigma \times 4\pi R^2}{4\pi \varepsilon_0 r^2}\]
\[\therefore \; E = \frac{\sigma R^2}{\varepsilon_0 r^2} \qquad \text{--- (8.4)}\]
In these results \(E\) is the electric field intensity in N C⁻¹, \(q\) is the total charge on the shell in coulomb (C), \(\sigma\) is the surface charge density in C m⁻², \(R\) is the radius of the charged shell and \(r\) is the distance of the point from its centre, both in metre (m).
Reading the Result
From Eq. (8.3) it can be seen that the electric field at a point outside the shell is the same as that due to a point charge. Notice that the radius \(R\) of the shell has disappeared completely from the expression — only the total charge \(q\) and the distance \(r\) from the centre survive.
Thus it can be concluded that a uniformly charged sphere is equivalent to a point charge at its centre.
The Two Special Cases
Case (i) If point P lies on the surface of the charged sphere: here \(r = R\), and Eq. (8.3) gives
\[\therefore \; E = \frac{q}{4\pi \varepsilon_0 R^2} = \frac{\sigma}{\varepsilon_0}\]
The second form follows by substituting \(q = \sigma \times 4\pi R^2\) and cancelling. It is worth remembering, because it shows that the field just at the surface depends only on how densely the charge is packed there, and not on how large the sphere is.
Case (ii) If point P lies inside the sphere: the Gaussian sphere of radius \(r < R\) now lies entirely within the hollow region. Since there are no charges inside, the charge enclosed is zero,
\[\therefore \; E = 0\]
The complete behaviour of the field is easiest to grasp as a graph: zero everywhere inside, a sudden jump to \(\sigma/\varepsilon_0\) at the surface, and then a steady \(1/r^2\) fall outside, as shown in Fig. 8.1b.
Diagram
Fig. 8.1b – Graph of E against r for a charged spherical shell: zero from r = 0 to r = R, a vertical jump at r = R, then a curve falling as 1/r².
8.2.2 Electric Field Intensity due to an Infinitely Long Straight Charged Wire
Setting Up the Problem
Consider a uniformly charged wire of infinite length having a constant linear charge density \(\lambda\) (charge per unit length), kept in a medium of permittivity \(\varepsilon\), where \(\varepsilon = \varepsilon_0 k\).
Linear charge density is the counterpart, for a thin wire, of the surface charge density used for the shell. It tells us how much charge each metre of the wire carries, and its SI unit is C m⁻¹. If a length \(l\) of the wire carries a charge \(q\), then \(\lambda = q/l\).
The phrase “infinite length” is an idealisation that keeps the symmetry perfect. In practice the result is used for a wire that is long compared with the distance of the point from it, with the point taken near the middle of the wire so that the two ends are effectively far away.
Choosing the Gaussian Surface
To find the electric field intensity at P, at a distance \(r\) from the axis of the charged wire, imagine a coaxial Gaussian cylinder of length \(l\) and radius \(r\), closed at each end by plane caps normal to the axis, passing through the point P. This arrangement is shown in Fig. 8.2. Consider a very small area \(ds\) at the point P on the Gaussian surface.
Diagram
Fig. 8.2 – An infinitely long straight charged wire with a coaxial Gaussian cylinder of radius r and length l around it, showing a small area ds at point P on the curved surface.
A cylinder is the natural choice here because the wire looks exactly the same from every direction perpendicular to it, and looks the same all along its length. Any other shape would break that match and make the integral impossible to simplify.
By symmetry, the magnitude of the electric field will be the same at all the points on the curved surface of the cylinder, and will be directed radially outward. The angle between the direction of \(E\) and the normal to the curved or flat surface of the cylinder (\(ds\)) is zero or \(\pi/2\), i.e. \(\cos\theta = 1\) or \(\cos(\pi/2) = 0\).
\[\therefore \; E\cdot ds = E\,ds\cos\theta = E\,ds\]
\[\text{Flux } d\phi \text{ through the area } ds = E\,ds\]
Carrying Out the Integration
The total electric flux through the Gaussian cylindrical surface is
\[\phi = \oint \vec{E}\cdot \overrightarrow{ds} = \oint E\,ds = E \oint ds\]
Here \(\oint ds\) is the area of the curved surface of the cylinder, which is the circumference \(2\pi r\) multiplied by the length \(l\).
\[\therefore \; \phi = E \cdot 2\pi r l \qquad \text{--- (8.5)}\]
From equations (8.1) and (8.5),
\[\frac{q}{\varepsilon_0} = E\,2\pi r l\]
Since \(\lambda = q/l\), we have \(q = \lambda l\), and the charge enclosed by the Gaussian cylinder can be written in terms of the charge density:
\[\therefore \; \frac{\lambda l}{\varepsilon_0} = E\,2\pi r l\]
The length \(l\) appears on both sides and cancels, which is exactly what we should expect — the answer cannot depend on how long a stretch of wire we chose to enclose.
\[E = \frac{\lambda}{2\pi \varepsilon_0 r} \qquad \text{--- (8.6)}\]
In this result \(E\) is the electric field intensity in N C⁻¹, \(\lambda\) is the linear charge density in C m⁻¹, \(r\) is the perpendicular distance of the point from the axis of the wire in metre, and \(\varepsilon_0\) is the permittivity of free space.
Direction and Behaviour of the Field
The direction of the electric field \(E\) is directed outward if \(\lambda\) is positive and inward if \(\lambda\) is negative, as shown in Fig. 8.3.
Diagram
Fig. 8.3 – Two charged wires side by side, one positive with field arrows pointing radially outward and one negative with arrows pointing radially inward.
8.2.3 Electric Field due to a Charged Infinite Plane Sheet
Setting Up the Problem
Consider a uniformly charged infinite plane sheet with surface charge density \(\sigma\). By symmetry, the electric field is perpendicular to the plane sheet and directed outwards, having the same magnitude at a given distance on either side of the sheet. Let P be a point at a distance \(r\) from the sheet, and \(E\) be the electric field at P.
It is worth pausing on why the field must be perpendicular to the sheet. If the field had any component running parallel to the surface, that component would have to point in some particular direction along the sheet. But an infinite uniformly charged sheet looks exactly the same in every direction along its surface, so there is no reason for the field to prefer one direction over another. The only possibility left is that the field points straight out, at right angles to the sheet.
Choosing the Gaussian Surface
To find the electric field due to a charged infinite plane sheet at P, we consider a Gaussian surface around P in the form of a cylinder having cross-sectional area \(A\) and length \(2r\), with its axis perpendicular to the plane sheet. The plane sheet passes through the middle of the length of the cylinder, such that the ends of the cylinder (called end caps P and P’) are equidistant, at a distance \(r\), from the plane sheet. This arrangement is shown in Fig. 8.4.
Diagram
Fig. 8.4 – A charged infinite plane sheet passing through the middle of a cylindrical Gaussian surface of cross-section A, with end caps P and P’ at distance r on either side and field arrows pointing outward from both faces.
Note that this cylinder is placed differently from the one used for the charged wire. There, the axis lay along the charge; here, the axis is perpendicular to the sheet. In each case the surface is chosen so that the field is either exactly along the outward normal or exactly perpendicular to it, never at some awkward angle in between.
By symmetry, the electric field is at right angles to the end caps and away from the plane. Its magnitude is the same at P and P’. The flux passing through the curved surface is zero, as the electric field is tangential to this surface.
Carrying Out the Integration
The total flux through the closed Gaussian surface is therefore the sum of three parts:
\[\phi = \left[\oint E\,ds\right]_{P} + \left[\oint E\,ds\right]_{P'} + \left[\oint E\,ds\right]_{\text{curved surface}}\]
For both end caps the field is along the outward normal, so \(\theta = 0\) and \(\cos\theta = 1\); for the curved surface the contribution vanishes as explained above.
\[= EA + EA\]
\[\therefore \; \phi = 2EA \qquad \text{--- (8.7)}\]
If \(\sigma\) is the surface charge density, then
\[\sigma = q/A, \qquad q = \sigma A\]
Here \(q\) is the charge on that patch of the sheet which is cut out by the Gaussian cylinder, and this is the only charge enclosed by the surface. Hence Eq. (8.1) can be written as
\[\phi = \frac{\sigma A}{\varepsilon_0} \qquad \text{--- (8.8)}\]
From Eq. (8.7) and Eq. (8.8),
\[2EA = \frac{\sigma A}{\varepsilon_0} \qquad \therefore \; E = \frac{\sigma}{2\varepsilon_0}\]
Here \(E\) is the electric field intensity in N C⁻¹, \(\sigma\) is the surface charge density of the sheet in C m⁻², and \(\varepsilon_0\) is the permittivity of free space. The area \(A\) cancels out, just as the length \(l\) cancelled in the case of the wire — the answer cannot depend on the size of an imaginary surface that we invented for our own convenience.
A Field That Does Not Fade
The most striking feature of this result is what is missing from it: the distance \(r\). Move twice as far from the sheet and the field is exactly the same. Compare this with the \(1/r^2\) fall for a point charge and the \(1/r\) fall for a long wire.
A second point deserves attention because it is a frequent source of error. In section 8.2.1 the field just at the surface of a charged sphere came out as \(\sigma/\varepsilon_0\), while here it is \(\sigma/2\varepsilon_0\). The factor of two is not a slip. For the sheet, the field spreads out on both sides, so the flux is shared between two end caps; for the charged shell, the interior field is zero and all the flux emerges on one side only.
8.3 Electric Potential and Potential Energy
What Electrostatic Potential Energy Means
We have studied earlier that the potential energy of a system is the stored energy that depends upon the relative positions of its constituents. Electrostatic potential energy is the work done against the electrostatic forces to achieve a certain configuration of charges in a given system. Since every system tries to attain the lowest potential energy, work is always required to be done to change the configuration.
This is the same idea you met with gravity. Lifting a stone against the earth’s pull stores energy in the stone–earth system, and the stone gives that energy back the moment it is released. Charges behave in the same way, except that the electric force can be either attractive or repulsive, so energy may have to be spent either to push charges together or to pull them apart.
Why Any Collection of Charges Stores Energy
We know that like charges repel and unlike charges attract each other. A charge exerts a force on any other charge in its vicinity. Some work is always done to move a charge in the presence of another charge. Thus, any collection of charges possesses potential energy.
Consider a positive charge \(Q\) fixed at some point in space. For bringing any other positive charge close to it, work is necessary, because the two charges push each other apart the whole way. This work is equal to the change in the potential energy of the system of two charges.
Thus, work done against an electrostatic force = increase in the potential energy of the system.
\[\therefore \; \vec{F}\cdot d\vec{r} = dU\]
where \(dU\) is the increase in potential energy when the charge is displaced through \(d\vec{r}\), and \(\vec{F}\) is the force exerted on the charge.
Note carefully that the energy belongs to the system, not to either charge separately. It is the arrangement of the two charges that stores the energy, in the same way that it is the stone-and-earth arrangement, and not the stone by itself, that stores gravitational potential energy.
Expression for Potential Energy
Let us consider the electrostatic field due to a source charge \(+Q\) placed at the origin O. Let a small charge \(+q_0\) be brought from point A to point B, at respective distances \(r_1\) and \(r_2\) from O, against the repulsive forces acting on it. The arrangement is shown in Fig. 8.5.
Diagram
Fig. 8.5 – A fixed charge +Q at origin O and a test charge +q₀ being displaced through a small distance dr towards Q, with the repulsive force and the displacement marked in opposite directions.
Work done against the electrostatic force \(\vec{F}_E\), in displacing the charge \(q_0\) through a small displacement \(d\vec{r}\), appears as an increase in the potential energy of the system.
\[dU = -\vec{F}_E \cdot d\vec{r} = -F_E\,dr\]
The negative sign appears because the displacement \(d\vec{r}\) is against the electrostatic force \(\vec{F}_E\). In other words, the external force we must apply is equal in magnitude to \(\vec{F}_E\) but opposite in direction, and it is this applied force that does positive work.
For the displacement of the charge from the initial position A to the final position B, the change in potential energy \(\Delta U\) can be obtained by integrating \(dU\):
\[\therefore \; \Delta U = \int_{A}^{B} dU = -\int_{A}^{B} \vec{F}_E \cdot d\vec{r}\]
The electrostatic force (Coulomb force) between the two charges separated by distance \(r\) is
\[\vec{F}_E = \frac{1}{4\pi\varepsilon_0}\,\frac{Qq_0}{r^2}\,\hat{r}\]
where \(\hat{r}\) is the unit vector in the direction of \(\vec{r}\). Here \(Q\) and \(q_0\) are the two charges in coulomb, \(r\) is their separation in metre, and \(F_E\) is in newton.
\[\therefore \; \Delta U = \int_{A}^{B} dU = -\int_{r_1}^{r_2} \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{r^2}\,\hat{r}\cdot d\vec{r}\]
Performing the integration, using \(\displaystyle\int \frac{dr}{r^2} = -\frac{1}{r}\), we obtain
\[\Delta U = \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{r_2} - \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{r_1}\]
\[\therefore \; \Delta U = \frac{Qq_0}{4\pi\varepsilon_0}\left(\frac{1}{r_2} - \frac{1}{r_1}\right)\]
The change in the potential energy depends only upon the end points and is independent of the path taken by the charge. Only \(r_1\) and \(r_2\) appear in the answer; nothing about the route between A and B survives. This is precisely what we mean by calling the electrostatic force conservative.
The change in potential energy is equal to the work done \(W_{AB}\) against the electrostatic force.
\[W_{AB} = \Delta U = \frac{Qq_0}{4\pi\varepsilon_0}\left(\frac{1}{r_2} - \frac{1}{r_1}\right)\]
Fixing the Zero of Potential Energy
So far we have defined and calculated only the change in the potential energy for a system of charges. To speak of the potential energy at a single point, we need a reference position where the energy is agreed to be zero. It is convenient to choose infinity to be the point of zero potential energy, as the electrostatic force is zero at \(r \to \infty\).
Thus, the potential energy \(U\) of the system of two point charges \(q_1\) and \(q_2\) separated by \(r\) can be obtained from the above equation by using \(r_1 = \infty\) and \(r_2 = r\). It is then given by
\[U(r) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1 q_2}{r}\right) \qquad \text{--- (8.9)}\]
Here \(U\) is in joule, \(q_1\) and \(q_2\) are in coulomb and \(r\) is in metre. Note that the sign takes care of itself: for two like charges \(U\) is positive, while for a pair of unlike charges \(U\) is negative, indicating that work must be done on the system to pull them apart to infinity.
Units of Potential Energy
The SI unit of potential energy is the joule (J).
\[1\ \text{eV} = 1.6 \times 10^{-19}\ \text{joule}\]
Other related units are:
\[1\ \text{meV} = 1.6 \times 10^{-22}\ \text{J}\]
\[1\ \text{keV} = 1.6 \times 10^{-16}\ \text{J}\]
Concept of Potential
Equation (8.9) gives the potential energy of a two-particle system at a distance \(r\) from each other. Let us rewrite it by deliberately separating one charge from the rest of the expression:
\[U(r) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1 q_2}{r}\right) = q_1\left(\frac{q_2}{4\pi\varepsilon_0 r}\right) = q_2\left(\frac{q_1}{4\pi\varepsilon_0 r}\right)\]
The quantity
\[V(r) = \frac{q}{4\pi\varepsilon_0 r}\]
depends upon the charge \(q\) and the location of a point at a distance \(r\) from it. This is defined as the electrostatic potential of the charge \(q\) at a distance \(r\) from it.
The usefulness of this step is worth appreciating. The bracketed quantity describes what the source charge does to the space around it, regardless of what charge we may later place there. It is a property of the field, not of the pair.
In terms of potential, we can write the potential energy of the system of two charges as
\[U(r) = V_1(r)\,q_2 = V_2(r)\,q_1\]
where \(V_1(r)\) and \(V_2(r)\) are the respective potentials of charges \(q_1\) and \(q_2\) at distance \(r\) from either.
\[\therefore \; \text{Electrostatic potential energy } (U) = \text{electric potential } V \times \text{charge } q\]
Or, electrostatic potential \((V)\) = electrostatic potential energy per unit charge, i.e.
\[V = U/q\]
Electrostatic potential difference between any two points in an electric field can be written as
\[V_2 - V_1 = \frac{U_2 - U_1}{q} = \frac{dW}{q}\]
that is, the work done \(dW\) (or change in PE) per unit charge to move the charge from point 2 to point 1.
Relation Between Electric Field and Electric Potential
Consider the electric field produced by a charge \(+q\) kept at point O (see Fig. 8.6). Let us calculate the work done to move a unit positive charge from point M to point N, which is at a small distance \(dx\) from M.
The direction of the electric field at M is along \(\overrightarrow{OM}\). Thus the force acting on the unit positive charge is along \(\overrightarrow{OM}\). The work done is
\[dW = -F\,dx = -E\,dx\]
The negative sign indicates that we are moving the charge against the force acting on it. As it is the work done on a unit positive charge, \(dW = dV\) = difference in potential between M and N.
\[\therefore \; dV = -E\,dx\]
\[\therefore \; E = -\frac{dV}{dx}\]
Thus the electric field at a point is the negative gradient of the potential at that point.
This relation is useful in both directions. If the potential is known everywhere, the field can be obtained by differentiation; and it tells us that the field is strong wherever the potential changes rapidly over a short distance. It also gives us an alternative SI unit for electric field: since \(E = -dV/dx\), the field may be measured in volt per metre (V m⁻¹), which is exactly the same as N C⁻¹.
Zero Potential
The nature of potential is such that its zero point is arbitrary. This does not mean that the choice of zero point is insignificant. Once the zero point of the potential is set, then every potential is measured with respect to that reference. The zero potential is set conveniently.
In the case of a point charge or a localised collection of charges, the zero point is set at infinity. For electrical circuits, the earth is usually taken to be at zero potential.
Thus the potential at a point A in an electric field is the amount of work done to bring a unit positive charge from infinity to point A.
8.4 Electric Potential due to a Point Charge, a Dipole and a System of Charges
a) Electric potential due to a point charge
Here, we shall derive an expression for the electrostatic potential due to a point charge.
Figure 8.6 shows a point charge \(+q\), located at point O. We need to determine its potential at a point A, at a distance \(r\) from it.
Diagram
Fig. 8.6 – A point charge +q at O, a point A at distance r, and points N and M further out on the line OA extended, with the small displacement dx between M and N marked.
As seen above, the electric potential at a point A is the amount of work done per unit positive charge, which is displaced from \(\infty\) to point A. As the work done is independent of the path, we choose a convenient path along the line extending OA to \(\infty\). Choosing this straight radial path is what makes the calculation easy — along it, the force and the displacement lie along the same line, so no components need to be resolved.
Let M be an intermediate point on this path where OM \(= x\). The electrostatic force on a unit positive charge at M is of magnitude
\[F = \frac{1}{4\pi\epsilon_0}\times \frac{q}{x^{2}} \qquad \text{--- (8.10)}\]
It is directed away from O, along OM. For an infinitesimal displacement \(dx\) from M to N, the amount of work done is given by
\[\therefore \; dW = -F\,dx \qquad \text{--- (8.11)}\]
The negative sign appears as the displacement is directed opposite to that of the force.
\(\therefore\) Total work done in displacing the unit positive charge from \(\infty\) to point A is given by
\[W = \int_{\infty}^{r} -F\,dx = \int_{\infty}^{r} -\frac{1}{4\pi\epsilon_0}\frac{q}{x^{2}}\,dx\]
\[= \frac{-q}{4\pi\epsilon_0}\int_{\infty}^{r} x^{-2}\,dx\]
\[= \frac{-q}{4\pi\epsilon_0}\left[\frac{-1}{x}\right]_{\infty}^{r} \qquad \left(\because \int x^{-2}\,dx = \frac{-1}{x}\right)\]
\[= \frac{q}{4\pi\epsilon_0}\left[\frac{1}{r}-\frac{1}{\infty}\right] \qquad \left(\because \frac{1}{\infty}=0\right)\]
\[W = \frac{q}{4\pi\epsilon_0 r} \qquad \text{--- (8.12)}\]
By definition this is the electrostatic potential at A due to charge \(q\).
\[\therefore V = W = \frac{q}{4\pi\epsilon_0 r} \qquad \text{--- (8.13)}\]
Here \(V\) is in volt, \(q\) in coulomb and \(r\) in metre.
What the Expression Tells Us
A positively charged particle produces a positive electric potential and a negatively charged particle produces a negative electric potential. Unlike the electric field, potential is a scalar — it carries a sign but no direction, and the sign comes directly from the sign of \(q\).
\[\text{At } r = \infty, \; V = \frac{q}{\infty} = 0\]
This shows that the electrostatic potential is zero at infinity, which is consistent with the reference we chose in the previous section.
Equation (8.13) shows that for any point at a distance \(r\) from the point charge \(q\), the value of \(V\) is the same and is independent of the direction of \(r\). Hence electrostatic potential due to a single charge is spherically symmetric.
Figure 8.7 shows how electric potential (\(V \propto \frac{1}{r}\)) and electric field (\(E \propto \frac{1}{r^{2}}\)) vary with \(r\), the distance from the charge.
Diagram
Fig. 8.7 – Two curves on the same axes against distance r: a dashed 1/r curve for potential and a solid 1/r² curve for electric field, both falling steeply near the origin.
b) Electric potential due to an electric dipole
We have studied electric and magnetic dipoles in Std XI. Figure 8.8 shows an electric dipole AB consisting of two charges \(+q\) and \(-q\) separated by a finite distance \(2l\). Its dipole moment is \(\vec{p}\), of magnitude \(p = q \times 2l\), directed from \(-q\) to \(+q\). The line joining the centres of the two charges is called the dipole axis. A straight line drawn perpendicular to the axis and passing through the centre O of the electric dipole is called the equator of the dipole.
In order to determine the electric potential due to a dipole, let the origin be at the centre (O) of the dipole.
Let C be any point near the electric dipole at a distance \(r\) from the centre O, inclined at an angle \(\theta\) with the axis of the dipole. \(r_1\) and \(r_2\) are the distances of point C from charges \(+q\) and \(-q\), respectively.
Diagram
Fig. 8.8 – An electric dipole with +q at A and −q at B separated by 2l, centre O, and a point C at distance r making an angle θ with the axis, with r₁ and r₂ marked from the two charges.
Potential at C due to charge \(+q\) at A is
\[V_1 = +\frac{q}{4\pi\varepsilon_0 r_1}\]
Potential at C due to charge \(-q\) at B is
\[V_2 = -\frac{q}{4\pi\varepsilon_0 r_2}\]
The electrostatic potential is the work done by the electric field per unit charge, \(V = \left(\dfrac{W}{Q}\right)\).
The potential at C due to the dipole is the algebraic sum of these two, since potential is a scalar:
\[V_C = V_1 + V_2 = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{r_1}-\frac{1}{r_2}\right]\]
Simplifying for a Short Dipole
By geometry,
\[r_1^{2} = r^{2}+l^{2}-2rl\cos\theta\]
\[r_2^{2} = r^{2}+l^{2}+2rl\cos\theta\]
These are simply the cosine rule applied to the two triangles OAC and OBC; the sign of the last term differs because the angle at O is \(\theta\) in one case and \(180^\circ - \theta\) in the other. Taking \(r^2\) common,
\[r_1^{2} = r^{2}\left(1+\frac{l^{2}}{r^{2}}-\frac{2l\cos\theta}{r}\right)\]
\[r_2^{2} = r^{2}\left(1+\frac{l^{2}}{r^{2}}+\frac{2l\cos\theta}{r}\right)\]
For a short dipole, \(2l \ll r\), and if \(r \gg l\) then \(l/r\) is small, so \(\dfrac{l^{2}}{r^{2}}\) can be neglected. A small quantity squared is very much smaller still, which is why the middle term goes first.
\[\therefore \; r_1^{2} = r^{2}\left(1-\frac{2l\cos\theta}{r}\right), \qquad r_2^{2} = r^{2}\left(1+\frac{2l\cos\theta}{r}\right)\]
Taking square roots,
\[r_1 = r\left(1-\frac{2l\cos\theta}{r}\right)^{1/2}, \qquad r_2 = r\left(1+\frac{2l\cos\theta}{r}\right)^{1/2}\]
\[\therefore \; \frac{1}{r_1} = \frac{1}{r}\left(1-\frac{2l\cos\theta}{r}\right)^{-1/2}, \qquad \frac{1}{r_2} = \frac{1}{r}\left(1+\frac{2l\cos\theta}{r}\right)^{-1/2}\]
and
\[V_C = V_1 + V_2 = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{r}\left(1-\frac{2l\cos\theta}{r}\right)^{-1/2} - \frac{1}{r}\left(1+\frac{2l\cos\theta}{r}\right)^{-1/2}\right]\]
Using binomial expansion, \((1+x)^{n} = 1+nx\) for \(x \ll 1\), and retaining terms up to the first order of \(\dfrac{l}{r}\) only, we get
\[V_C = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{r}\left(1+\frac{l\cos\theta}{r}\right)-\frac{1}{r}\left(1-\frac{l\cos\theta}{r}\right)\right]\]
\[= \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{r}+\frac{l\cos\theta}{r^{2}}-\frac{1}{r}+\frac{l\cos\theta}{r^{2}}\right]\]
\[= \frac{q}{4\pi\varepsilon_0}\left[\frac{2l\cos\theta}{r^{2}}\right]\]
\[\therefore \; V_C = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^{2}} \qquad (\because p = 2ql)\]
Electric potential at C can also be expressed as
\[V_C = \frac{1}{4\pi\varepsilon_0}\frac{\vec{p}\cdot\vec{r}}{r^{3}}\]
\[V_C = \frac{1}{4\pi\varepsilon_0}\left(\frac{\vec{p}\cdot\hat{r}}{r^{2}}\right)\]
where \(\hat{r}\) is a unit vector along the position vector \(\overrightarrow{OC} = \vec{r}\).
The Two Important Directions
i) Potential at an axial point, \(\theta = 0^\circ\) (towards \(+q\)) or \(180^\circ\) (towards \(-q\)):
\[V_{\text{axial}} = \pm\frac{1}{4\pi\varepsilon_0}\frac{p}{r^{2}}\]
i.e. this is the maximum value of the potential, since \(\cos\theta\) can be no larger than 1 in magnitude.
ii) Potential at an equatorial point, \(\theta = 90^\circ\) and \(V = 0\), because \(\cos 90^\circ = 0\).
Hence, the potential at any point on the equatorial line of a dipole is zero. This is the minimum value of the magnitude of the potential of a dipole. The reason is easy to see physically: every point on the equator is equidistant from \(+q\) and \(-q\), so the positive and negative contributions are equal in size and cancel exactly.
Thus the plane perpendicular to the line between the charges at the midpoint is an equipotential plane with potential zero. The work done to move a charge anywhere in this plane (the potential difference being zero) will be zero.
c) Electrostatic potential due to a system of charges
We now extend the analysis to a system of charges.
Consider a system of charges \(q_1, q_2, \ldots q_n\) at distances \(r_1, r_2, \ldots r_n\) respectively from point P, as shown in Fig. 8.9. The potential \(V_1\) at P due to the charge \(q_1\) is
\[V_1 = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_1}\]
Diagram
Fig. 8.9 – Several point charges q₁, q₂, q₃ … qₙ scattered in space with their distances r₁, r₂, r₃ … rₙ drawn to a common point P.
Similarly the potentials \(V_2, V_3, \ldots V_n\) at P due to the individual charges \(q_2, q_3, \ldots q_n\) are given by
\[V_2 = \frac{1}{4\pi\varepsilon_0}\frac{q_2}{r_2}, \quad V_3 = \frac{1}{4\pi\varepsilon_0}\frac{q_3}{r_3}, \quad \ldots \quad V_n = \frac{1}{4\pi\varepsilon_0}\frac{q_n}{r_n}\]
By the superposition principle, the potential \(V\) at P due to the system of charges is the algebraic sum of the potentials due to the individual charges.
\[\therefore V = V_1 + V_2 + \cdots + V_n = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+ \cdots +\frac{q_n}{r_n}\right)\]
\[\text{Or, } \; V = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_i}\]
The word algebraic is the crucial one. Since potential is a scalar, negative charges enter with a minus sign and the whole calculation is ordinary arithmetic — no components, no angles, no parallelogram law. This is why potential is often an easier route to a problem than the field.
For a continuous charge distribution, the summation should be replaced by integration.
8.5 Equipotential Surfaces
What an Equipotential Surface Is
An equipotential surface is that surface, at every point of which the electric potential is the same.
We know that the potential \(V\) for a single charge \(q\) is given by
\[V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}\]
If \(r\) is constant then \(V\) will be constant. Hence, equipotential surfaces of a single point charge are concentric spherical surfaces centered at the charge. Since the potential of a point charge depends on nothing except the distance \(r\), the set of all points sharing one value of \(V\) is simply the set of all points at one fixed distance — that is, a sphere.
For a line charge, the shape of the equipotential surface is cylindrical, for exactly the same reason: the potential there depends only on the perpendicular distance from the line. A family of such surfaces around a charge is shown in Fig. 8.10, drawn together with the lines of force.
Diagram
Fig. 8.10 – Concentric circular equipotential surfaces around a central charge, with radial lines of force cutting them at right angles.
Equipotential surfaces can be drawn through any region in which there is an electric field. There is nothing special about the particular surfaces that are drawn — through every single point of the region one such surface passes, and we draw only a selection of them, just as a map shows only a few contour lines out of infinitely many possible ones.
No Work Is Done Along an Equipotential Surface
By definition, the potential difference between two points P and Q is the work done per unit positive charge displaced from Q to P.
\[\therefore \; V_P - V_Q = W_{QP}\]
If points P and Q lie on an equipotential surface, \(V_P = V_Q\).
\[\therefore \; W_{QP} = 0\]
Thus, no work is required to move a test charge along an equipotential surface.
Why the Field Must Be Perpendicular to the Surface
a) If \(dx\) is the small distance over the equipotential surface through which a unit positive charge is carried, then
\[dW = \vec{E}\cdot d\vec{x} = E\,dx\cos\theta = 0\]
Since neither \(E\) nor \(dx\) is zero, the only way this product can vanish is through the cosine term.
\[\therefore \; \cos\theta = 0 \; \text{ or } \; \theta = 90^{\circ}\]
i.e. \(\vec{E} \perp d\vec{x}\), as shown in Fig. 8.11.
Diagram
Fig. 8.11 – A curved equipotential surface with a small displacement dx lying in the surface and the field vector E drawn perpendicular to it at that point.
Hence electric field intensity \(\vec{E}\) is always normal to the equipotential surface, i.e. for any charge distribution, the equipotential surface through a point is normal to the electric field at that point.
b) If the field is not normal, it would have a nonzero component along the surface. So to move a test charge against this component, work would have to be done. But by the definition of equipotential surfaces, there is no potential difference between any two points on an equipotential surface, and hence no work is required to displace the charge on the surface. Therefore, we can conclude that the electrostatic field must be normal to the equipotential surface at every point, and vice versa.
Reading a Picture of Equipotential Surfaces
For a uniform electric field, the equipotential surfaces are a set of parallel planes at right angles to the field, as shown in Fig. 8.12.
Diagram
Fig. 8.12 – Equally spaced parallel plane equipotential surfaces cutting across horizontal field lines of a uniform electric field.
For an electric dipole the surfaces are no longer simple, closing around each charge and flattening into the zero-potential plane through the centre, as shown in Fig. 8.13. For two identical positive charges the pattern is different again, as shown in Fig. 8.14 — the surfaces wrap each charge separately when close in, and merge into a single surface enclosing both when far away.
Diagram
Fig. 8.13 – Equipotential surfaces around an electric dipole, drawn as closed curves about each charge with dashed field lines crossing them at right angles.
Diagram
Fig. 8.14 – Equipotential surfaces for two identical positive charges, showing separate closed curves near each charge merging into common curves further out.
Practical arrangements show the same behaviour. Between two plane metallic sheets the surfaces are parallel planes, as in Fig. 8.15 (a); when one of the sheets is replaced by a charged metallic sphere, the surfaces curve smoothly from spherical near the sphere to flat near the plate, as in Fig. 8.15 (b).
Diagram
Fig. 8.15 – (a) Parallel plane equipotential surfaces between two plane metallic sheets; (b) the same arrangement with one sheet replaced by a charged metallic sphere, the surfaces curving from spherical to planar.
Like the lines of force, the equipotential surfaces give a visual picture of both the direction and the magnitude of the electric field in a region of space. The direction is read off at once, since the field is perpendicular to the surfaces and points from higher to lower potential. The magnitude is read from the spacing: because \(E = -dV/dx\), a large field means the potential changes rapidly over a short distance, so surfaces drawn at equal potential intervals crowd together where the field is strong and spread apart where it is weak.
8.6 Electrostatic Potential Energy of Two Point Charges and of a Dipole in an Electrostatic Field
The Idea of Assembling a Configuration
When two like charges lie an infinite distance apart, their potential energy is zero, because no work has to be done in moving one charge at an infinite distance from the other. But when they are brought closer to one another, work has to be done against the force of repulsion. As the electrostatic force is conservative, this work gets stored as the potential energy of the two charges.
The phrase “bringing them from infinity” is the key to every calculation in this section. We imagine the charges starting infinitely far apart, where they neither help nor hinder one another, and we count up the work done as they are put in place one at a time.
a) Potential energy of a system of two point charges
Let us consider 2 charges \(q_1\) and \(q_2\) with position vectors \(\vec{r_1}\) and \(\vec{r_2}\) relative to the origin (O), as shown in Fig. 8.16.
Diagram
Fig. 8.16 – Two point charges q₁ at A and q₂ at B with position vectors r₁ and r₂ drawn from a common origin O, and their mutual separation r₁₂ marked.
To calculate the electric potential energy of the two-charge system, we assume that the two charges \(q_1\) and \(q_2\) are initially at infinity. We then determine the work done in bringing the charges to the given location by an external agency.
In bringing the first charge \(q_1\) to position A\((\vec{r_1})\), no work is done, because there is no external field against which work is required to be done, as charge \(q_2\) is still at infinity, i.e. \(W_1 = 0\). This charge produces a potential in space given by
\[V_1 = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_1} \qquad \text{--- (8.14)}\]
where \(r_1\) is the distance of point A from the origin.
When we bring charge \(q_2\) from infinity to B\((\vec{r_2})\), at a distance \(r_{12}\) from \(q_1\), the work done is
\[W_2 = (\text{potential at B due to charge } q_1)\times q_2 = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_{12}}\,q_2, \quad (\text{where AB}=r_{12}) \qquad \text{--- (8.15)}\]
This work done in bringing the two charges to their respective locations is stored as the potential energy of the configuration of two charges.
\[\therefore \; U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}} \qquad \text{--- (8.16)}\]
Here \(U\) is in joule, \(q_1\) and \(q_2\) in coulomb, and \(r_{12}\), the separation between the two charges, in metre. Equation (8.16) can be generalised for a system of any number of point charges.
Notice that only the separation \(r_{12}\) appears in the final answer, and not the individual position vectors \(\vec{r_1}\) and \(\vec{r_2}\). The energy of a pair depends on how far apart the charges are, not on where the pair happens to sit.
b) Potential energy for a system of N point charges
Equation (8.16) gives an expression for potential energy for a system of two charges. We now analyse the situation for a system of \(N\) point charges.
In bringing a charge \(q_3\) from \(\infty\) to C\((\vec{r_3})\), work has to be done against the electrostatic forces of both \(q_1\) and \(q_2\), since both are now already in place.
\[\therefore \; W_3 = (\text{potential at C due to } q_1 \text{ and } q_2)\times q_3 = \frac{1}{4\pi\varepsilon_0}\left[\frac{q_1}{r_{13}}+\frac{q_2}{r_{23}}\right]\times q_3\]
\[= \frac{1}{4\pi\varepsilon_0}\left[\frac{q_1 q_3}{r_{13}}+\frac{q_2 q_3}{r_{23}}\right]\]
Similarly, in bringing a charge \(q_4\) from \(\infty\) to D\((\vec{r_4})\), work has to be done against the electrostatic forces of \(q_1\), \(q_2\) and \(q_3\):
\[W_4 = \frac{1}{4\pi\varepsilon_0}\left[\frac{q_1 q_4}{r_{14}}+\frac{q_2 q_4}{r_{24}}+\frac{q_3 q_4}{r_{34}}\right]\]
Proceeding in the same way, we can write the electrostatic potential energy of a system of \(N\) point charges at \(\vec{r_1}, \vec{r_2}, \ldots \vec{r_N}\) as
\[U = \frac{1}{4\pi\varepsilon_0}\sum_{\text{all pairs}}\frac{q_j q_k}{r_{jk}}\]
The instruction “all pairs” is what makes this formula easy to use and easy to get wrong. Every pair of charges is counted once and only once. For three charges there are three pairs; for four charges there are six pairs; and in general for \(N\) charges there are \(N(N-1)/2\) pairs. Listing the pairs systematically before substituting numbers saves a great deal of trouble.
c) Potential energy of a single charge in an external electric field
Above, we have obtained an expression for the potential energy of a system of charges when the source of the electric field, i.e. the charges and their locations, were specified. In this section, we determine the potential energy of a charge (or charges) in an external field \(\vec{E}\), which is not produced by the given charge (or charges) whose potential energy we wish to calculate. The external sources could be known, unknown or unspecified, but what is known is the electric field \(E\) or the electrostatic potential \(V\) due to the external sources.
Here we assume that the external field is not affected by the charge \(q\), if \(q\) is very small. The external electric field \(E\) and the corresponding external potential \(V\) may vary from point to point.
This is the practical situation in most laboratory and circuit problems: some arrangement of plates or electrodes sets up a field, and we drop a small charge into it without disturbing that field.
If \(V(\vec{r})\) is the external potential at any point P having position vector \(\vec{r}\), then by definition, work done in bringing a unit positive charge from \(\infty\) to the point P is equal to \(V\).
\(\therefore\) Work done in bringing a charge \(q\) from \(\infty\) to the given point in the external field is \(qV(\vec{r})\).
This work is stored in the form of potential energy of a charge \(q\) at a given point in the external electric field.
\(\therefore\) PE of a system of a single charge \(q\) at \(\vec{r}\) in an external field is given by
\[\text{PE} = qV(\vec{r}) \qquad \text{--- (8.17)}\]
d) Potential energy of a system of two charges in an external electric field
In order to find the potential energy of a system of two charges \(q_1\) and \(q_2\) located at \(\vec{r_1}\) and \(\vec{r_2}\) respectively in an external electric field, we calculate the work done in bringing the charge \(q_1\) from \(\infty\) to \(\vec{r_1}\).
From Eq. (8.17), in the said process the work done is
\[W = q_1 V(\vec{r_1}) \qquad \text{--- (8.18)}\]
To bring the charge \(q_2\) to \(\vec{r_2}\), the work is done not only against the external field \(E\) but also against the field due to \(q_1\), which is by now already in position.
\(\therefore\) Work done on \(q_2\) against the external field \(= q_2 V(\vec{r_2})\), and work done on \(q_2\) against the field due to \(q_1\) is
\[\frac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}\]
where \(r_{12}\) = distance between \(q_1\) and \(q_2\).
By the principle of superposition for fields, we add up the work done on \(q_2\) against the two fields.
\(\therefore\) Work done in bringing \(q_2\) to \(\vec{r_2}\) is
\[q_2 V(\vec{r_2}) + \frac{q_1 q_2}{4\pi\varepsilon_0 r_{12}} \qquad \text{--- (8.19)}\]
Thus from (8.18) and (8.19), potential energy of the system = total work done in assembling the configuration
\[= q_1 V(\vec{r_1}) + q_2 V(\vec{r_2}) + \frac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}\]
Read the three terms as a story: the first charge pays only for fighting the external field, the second charge pays for fighting the external field and for fighting the first charge. Missing out the last term is the usual error.
e) Potential energy of a dipole in an external field
Consider a dipole with charges \(-q\) and \(+q\) separated by a finite distance \(2l\), placed in a uniform electric field \(\vec{E}\). It experiences a torque \(\vec{\tau}\) which tends to rotate it. The two forces on the two charges are equal in magnitude and opposite in direction, so they produce no net force but do form a couple, as shown in Fig. 8.17.
Diagram
Fig. 8.17 – A dipole inclined at angle θ to a uniform field, with equal and opposite forces qE acting on the two charges forming a couple.
\[\vec{\tau} = \vec{p}\times\vec{E} \quad \text{or} \quad \tau = pE\sin\theta\]
In order to neutralize this torque, let us assume an external torque \(\vec{\tau}_{ext}\) is applied, which rotates it in the plane of the paper from angle \(\theta_0\) to angle \(\theta\), without angular acceleration and at an infinitesimal angular speed. The two conditions in that sentence matter: rotating slowly and without acceleration ensures that no kinetic energy is produced, so all the work done goes into potential energy.
Work done by the external torque
\[W = \int_{\theta_0}^{\theta}\tau_{ext}\,d\theta = \int_{\theta_0}^{\theta} pE\sin\theta\,d\theta\]
\[= pE\left[-\cos\theta\right]_{\theta_0}^{\theta}\]
\[= pE\left[-\cos\theta - (-\cos\theta_0)\right]\]
\[= pE\left[-\cos\theta + \cos\theta_0\right]\]
\[= pE\left[\cos\theta_0 - \cos\theta\right]\]
This work done is stored as the potential energy of the system in the position when the dipole makes an angle \(\theta\) with the electric field. The zero potential energy can be chosen as per convenience. We can choose \(U(\theta_0) = 0\), giving
\[\therefore \; U(\theta) = U(\theta) - U(\theta_0) = pE\left[\cos\theta_0 - \cos\theta\right]\]
a) If initially the dipole is perpendicular to the field \(E\), i.e. \(\theta_0 = \dfrac{\pi}{2}\), then
\[U(\theta) = pE\left[\cos\frac{\pi}{2}-\cos\theta\right] = -pE\cos\theta\]
\[U = -\vec{p}\cdot\vec{E}\]
b) If initially the dipole is parallel to the field \(E\), then \(\theta_0 = 0\)
\[U(\theta) = pE\left[\cos 0 - \cos\theta\right] = pE\left[1-\cos\theta\right]\]
8.7 Conductors and Insulators, Free Charges and Bound Charges Inside a Conductor
a) Conductors and Insulators
When you come in contact with wires in a wet condition, or while opening the window of your car, you might have experienced a feeling of electric shock. Why don’t you get similar experiences with wooden materials?
The reason you get a shock is that there occurs a flow of electrons from one body to another when they come in contact via rubbing or moving against each other. Shock is basically a wild feeling of current passing through your body.
Conductors are materials or substances which allow electricity to flow through them. This is because they contain a large number of free charge carriers (free electrons). In a metal, the electrons in the outermost orbit (valence electrons) are loosely bound to the nucleus and are thus free for conduction, when an external electric field is applied.
Metals, humans, earth and animal bodies are all conductors. The main reason we get electric shocks is that, being a good conductor, our human body allows a resistance-free path for the current to flow from the wire to our body.
The Five Electrostatic Properties of a Conductor
Under electrostatic conditions the conductors have the following properties. The phrase “electrostatic conditions” is important: it means that the charges have already settled down and nothing is moving any longer.
1. In the interior of a conductor, the net electrostatic field is zero.
The reason lies in the free electrons. If any field did exist inside the metal, it would push the free electrons along, and they would keep moving until their own displacement set up an opposing field that exactly cancelled the original one. Only when the interior field has been reduced to zero does the movement stop — and it is that final settled state we call electrostatic.
2. Potential is constant within and on the surface of a conductor.
This follows immediately from the first property together with \(E = -dV/dx\). If \(E = 0\) everywhere inside, then \(dV/dx = 0\) everywhere inside, so \(V\) cannot change from point to point. The whole body of a conductor, surface included, is therefore a single equipotential volume.
3. The interior of a conductor does not have any charge.
Apply Gauss’ law to any closed surface drawn just inside the conductor. Since \(E = 0\) at every point of that surface, the flux through it is zero, and hence the charge enclosed must be zero. Any charge given to a conductor therefore has nowhere to go but the outer surface.
4. The electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.
If the field had a component along the surface, that component would drive the surface charges sideways, and again the arrangement would not be static. Note also that this is exactly the statement that the conductor’s surface is an equipotential surface, as established in section 8.5.
5. Surface charge density of a conductor could be different at different points.
The charge does not spread itself evenly unless the conductor happens to be a sphere. On an irregular body it crowds towards the sharply curved regions, so the surface charge density — and with it the field just outside — is greatest at points and edges, and least on broad flat portions.
Insulators
In insulators, the electrons are tightly bound to the nucleus and are thus not available for conduction, and hence they are poor conductors of electricity. There are no free charges, since all the charges are bound to the nucleus.
An insulating material can be considered as a collection of molecules that are not easily ionized. An insulator can carry any distribution of external electric charges on its surface or in its interior, and the electric field in the interior can have non-zero values, unlike conductors.
This last point is worth dwelling on, because it explains a difference you may have noticed in practice. Charge given to a metal spreads instantly over its whole outer surface; charge given to a plastic rod stays exactly where it was placed, because nothing inside the plastic is free to carry it away.
b) Free charges and bound charges inside materials
The electrical behaviour of conductors and insulators can be understood on the basis of free and bound charges.
In metallic conductors, the electrons in the outermost shells of the atoms are loosely bound to the nucleus and hence can easily get detached and move freely inside the metal. When an external electric field is applied, they drift in a direction opposite to the direction of the applied electric field. These charges are called free charges.
The drift is opposite to the field because these carriers are electrons, which are negatively charged; the force \(\vec{F} = q\vec{E}\) on a negative charge points against the field.
The nucleus, which consists of the positively charged protons, and the inner shell electrons, keeps the charges fixed in their positions. These immobile charges are called bound charges.
The word “slightly” in that definition carries a great deal of weight, and the next section is built on it. Bound charges cannot travel through the material, but they can shift a little within each atom, and that small shift is what gives insulators their most useful electrical property.
In electrolytic conductors, positive and negative ions act as charge carriers, but their movements are restricted by the electrostatic force between them and the external electric field.
8.8 Dielectrics and Electric Polarisation
What a Dielectric Is
Dielectrics are insulators which can be used to store electrical energy. This is because when such substances are placed in an external field, their positive and negative charges get displaced in opposite directions and the molecules develop a net dipole moment. This is called polarization of the material, and such materials are called dielectrics.
In every atom there is a positively charged nucleus and there are negatively charged electrons surrounding it. The negative charges form an electron cloud around the positive charge. These two oppositely charged regions have their own centres of charge, that is, the points where the effective charge may be taken to be located. The centre of negative charge is the centre of mass of the negatively charged electrons, and that of positive charge is the centre of mass of the positively charged protons in the nucleus.
Thus, dielectrics are insulating materials or non-conducting substances which can be polarized through small localized displacement of charges, e.g. glass, wax, water, wood, mica, rubber, stone, plastic etc. Dielectric constants of various materials are given in Table 1.
Note carefully what separates a dielectric from a conductor. In a conductor, charges travel right across the body. In a dielectric, nothing travels; charges merely shift within their own atoms, over distances smaller than an atomic diameter. Yet, as we shall see, that tiny shift has large consequences.
Dielectrics can be classified as polar dielectrics and non-polar dielectrics, as described below.
Polar Dielectrics
A molecule in which the centre of mass of positive charges (protons) does not coincide with the centre of mass of negative charges (electrons), because of the asymmetric shape of the molecule, is called a polar molecule, as shown in Fig. 8.18 (a).
Diagram
Fig. 8.18 – (a) A polar molecule with the centres of positive and negative charge separated; (b) structure of HCl; (c) bent structure of the H₂O molecule with its two O–H bonds.
They have permanent dipole moments of the order of \(10^{-30}\) C m. They act as tiny electric dipoles, as the charges are separated by a small distance. The dielectrics like HCl, water, alcohol, NH₃ etc. are made of polar molecules and are called polar dielectrics.
The water molecule has a bent shape, with its two O–H bonds inclined at an angle of about 105°. It has a very high dipole moment of \(6.1 \times 10^{-30}\) C m. It is precisely this bent shape that makes water polar: if the molecule were straight, the two bond dipoles would point in exactly opposite directions and cancel. Figures 8.18 (b) and (c) show the structures of HCl and H₂O respectively.
Non-Polar Dielectrics
A molecule in which the centre of mass of the positive charges coincides with the centre of mass of the negative charges is called a non-polar molecule, as shown in Fig. 8.19 (a). These have symmetrical shapes and have zero dipole moment in the normal state.
Diagram
Fig. 8.19 – (a) A non-polar molecule with coinciding centres of positive and negative charge; (b) structure of H₂; (c) linear structure of CO₂.
The dielectrics like hydrogen, nitrogen, oxygen, CO₂, benzene and methane are made up of non-polar molecules and are called non-polar dielectrics. Structures of H₂ and CO₂ are shown in Fig. 8.19 (b) and (c) respectively.
Carbon dioxide deserves a second look, since it contains two polar C=O bonds and yet is non-polar overall. The molecule is straight, so the two bond dipoles point in exactly opposite directions and cancel. Symmetry, and not the type of bond, decides the matter.
Polarization of a Non-Polar Dielectric in an External Electric Field
In the presence of an external electric field \(E_0\), the centre of the positive charge in each molecule of a non-polar dielectric is pulled in the direction of \(E_0\), while the centre of the negative charges is displaced in the opposite direction. Therefore, the two centres are separated and the molecule gets distorted.
The displacement of the charges stops when the force exerted on them by the external field is balanced by the electric field of the induced dipole of the molecule. In other words, the atom stretches only until its own internal restoring pull matches the external tug, and then holds that stretched shape.
Each molecule becomes a tiny dipole having a dipole moment. The induced dipole moments of different molecules add up, giving a net dipole moment to the dielectric in the presence of the external field. This is shown in Fig. 8.20 (a) and (b).
Diagram
Fig. 8.20 – (a) A non-polar dielectric material in the absence of an electric field, with undistorted molecules; (b) the same material in the presence of an external field, with each molecule stretched into a small dipole aligned with the field.
Polarization of a Polar Dielectric in an External Electric Field
The molecules of a polar dielectric have tiny permanent dipole moments. Due to thermal agitation in the material in the absence of any external electric field, these dipole moments are randomly oriented, as shown in Fig. 8.21 (a). Hence the total dipole moment is zero — not because the individual dipoles are absent, but because they point in every direction and cancel one another.
Diagram
Fig. 8.21 – (a) A polar dielectric in the absence of an electric field, with randomly oriented molecular dipoles; (b) the same polar dielectric in the presence of a strong external field, with the dipoles largely aligned along the field.
When an external electric field is applied, the dipole moments of different molecules tend to align with the field. As a result the dielectric develops a net dipole moment in the direction of the external field. Hence the dielectric is polarized. The extent of polarization depends on the relative values of the following two opposing fields:
- The applied external electric field, which tends to align the dipole with the field.
- The electric field due to the induced dipole.
The polarization in the presence of a strong external electric field is shown in Fig. 8.21 (b).
Thus, both polar and non-polar dielectric materials develop a net dipole moment in the presence of an electric field. The mechanism differs — the polar material turns dipoles that were already there, while the non-polar material creates dipoles that did not exist before — but the outcome is the same.
Measuring Polarization
The dipole moment per unit volume is called polarization and is denoted by \(\vec{P}\). For linear isotropic dielectrics
\[\vec{P} = \chi_e \vec{E}\]
\(\chi_e\) is a constant called the electric susceptibility of the dielectric medium. It describes the electrical behaviour of a dielectric. It has different values for different dielectric materials. For vacuum \(\chi_e = 0\), since there are no molecules present to be polarized at all.
Reduction of Electric Field due to Polarization of a Dielectric
When a dielectric is placed in an external electric field, the value of the field inside the dielectric is less than the external field as a result of polarization.
Consider a rectangular dielectric slab placed in a uniform electric field \(\vec{E}\) acting parallel to two of its faces. Since the electric charges are not free to move about in a dielectric, no current results when it is placed in an electric field. Instead of moving the charges, the electric field produces a slight rearrangement of charges within the atoms, resulting in aligning them with the field. This is shown in Fig. 8.20 and Fig. 8.21. During the process of alignment, charges move only over distances that are less than an atomic diameter.
As a result of the alignment of the dipole moments, there is an apparent sheet of positive charges on the right side and negative charges on the left side of the dielectric. Inside the material the head of one dipole sits next to the tail of the next, so their charges cancel in pairs; only at the two end faces is there nothing to cancel against, and that is why the uncompensated charge appears there.
These two sheets of induced surface charges produce an electric field \(\vec{E_0}\), called the polarization field, in the insulator, which opposes the applied electric field \(\vec{E}\). The net field \(\vec{E}'\) inside the dielectric is the vector sum of the applied field \(\vec{E}\) and the polarization field \(\vec{E_0}\).
\[\therefore \; E' = E - E_0 \quad \text{(in magnitude)}\]
This is shown in Fig. 8.22 (a), (b) and (c).
Diagram
Fig. 8.22 – (a) A dielectric slab placed between the plates of a capacitor; (b) the induced surface charges on the two faces of the slab and the polarization field they produce; (c) the net field E′ shown as the vector sum of E and E₀.
The greater the applied field, the greater is the degree of alignment of the dipoles, and hence the greater is the polarization field.
When the Field Is Removed
The induced dipole moment disappears when the field is removed. The material returns quietly to its unpolarized state, which is what makes the effect usable again and again in devices.
The induced dipole moment is often responsible for the attraction of a charged object towards an uncharged insulator, such as a charged comb and uncharged bits of paper.
Table 1: Dielectric constants of various materials
| Material | Min | Max |
|---|---|---|
| Air | 1 | 1 |
| Ebonite | 2.7 | 2.7 |
| Glass | 3.8 | 14.5 |
| Mica | 4 | 9 |
| Paper | 1.5 | 3 |
| Paraffin | 2 | 3 |
| Porcelain | 5 | 6.5 |
| Quartz | 5 | 5 |
| Rubber | 2 | 4 |
| Wood dry | 1.4 | 2.9 |
| Metals | ∞ | ∞ |
8.9 Capacitors and Capacitance, Combination of Capacitors in Series and Parallel
Why a Charge-Storing Device Was Needed
In Std XI you have studied about resistors, resistance and conductance. A resistor is an electrical component which allows current to pass through it and dissipates heat, but can’t store electrical energy. So there was a need to develop a device that can store electrical energy.
The most common arrangement for this consists of a set of conductors (conducting plates) having opposite charges on them and separated by a dielectric material.
The conductors 1 and 2 shown in Fig. 8.23 have charges \(+Q\) and \(-Q\), with potential difference \(V = V_1 - V_2\) between them. The electric field in the region between them is proportional to the magnitude of charge \(Q\).
Diagram
Fig. 8.23 – Two conductors carrying charges +Q and −Q at potentials V₁ and V₂, with field lines running from one to the other across the gap between them.
Defining Capacitance
The potential difference \(V\) is the work done to carry a unit positive test charge from the conductor 2 to conductor 1 against the field. As this work done will be proportional to \(Q\), then \(V \propto Q\) and the ratio \(\dfrac{Q}{V}\) is a constant.
\[\therefore \; C = \frac{Q}{V}\]
The constant \(C\) is called the capacitance of the capacitor, which depends on the size, shape and separation of the system of two conductors.
The SI unit of capacitance is the farad (F). Its dimensional formula is \([\text{M}^{-1}\text{L}^{-2}\text{T}^{4}\text{A}^{2}]\).
\[1\ \text{farad} = 1\ \text{coulomb}/1\ \text{volt}\]
A capacitor has a capacitance of one farad if the potential difference across it rises by 1 volt when 1 coulomb of charge is given to it. In practice the farad is a big unit; the most commonly used units are its submultiples.
\[1\ \mu\text{F} = 10^{-6}\ \text{F}\]
\[1\ \text{nF} = 10^{-9}\ \text{F}\]
\[1\ \text{pF} = 10^{-12}\ \text{F}\]
Principle of a Capacitor
To understand the principle of a capacitor, let us consider a metal plate P₁ having area \(A\). Let some positive charge \(+Q\) be given to this plate. Let its potential be \(V\). Its capacitance is given by
\[C_1 = \frac{Q}{V}\]
Now consider another insulated metal plate P₂ held near the plate P₁. By induction, a negative charge is produced on the nearer face and an equal positive charge develops on the farther face of P₂, as shown in Fig. 8.24 (a). The induced negative charge lowers the potential of plate P₁, while the induced positive charge raises its potential.
Diagram
Fig. 8.24 – (a) A charged plate P₁ with a neutral insulated plate P₂ nearby showing induced negative charge on the near face and positive charge on the far face; (b) the same arrangement with the outer surface of P₂ earthed, leaving only the induced negative charge.
As the induced negative charge is closer to P₁, it is more effective, and thus there is a net reduction in the potential of plate P₁.
If the outer surface of P₂ is connected to earth, the induced positive charges on P₂, being free, flow to earth. The induced negative charge on P₂ stays on it, as it is bound to the positive charge of P₁. This greatly reduces the potential of P₁, as shown in Fig. 8.24 (b).
If \(V_1\) is the potential on plate P₂ due to charge \((-Q)\), then the net potential difference between P₁ and P₂ will now be \(V - V_1\). Hence the capacitance is
\[C_2 = \frac{Q}{V-V_1}\]
\[\therefore \; C_2 > C_1\]
The inequality follows because the denominator has been made smaller while the numerator is unchanged. This is the whole principle in one line: earthing a nearby plate cuts down the potential for the same stored charge, and therefore raises the capacitance.
Thus the capacitance of metal plate P₁ is increased by placing an identical earth-connected metal plate P₂ near it.
Such an arrangement is called a capacitor. It is symbolically shown as \(\dashv\vdash\).
If the conductors are plane sheets then it is called a parallel plate capacitor. We also have spherical capacitors, cylindrical capacitors etc., based on the shape of the conductors.
Combination of Capacitors
When there is a combination of capacitors to be used in a circuit, we can sometimes replace it with an equivalent capacitor, that is, a single capacitor that has the same capacitance as the actual combination of capacitors. The effective capacitance depends on the way the individual capacitors are combined. Here we discuss two basic combinations of capacitors which can be replaced by a single equivalent capacitor.
a) Capacitors in series
When a potential difference \((V)\) is applied across several capacitors connected end to end in such a way that the sum of the potential differences across all the capacitors is equal to the applied potential difference \(V\), then the capacitors are said to be connected in series.
In the series arrangement as shown in Fig. 8.25, the second plate of the first conductor is connected to the first plate of the second conductor, and so on. The last plate is connected to earth. In a series combination, the charges on the plates \((\pm Q)\) are the same on each capacitor.
Diagram
Fig. 8.25 – Three capacitors C₁, C₂ and C₃ connected end to end across a supply V, with charges +Q and −Q marked on each pair of plates.
The potential difference across the series combination of capacitors is \(V\) volt, where
\[V = V_1 + V_2 + V_3\]
Since each capacitor carries the same charge \(Q\), each potential difference can be written as \(Q/C\):
\[\therefore \; V = \frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}\]
Let \(C_s\) represent the equivalent capacitance shown in Fig. 8.26, then \(V = \dfrac{Q}{C_s}\)
Diagram
Fig. 8.26 – The three series capacitors replaced by a single equivalent capacitor Cₛ carrying the same charge Q across the same potential difference V.
\[\therefore \; \frac{Q}{C_s} = \frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}\]
\[\therefore \; \frac{1}{C_s} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\]
(for 3 capacitors in series)
This argument can be extended to yield an equivalent capacitance for \(n\) capacitors connected in series. The reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances of the capacitors.
\[\therefore \; \frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2}+ \cdots +\frac{1}{C_n}\]
If all capacitors are equal then
\[\frac{1}{C_{eq}} = \frac{n}{C} \quad \text{or} \quad C_{eq} = \frac{C}{n}\]
Note that the equivalent capacitance of a series combination is always smaller than the smallest capacitance in the chain. Joining capacitors in series is effectively the same as increasing the plate separation, which is why the capacitance goes down.
b) Capacitors in parallel
The parallel arrangement of capacitors is as shown in Fig. 8.27 below, where the insulated plates are connected to a common terminal A which is joined to the source of potential, while the other plates are connected to another common terminal B which is earthed.
Diagram
Fig. 8.27 – Three capacitors C₁, C₂ and C₃ connected between common terminals A and B, carrying charges Q₁, Q₂ and Q₃ across the same potential difference V.
In this combination all the capacitors have the same potential difference, but the plate charges \((\pm Q_1)\) on capacitor 1, \((\pm Q_2)\) on capacitor 2 and \((\pm Q_3)\) on capacitor 3 are not necessarily the same. If charge \(Q\) is applied at point A, then it will be distributed to the capacitors depending on the capacitances.
\(\therefore\) Total charge \(Q\) can be written as
\[Q = Q_1 + Q_2 + Q_3 = C_1V + C_2V + C_3V\]
Let \(C_p\) be the equivalent capacitance of the combination, then \(Q = C_p V\)
\[\therefore \; C_p V = C_1 V + C_2 V + C_3 V\]
\[\therefore \; C_p = C_1 + C_2 + C_3\]
The general formula for effective capacitance \(C_p\) for a parallel combination of \(n\) capacitors follows similarly:
\[C_p = C_1 + C_2 + \cdots + C_n\]
If all capacitors are equal then \(C_{eq} = nC\).
Here the capacitances simply add, because connecting plates side by side amounts to enlarging the total plate area available for storing charge.
8.10 Capacitance of a Parallel Plate Capacitor Without and With Dielectric Medium Between the Plates
In section 8.8 we have studied the behaviour of dielectrics in an external field. Let us now see how the capacitance of a parallel plate capacitor is modified when a dielectric is introduced between its plates.
a) Capacitance of a parallel plate capacitor without a dielectric
A parallel plate capacitor consists of two thin conducting plates, each of area \(A\), held parallel to each other at a suitable distance \(d\) apart. One of the plates is isolated and charged and the other is earthed, as shown in Fig. 8.28.
Diagram
Fig. 8.28 – A parallel plate capacitor of plate area A and separation d, one plate charged +Q and the other earthed, with uniform field lines between the plates.
When a charge \(+Q\) is given to the isolated plate, then a charge \(-Q\) is induced on the inner face of the earthed plate and \(+Q\) is induced on its outer face. But as this face is earthed, the charge \(+Q\), being free, flows to earth.
In the outer regions, the electric fields due to the two charged plates cancel out. The net field is zero.
\[E = \frac{\sigma}{2\varepsilon_0}-\frac{\sigma}{2\varepsilon_0} = 0\]
In the inner region between the two capacitor plates, the electric fields due to the two charged plates add up. The net field is thus
\[E = \frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A\varepsilon_0} \qquad \text{--- (8.20)}\]
The direction of \(E\) is from the positive to the negative plate.
Let \(V\) be the potential difference between the 2 plates. Then the electric field between the plates is given by
\[E = \frac{V}{d} \quad \text{or} \quad V = Ed \qquad \text{--- (8.21)}\]
This simple relation may be used because the field between the plates is uniform; had it varied from point to point, an integration would have been needed instead.
Substituting Eq. (8.20) in Eq. (8.21), we get
\[V = \frac{Qd}{A\varepsilon_0}\]
Capacitance of the parallel plate capacitor is given by
\[C = \frac{Q}{V} = \frac{Q}{\left(\dfrac{Qd}{A\varepsilon_0}\right)} = \frac{A\varepsilon_0}{d} \qquad \text{--- (8.22)}\]
Here \(C\) is in farad, \(A\) is the plate area in m², \(d\) is the separation between the plates in metre, and \(\varepsilon_0 = 8.85\times10^{-12}\) C² N⁻¹ m⁻².
The charge \(Q\) has cancelled out, confirming what was said in the previous section: capacitance is set by geometry alone. To make \(C\) large, one needs large plates placed very close together — which is exactly how practical capacitors are built, using long metallised strips rolled up with a very thin insulating film between them.
b) Capacitance of a parallel plate capacitor with a dielectric slab between the plates
Let us now see how Eq. (8.22) gets modified with a dielectric slab in between the plates of the capacitor. Consider a parallel plate capacitor with the two plates each of area \(A\) separated by a distance \(d\). The capacitance of the capacitor is given by
\[C_0 = \frac{A\varepsilon_0}{d}\]
Let \(E_0\) be the electric field intensity between the plates before the introduction of the dielectric slab. Then the potential difference between the plates is given by \(V_0 = E_0 d\),
where
\[E_o = \frac{\sigma}{\varepsilon_o} = \frac{Q}{A\varepsilon_o}\]
and \(\sigma\) is the surface charge density on the plates.
Let a dielectric slab of thickness \(t\) \((t < d)\) be introduced between the plates of the capacitor. The field \(E_0\) polarizes the dielectric, inducing charge \(-Q_p\) on the left side and \(+Q_p\) on the right side of the dielectric, as shown in Fig. 8.29.
Diagram
Fig. 8.29 – A dielectric slab of thickness t placed between capacitor plates separated by d, showing induced charges −Q_p and +Q_p on its two faces and the opposing fields E₀ and E_p.
These induced charges set up a field \(E_p\) inside the dielectric in the opposite direction to \(E_0\). The induced field is given by
\[E_{\text{p}} = \frac{\sigma_{\text{p}}}{\varepsilon_o} = \frac{Q_{\text{p}}}{A\varepsilon_o} \qquad \left[\sigma_{\text{p}} = \frac{Q_{\text{p}}}{A}\right]\]
The net field \((E)\) inside the dielectric reduces to \(E_0 - E_p\).
Hence,
\[E = E_o - E_{\text{p}} = \frac{E_o}{k} \qquad \left[\because \frac{E_o}{E_o - E_{\text{p}}} = k\right]\]
where \(k\) is a constant called the dielectric constant.
\[\therefore \; E = \frac{Q}{A\varepsilon_0 k} \quad \text{or} \quad Q = Ak\varepsilon_0 E \qquad \text{--- (8.23)}\]
Working Out the New Capacitance
The field \(E_p\) exists over a distance \(t\) and \(E_0\) over the remaining distance \((d - t)\) between the capacitor plates. Hence the potential difference between the capacitor plates is the sum of the contributions from the two regions:
\[V = E_o\left(d-t\right) + E\left(t\right)\]
\[= E_o\left(d-t\right)+\frac{E_o}{k}\left(t\right) \qquad \left(\because E = \frac{E_0}{k}\right)\]
\[= E_o\left[\left(d-t\right)+\frac{t}{k}\right]\]
\[= \frac{Q}{A\varepsilon_o}\left[d-t+\frac{t}{k}\right]\]
The capacitance of the capacitor on the introduction of the dielectric slab becomes
\[C = \frac{Q}{V} = \frac{Q}{\dfrac{Q}{A\varepsilon_0}\left[d-t+\dfrac{t}{k}\right]} = \frac{A\varepsilon_0}{d-t+\dfrac{t}{k}}\]
Since \(k > 1\), the quantity \(t/k\) is smaller than \(t\), so the denominator is smaller than \(d\) and the capacitance is larger than \(C_0\). The slab has, in effect, made the plates behave as though they were closer together.
Special Cases
1. If the dielectric fills up the entire space then \(t = d\)
\[\therefore \; C = \frac{A\varepsilon_0 k}{d} = kC_0\]
\(\therefore\) the capacitance of a parallel plate capacitor increases \(k\) times, i.e.
\[k = \frac{C}{C_0}\]
This gives a second, entirely practical, meaning of the dielectric constant: it is the factor by which a material multiplies the capacitance of a capacitor when it completely fills the space between the plates.
2. If the capacitor is filled with \(n\) dielectric slabs of thickness \(t_1, t_2, \ldots t_n\), then this arrangement is equivalent to \(n\) capacitors connected in series, as shown in Fig. 8.30.
Diagram
Fig. 8.30 – A parallel plate capacitor whose gap is filled with n dielectric slabs of thicknesses t₁, t₂ … tₙ stacked one above the other between the plates.
\[\therefore \; C = \frac{A\varepsilon_0}{\dfrac{t_1}{k_1}+\dfrac{t_2}{k_2}+ \cdots +\dfrac{t_n}{k_n}}\]
The slabs are in series because the same charge sits on the plates while the potential difference divides itself among the layers.
3. If the arrangement consists of \(n\) capacitors in parallel with plate areas \(A_1, A_2, \ldots A_n\) and plate separation \(d\),
\[C = \frac{\varepsilon_0}{d}\left(A_1k_1 + A_2k_2 + \cdots + A_nk_n\right)\]
if \(A_1 = A_2 = \cdots = A_n = A\), then
\[C = \frac{A\varepsilon_0}{d}\left(k_1+k_2+\cdots+k_n\right)\]
Here the slabs sit side by side rather than stacked, so each spans the full gap and all share the same potential difference — the condition for a parallel combination.
4. If the capacitor is filled with a conducting slab \((k = \infty)\) then
\[C = \left(\frac{d}{d-t}\right)C_o \qquad \therefore \; C > C_o\]
The capacitance thus increases by a factor \(\left(\dfrac{d}{d-t}\right)\).
8.11 Displacement Current
The Problem of the Gap
We know that electric current in a DC circuit constitutes a flow of free electrons. In a circuit as shown in Fig. 8.31, a parallel plate capacitor with a dielectric is connected across a DC source. In the conducting part of the circuit, free electrons are responsible for the flow of current. But in the region between the plates of the capacitor, there are no free electrons available for conduction in the dielectric.
Diagram
Fig. 8.31 – A parallel plate capacitor with a dielectric connected to a DC source, showing conduction current i_c in the leads and displacement current i_d in the space between the plates.
This is a genuine difficulty. A current meter placed anywhere in the wires records a reading, yet the circuit is plainly broken at the capacitor, where an insulator sits between the plates. Either the current stops at the gap — which would contradict every measurement — or something is going on inside the gap that deserves to be called a current.
As the circuit is closed, the current flows through the circuit and grows to its maximum value \((i_c)\) in a finite time (the time constant of the circuit). The conduction current \(i_c\) is found to be the same everywhere in the circuit except inside the capacitor.
As the current passes through the leads of the capacitor, the electric field between the plates increases, and this in turn causes polarisation of the dielectric. Thus, there is a current in the dielectric due to the movement of the bound charges.
Recall from section 8.7 that bound charges cannot travel through the material, but can shift slightly within each molecule. While the field is growing, those small shifts are all happening in the same direction at the same time — and charge in motion, however short the journey, constitutes a current.
Relating the Two Currents
We can now derive an expression between \(i_c\) and \(i_d\).
From Eq. (8.23) we can infer that the charge produced on the plates of a capacitor is due to the electric field \(E\):
\[q = Ak\varepsilon_0 E\]
Differentiating the above equation with respect to time, since \(A\), \(k\) and \(\varepsilon_0\) are all constants for a given capacitor, we get
\[\frac{dq}{dt} = Ak\varepsilon_0\frac{dE}{dt} \qquad \text{--- (8.24)}\]
\(\dfrac{dq}{dt}\) is the conduction current \((i_c)\) in the conducting part of the circuit.
\[i_c = \frac{dq}{dt} = Ak\varepsilon_0\frac{dE}{dt}\]
\[\therefore \; \frac{dE}{dt} = \frac{i_c}{Ak\varepsilon_0} \quad \Rightarrow \quad \frac{dE}{dt} \propto i_c \qquad \text{(for a fixed value of } A\text{)}\]
The rate of change of electric field \(\left(\dfrac{dE}{dt}\right)\) across the capacitor is directly proportional to the current \((i_c)\) flowing in the conducting part of the circuit.
The quantity on the RHS of Eq. (8.24) has the dimension of electric current, and is caused by the displacement of bound charges in the dielectric of the capacitor under the influence of the electric field. This current, called displacement current \((i_d)\), is equivalent to the rate of flow of charge \(\left(\dfrac{dq}{dt} = i_c\right)\) in the conducting part of the circuit.
In the absence of any dielectric between the plates of the capacitor, \(k = 1\) (for air or vacuum), and the displacement current is
\[i_d = A\varepsilon_0\frac{dE}{dt}\]
The word “equivalent” is the important one. The displacement current is exactly as large as the conduction current in the wires, so the current is continuous all the way round the circuit after all. The break at the capacitor was only apparent.
A Wider Meaning
As a broad generalization of displacement current in a circuit containing a capacitor, it can be stated that displacement currents do not remain confined to the space between the plates of a capacitor. A displacement current \((i_d)\) exists at any point in space where a time-varying electric field \((E)\) exists, i.e. where \(\dfrac{dE}{dt} \neq 0\).
Notice how far this statement reaches. It applies not only inside a capacitor, and not only inside a dielectric, but at any point in empty space where the electric field is changing.
8.12 Energy Stored in a Capacitor
Why Charging a Capacitor Costs Energy
A capacitor is a device used to store energy. Charging a capacitor means transferring electrons from one plate of the capacitor to the other. Hence work will have to be done by the battery in order to remove the electrons against the opposing forces.
These opposing forces arise since the electrons are being pushed to the negative plate, which repels them, and electrons are removed from the positive plate, which tends to attract them. In both cases, the coulombian forces oppose the transfer of charges from one plate to another. As the charge on the plate increases, its opposing force also increases.
That last sentence is the reason a simple product will not do for the energy. The first electron moves across almost freely, while the last one has to be forced across against a fully charged pair of plates. The cost of transfer rises steadily as charging proceeds, so the total must be found by integration.
This work done is stored in the form of electrostatic energy in the electric field between the plates, which can later be recovered by discharging the capacitor.
Deriving the Expression
Consider a capacitor of capacitance \(C\) being charged by a DC source of \(V\) volts, as shown in Fig. 8.32.
Diagram
Fig. 8.32 – A capacitor of capacitance C connected across a DC source of V volts, with the polarity of the plates marked.
During the process of charging, let \(q'\) be the charge on the capacitor and \(V\) be the potential difference between the plates. Hence
\[C = \frac{q'}{V}\]
A small amount of work is done if a small charge \(dq\) is further transferred between the plates.
\[\therefore \; dW = V\,dq = \frac{q'}{C}\,dq\]
Total work done in transferring the charge, as \(q'\) grows from 0 to the final value \(Q\), is
\[W = \int dW = \int_{0}^{Q}\frac{q'}{C}\,dq = \frac{1}{C}\int_{0}^{Q} q'\,dq\]
\[= \frac{1}{C}\left[\frac{(q')^{2}}{2}\right]_{0}^{Q} = \frac{1}{2}\frac{Q^{2}}{C}\]
This work done is stored as electrical potential energy \(U\) of the capacitor. This work done can be expressed in different forms as follows:
\[\therefore \; U = \frac{1}{2}\frac{Q^{2}}{C} = \frac{1}{2}CV^{2} = \frac{1}{2}QV \qquad (\because Q = CV)\]
Here \(U\) is in joule, \(Q\) in coulomb, \(C\) in farad and \(V\) in volt. All three forms describe the same energy; which one is convenient depends on which two quantities a problem happens to give.
8.13 Van de Graaff Generator
Purpose of the Machine
The Van de Graaff generator is a device used to develop very high potentials, of the order of \(10^{7}\) volts. The resulting large electric fields are used to accelerate charged particles (electrons, protons, ions) to high energies, needed for experiments to probe the small scale structure of matter and for various experiments in nuclear physics.
It was designed by Van de Graaff (1901–1967) in the year 1931.
To see why such a machine was wanted, recall that a charge \(q\) accelerated through a potential difference \(V\) gains an energy \(qV\). Reaching the energies required to break into a nucleus therefore means building the largest steady potential difference that can be maintained without the surrounding air or gas breaking down.
Principle
This generator is based on
the phenomenon of corona discharge (action of sharp points),
the property that charge given to a hollow conductor is transferred to its outer surface and is distributed uniformly over it,
the fact that if a charge is continuously supplied to an insulated metallic conductor, the potential of the conductor goes on increasing.
All three ideas have already appeared in this chapter. The first is the consequence of surface charge density being greatest at sharp points, so that the field there is strong enough to ionise the surrounding air. The second is the property of conductors established in section 8.7. The third follows from \(V = q/4\pi\varepsilon_0 R\) — for a dome of fixed radius, the potential climbs in step with the accumulated charge.
Construction
Figure 8.33 shows the schematic diagram of the Van de Graaff generator.
Diagram
Fig. 8.33 – Schematic diagram of a Van de Graaff generator showing the belt over two pulleys, spray brush, collector brush, dome, evacuated accelerating tube and the surrounding steel vessel.
P₁, P₂ = Pulleys BB = Conveyer belt A = Spray brush C = Collector brush D = Dome shaped hollow conductor E = Evacuated accelerating tube I = Ion source P = DC power supply S = Steel vessel filled with nitrogen M = Earthed metal plate
An endless conveyor belt BB, made of an insulating material such as reinforced rubber or silk, can move over two pulleys P₁ and P₂. The belt is kept continuously moving by a motor (not shown in the figure) driving the lower pulley (P₁). The belt must be an insulator, so that the charge sprayed on it stays where it is put and is carried bodily upwards rather than leaking back down.
The spray brush A, consisting of a large number of pointed wires, is connected to the positive terminal of a high voltage DC power supply. From this brush, positive charge can be sprayed on the belt, which can be collected by another similar brush C. Both brushes are made of many fine points, because it is at sharp points that the field becomes large enough to ionise the air and let charge pass across the small gap to the belt.
This brush C is connected to a large, dome-shaped, hollow metallic conductor D, which is mounted on insulating pillars (not shown in the figure). E is an evacuated accelerating tube having an electrode I at its upper end, connected to the dome-shaped conductor. The tube is evacuated so that the accelerated ions travel without colliding with air molecules, which would scatter them and rob them of energy.
To prevent the leakage of charge from the dome, the pulley and belt arrangement, the dome and a part of the evacuated tube are enclosed inside a large steel vessel S, filled with nitrogen at high pressure. A small quantity of freon gas is mixed with nitrogen to ensure better insulation between the vessel S and its contents. A metal plate M, held opposite to the brush A on the other side of the belt, is connected to the vessel S, which is earthed.
Working
The electric motor connected to the pulley P₁ is switched on, which begins to rotate, setting the conveyor belt into motion. The DC supply is then switched on. From the pointed ends of the spray brush A, positive charge is continuously sprayed on the belt B.
The belt carries this charge in the upward direction, where it is collected by the collector brush C and sent to the dome-shaped conductor. The belt therefore acts as a mechanical carrier of charge — the motor does the work of transporting charge against the electric forces, and it is this mechanical energy that ends up as electrical energy stored on the dome.
As the dome is hollow, the charge is distributed over the outer surface of the dome. Its potential rises to a very high value due to the continuous accumulation of charges on it. The potential of the electrode I also rises to this high value.
The positive ions, such as protons or deuterons, from a small vessel (not shown in the figure) containing ionised hydrogen or deuterium, are then introduced in the upper part of the evacuated accelerator tube. These ions, repelled by the electrode I, are accelerated in the downward direction due to the very high fall of potential along the tube. These ions acquire very high energy. These high energy charged particles are then directed so as to strike a desired target.
Note that the ions are positive and the electrode is positive, so the repulsion drives them away from the top of the tube and down towards the earthed lower end — the whole potential difference of the machine is available to accelerate each one.
Uses
The main use of the Van de Graaff generator is to produce very high energy charged particles having energies of the order of 10 MeV. Such high energy particles are used
- to carry out the disintegration of nuclei of different elements,
- to produce radioactive isotopes,
- to study the nuclear structure,
- to study different types of nuclear reactions,
- accelerating electrons to sterilize food and to process materials.
8.14 Summary
Gauss’ Law and Its Applications
- Gauss’ law relates the total electric flux through a closed surface to the charge enclosed by it:
\[\phi = \oint \vec{E}\cdot \overrightarrow{ds} = \frac{q}{\varepsilon_0}\]
where \(\phi\) is the total flux out of the closed surface (N m² C⁻¹), \(q\) is the total charge inside it (C), and \(\varepsilon_0 = 8.85\times10^{-12}\) C² N⁻¹ m⁻². Only the charge enclosed contributes; charges outside give zero net flux.
The law holds for every closed surface, but it yields \(E\) only when the charge distribution is symmetric enough for \(E\) to be constant over a suitably chosen Gaussian surface.
Uniformly charged spherical shell of radius \(R\), surface charge density \(\sigma\), total charge \(q = \sigma \times 4\pi R^{2}\):
\[E = \frac{q}{4\pi\varepsilon_0 r^{2}} = \frac{\sigma R^{2}}{\varepsilon_0 r^{2}} \quad (r > R), \qquad E = \frac{q}{4\pi\varepsilon_0 R^{2}} = \frac{\sigma}{\varepsilon_0} \quad (r = R), \qquad E = 0 \quad (r < R)\]
Outside, a uniformly charged sphere is equivalent to a point charge at its centre.
- Infinitely long straight charged wire of linear charge density \(\lambda\):
\[E = \frac{\lambda}{2\pi\varepsilon_0 r}\]
directed radially outward for positive \(\lambda\) and inward for negative \(\lambda\).
- Charged infinite plane sheet of surface charge density \(\sigma\):
\[E = \frac{\sigma}{2\varepsilon_0}\]
perpendicular to the sheet and independent of the distance from it.
Electrostatic Potential Energy and Potential
- Electrostatic potential energy of a configuration is the work done against electrostatic forces in assembling it, the charges being brought from infinity. For two point charges:
\[U(r) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_2}{r}\right)\]
The work done in moving a charge between two points depends only on the end points, not on the path, because the electrostatic force is conservative.
Electric potential is potential energy per unit charge, \(V = U/q\); its SI unit is the volt (1 V = 1 J C⁻¹). Potential difference is the work done per unit positive charge in moving between two points.
SI unit of energy is the joule. 1 eV \(= 1.6\times10^{-19}\) J, 1 meV \(= 1.6\times10^{-22}\) J, 1 keV \(= 1.6\times10^{-16}\) J.
Field and potential are related by
\[E = -\frac{dV}{dx}\]
The electric field at a point is the negative gradient of the potential there. Field may be expressed in N C⁻¹ or V m⁻¹.
- The zero of potential is arbitrary; it is taken at infinity for point charges and localised charge groups, and at the earth for circuits.
Potential due to Charge Configurations
- Point charge:
\[V = \frac{q}{4\pi\epsilon_0 r}\]
The potential is spherically symmetric and independent of direction. For a single charge, \(F \propto 1/r^{2}\), \(E \propto 1/r^{2}\), but \(V \propto 1/r\).
- Short electric dipole of moment \(p = q\times 2l\), at distance \(r\) and angle \(\theta\) from the axis:
\[V_C = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^{2}} = \frac{1}{4\pi\varepsilon_0}\frac{\vec{p}\cdot\vec{r}}{r^{3}}\]
Axial point (\(\theta = 0^\circ\) or \(180^\circ\)): \(V = \pm\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^{2}}\), the maximum value. Equatorial point (\(\theta = 90^\circ\)): \(V = 0\), so the equatorial plane is an equipotential plane of zero potential.
- System of charges (superposition principle, algebraic sum):
\[V = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_i}\]
For a continuous distribution the summation is replaced by integration.
Equipotential Surfaces
- An equipotential surface has the same potential at every point on it. No work is done in moving a charge along such a surface.
- The electric field is always normal to the equipotential surface at every point, and vice versa.
- Equipotential surfaces never intersect, since that would give two field directions at one point.
- They are concentric spheres for a point charge, coaxial cylinders for a line charge, and parallel planes for a uniform field. Closely spaced surfaces indicate a strong field.
Potential Energy of Systems and of a Dipole
- For \(N\) point charges, summing over every distinct pair:
\[U = \frac{1}{4\pi\varepsilon_0}\sum_{\text{all pairs}}\frac{q_jq_k}{r_{jk}}\]
- Single charge \(q\) in an external field of potential \(V(\vec{r})\): PE \(= qV(\vec{r})\).
- Two charges in an external field:
\[U = q_1V(\vec{r_1}) + q_2V(\vec{r_2}) + \frac{q_1q_2}{4\pi\varepsilon_0 r_{12}}\]
- Dipole in a uniform field: \(\vec{\tau} = \vec{p}\times\vec{E}\), \(\tau = pE\sin\theta\), and
\[U(\theta) = pE\left[\cos\theta_0 - \cos\theta\right]\]
Taking \(\theta_0 = \pi/2\) gives \(U = -pE\cos\theta = -\vec{p}\cdot\vec{E}\); taking \(\theta_0 = 0\) gives \(U = pE\left[1-\cos\theta\right]\).
Conductors, Insulators and Dielectrics
- Conductors contain free charge carriers. Under electrostatic conditions: the net field inside is zero; the potential is constant within and on the surface; the interior carries no charge; the field just outside is perpendicular to the surface; and the surface charge density may differ from point to point.
- Placing an instrument inside a hollow conductor, where \(E = 0\), shields it from outside fields — electrostatic shielding, as in a Faraday cage.
- In insulators the electrons are tightly bound; free charges can move through a material, bound charges cannot, though they may be displaced slightly within an atom or molecule.
- Dielectrics are insulators that become polarized in an external field. Polar molecules (HCl, H₂O, NH₃) have permanent dipole moments of the order of \(10^{-30}\) C m; the water molecule is bent at about 105° and has a dipole moment of \(6.1\times10^{-30}\) C m. Non-polar molecules (H₂, N₂, O₂, CO₂, benzene, methane) have zero dipole moment until a field distorts them.
- Polarization is the dipole moment per unit volume, \(\vec{P} = \chi_e\vec{E}\) for linear isotropic dielectrics, where \(\chi_e\) is the electric susceptibility (\(\chi_e = 0\) for vacuum).
- Induced surface charges set up a polarization field opposing the applied field, so that inside the dielectric \(E' = E - E_0\) in magnitude. The field at which a dielectric breaks down and conducts is its dielectric strength.
Capacitors
- Capacitance is \(C = Q/V\), measured in farad (F), with dimensional formula \([\text{M}^{-1}\text{L}^{-2}\text{T}^{4}\text{A}^{2}]\); 1 farad = 1 coulomb/volt. Submultiples: 1 μF \(= 10^{-6}\) F, 1 nF \(= 10^{-9}\) F, 1 pF \(= 10^{-12}\) F. Capacitance depends on size, shape and separation of the conductors, not on \(Q\) or \(V\).
- Bringing an earthed plate near a charged plate lowers its potential and therefore raises its capacitance — the principle of the capacitor.
- Series: the charge on each capacitor is the same, and
\[\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2}+\cdots+\frac{1}{C_n}, \qquad C_{eq} = \frac{C}{n} \text{ if all are equal}\]
Used to divide a high voltage; the smallest capacitance takes the largest potential difference.
- Parallel: the potential difference across each capacitor is the same, and
\[C_p = C_1 + C_2 + \cdots + C_n, \qquad C_{eq} = nC \text{ if all are equal}\]
Used when a large capacitance is required at small potentials.
- Parallel plate capacitor, plate area \(A\), separation \(d\): the field between the plates is \(E = \sigma/\varepsilon_0 = Q/A\varepsilon_0\) and
\[C = \frac{A\varepsilon_0}{d}\]
For \(n\) parallel plates, \(C = (n-1)A\varepsilon_0/d\). For a spherical capacitor of radii \(a\) and \(b\), \(C = 4\pi\varepsilon_0\left(\dfrac{ab}{b-a}\right)\). For a cylindrical capacitor of length \(l\) and radii \(a\) and \(b\), \(C = \dfrac{2\pi\varepsilon_0 l}{\log_e (b/a)}\).
With a dielectric slab of thickness \(t\) and dielectric constant \(k\) (\(t < d\)), the field inside the slab is reduced to \(E = E_0/k\) and
\[C = \frac{A\varepsilon_0}{d-t+\dfrac{t}{k}}\]
Special cases: if the dielectric fills the gap, \(C = kC_0\), so \(k = C/C_0\); for \(n\) slabs stacked in the gap, \(C = \dfrac{A\varepsilon_0}{\dfrac{t_1}{k_1}+\dfrac{t_2}{k_2}+\cdots+\dfrac{t_n}{k_n}}\); for slabs side by side, \(C = \dfrac{\varepsilon_0}{d}\left(A_1k_1+A_2k_2+\cdots+A_nk_n\right)\); and for a conducting slab (\(k = \infty\)), \(C = \left(\dfrac{d}{d-t}\right)C_0\).
Displacement Current
- In the dielectric of a capacitor the bound charges shift as the field changes, producing a displacement current \(i_d\), also called charge-separation current, equal to the conduction current in the leads.
- From \(q = Ak\varepsilon_0E\),
\[\frac{dq}{dt} = Ak\varepsilon_0\frac{dE}{dt}\]
so \(dE/dt \propto i_c\) for fixed \(A\). In air or vacuum, where \(k = 1\), \(i_d = A\varepsilon_0\left(dE/dt\right)\). 3. A displacement current exists at any point in space where the electric field varies with time, i.e. where \(dE/dt \neq 0\); it is not confined to the space between capacitor plates.
Energy Stored in a Capacitor
- The work done in charging a capacitor is stored as electrostatic energy in the field between the plates:
\[U = \frac{1}{2}\frac{Q^{2}}{C} = \frac{1}{2}CV^{2} = \frac{1}{2}QV\]
- The battery supplies \(QV\), of which \(\frac{1}{2}QV\) is stored and the remaining \(\frac{1}{2}QV\) is dissipated as heat in the connecting wires and the battery itself.
Van de Graaff Generator
- A device that develops very high potentials, of the order of \(10^{7}\) volts, designed by Van de Graaff (1901–1967) in 1931.
- It works on corona discharge at sharp points, the transfer of charge given to a hollow conductor to its outer surface, and the continuous rise of potential as charge is supplied to an insulated conductor.
- A moving insulating belt carries sprayed positive charge from the spray brush to a collector brush and on to a hollow dome, whose potential rises steadily. Positive ions introduced at the top of an evacuated tube are repelled by the dome electrode and accelerated down the tube to high energy.
- The whole assembly is enclosed in a steel vessel filled with nitrogen at high pressure, mixed with a little freon, to prevent leakage of charge.
- It produces particles of energy of the order of 10 MeV, used for nuclear disintegration, producing radioactive isotopes, studying nuclear structure and nuclear reactions, and accelerating electrons to sterilize food and process materials.
8.15 Objective Questions
- Choose the correct option.
- A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential, capacitance respectively are
- Constant, decreases, decreases
- Increases, decreases, decreases
- Constant, decreases, increases
- Constant, increases, decreases
- A slab of material of dielectric constant \(k\) has the same area \(A\) as the plates of a parallel plate capacitor and has thickness \((3/4)d\), where \(d\) is the separation of the plates. The change in capacitance when the slab is inserted between the plates is
- \(C = \dfrac{A\varepsilon_0}{d}\left(\dfrac{k+3}{4k}\right)\)
- \(C = \dfrac{A\varepsilon_0}{d}\left(\dfrac{2k}{k+3}\right)\)
- \(C = \dfrac{A\varepsilon_0}{d}\left(\dfrac{k+3}{2k}\right)\)
- \(C = \dfrac{A\varepsilon_0}{d}\left(\dfrac{4k}{k+3}\right)\)
- Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio.
- 1:1 (B) 1:2 (C) 2:1 (D) 1:3
- Charge \(+q\) and \(-q\) are placed at points A and B respectively, which are a distance \(2L\) apart. C is the mid point of A and B. The work done in moving a charge \(+Q\) along the semicircle CRD as shown in the figure below is
- \(\dfrac{-qQ}{6\pi\varepsilon_0 L}\) (B) \(\dfrac{qQ}{2\pi\varepsilon_0 L}\) (C) \(\dfrac{qQ}{6\pi\varepsilon_0 L}\) (D) \(\dfrac{-qQ}{4\pi\varepsilon_0 L}\)
Diagram
Fig. 8.34 – Charges +q at A and −q at B a distance 2L apart, C the midpoint, D a point beyond B on the same line, and the semicircular path CRD drawn above the line.
- A parallel plate capacitor has circular plates of radius 8 cm and plate separation 1 mm. What will be the charge on the plates if a potential difference of 100 V is applied?
- \(1.78\times10^{-8}\) C (B) \(1.78\times10^{-5}\) C (C) \(4.3\times10^{4}\) C (D) \(2\times10^{-9}\) C
8.16 Check Your Concepts
A closed surface encloses no charge at all, yet the electric field is found to be non-zero at several points on the surface. Is this possible? Explain your reasoning using Gauss’ law.
Gauss’ law is valid for any closed surface whatsoever, but it can be used to calculate the electric field only in a few situations. What property must the charge distribution possess, and why is it essential?
The electric field of a point charge falls as \(1/r^{2}\), that of a long charged wire as \(1/r\), and that of a large charged sheet not at all. Explain the physical reason behind this pattern.
Why is the field just outside the surface of a charged conducting sphere equal to \(\sigma/\varepsilon_0\), while that near a charged infinite plane sheet is only \(\sigma/2\varepsilon_0\)? Both involve the same surface charge density.
A student says that since no work is done in moving a charge along an equipotential surface, no force can be acting on the charge there. Identify the error in this statement.
Two points in space have the same electric potential. Does it follow that the electric field is the same at both points? Does it follow that no work is needed to move a charge between them? Answer each part separately, with justification.
Why can two equipotential surfaces never intersect each other?
The electrostatic potential energy of a pair of charges comes out negative in a certain arrangement. What does the negative sign tell you about the two charges and about the work required to separate them?
An electric dipole is placed in a uniform electric field at an angle to it. Explain why it experiences a torque but no net force. At which orientations is it in stable and in unstable equilibrium?
Explain, using the properties of conductors, why any charge given to a hollow metal sphere ends up entirely on its outer surface.
The potential inside a charged conductor is constant. Does this mean that it is zero? Explain the difference between the two statements.
Carbon dioxide contains two polar C=O bonds, yet CO₂ is classified as a non-polar dielectric, while water with two O–H bonds is strongly polar. Account for this difference.
A polar dielectric is placed in an external electric field and then heated. What happens to the extent of its polarization, and why?
When a dielectric slab is introduced between the plates of a capacitor, the electric field inside the slab is reduced. Explain the mechanism, and state why the reduced field can never be reversed in direction by the polarization field.
Capacitors in series combine like resistors in parallel, and capacitors in parallel like resistors in series. Rather than memorising this, explain each rule from the geometry of the plates.
A capacitor is charged by a battery, which is then disconnected, and a dielectric slab is inserted. State what happens to the charge, the potential difference, the capacitance and the stored energy. Repeat the exercise for the case in which the battery is left connected.
A conducting slab of thickness \(t\) is inserted between the plates of a parallel plate capacitor of separation \(d\). Why does the position of the slab within the gap make no difference to the new capacitance?
There are no free charges in the dielectric of a capacitor, and yet a current is said to exist there while the capacitor is charging. Explain what this current consists of, and why the current in the circuit can be regarded as continuous.
A battery supplies an energy \(QV\) while charging a capacitor, of which only half is stored. Explain where the rest goes, and why the answer does not depend on the resistance of the connecting wires.
In a Van de Graaff generator, charge is delivered to the inner surface of the dome rather than sprayed onto the outside. Explain why this arrangement allows the potential of the dome to be raised to millions of volts.
Why is the vessel of a Van de Graaff generator filled with nitrogen at high pressure, and why are the spray and collector brushes made of many fine pointed wires?
Answer in brief.
- A charge \(q\) is moved from a point A above a dipole of dipole moment \(p\) to a point B below the dipole in the equatorial plane without acceleration. Find the work done in this process.
- If the difference between the radii of the two spheres of a spherical capacitor is increased, state whether the capacitance will increase or decrease.
- A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?
- The safest way to protect yourself from lightning is to be inside a car. Justify.
- A spherical shell of radius \(b\) with charge \(Q\) is expanded to a radius \(a\). Find the work done by the electrical forces in the process.
- A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, charge stored and voltage will increase, decrease or remain constant.
8.17 Practice with Numericals
A point charge of \(3\ \mu\)C is placed at the centre of a cube of side 15 cm. Calculate the total electric flux emerging from the whole cube, and the flux through any one of its faces.
A hollow metallic sphere of radius 15 cm carries a charge of \(2.5\ \mu\)C distributed uniformly over its surface. Calculate the electric field intensity at points 25 cm, 15 cm and 8 cm from its centre.
A uniformly charged spherical shell of radius 10 cm has a surface charge density of \(5\times10^{-8}\) C m⁻². Calculate the total charge on the shell and the electric field intensity just outside its surface.
A long straight wire carries a linear charge density of \(4\times10^{-7}\) C m⁻¹. At what distance from the wire will the electric field intensity be \(3.6\times10^{3}\) N C⁻¹?
Two large parallel plane sheets carry surface charge densities of \(+2\times10^{-7}\) C m⁻² and \(-2\times10^{-7}\) C m⁻² respectively. Calculate the electric field intensity in the region between the sheets and in the region outside them.
A charge of \(4\ \mu\)C is moved from a point at a potential of 120 V to a point at a potential of 45 V. Calculate the work done and the change in the potential energy of the charge.
Two point charges of \(+5\ \mu\)C and \(-3\ \mu\)C are separated by 25 cm in air. Calculate the electrostatic potential energy of the system, and the work needed to separate the charges to a distance of 50 cm.
Calculate the electric potential at a point 40 cm from a point charge of \(-8\ \mu\)C. How much work must an external agent do to bring an electron from infinity to that point?
A short electric dipole of dipole moment \(2.5\times10^{-9}\) C m is placed in air. Calculate the electric potential at a distance of 0.5 m from its centre, (a) on the axial line, (b) on the equatorial line, and (c) along a direction making 45° with the dipole axis.
Four point charges of \(+3\ \mu\)C, \(-3\ \mu\)C, \(+3\ \mu\)C and \(-3\ \mu\)C are placed in order at the four corners of a square of side 30 cm. Calculate the electric potential at the centre of the square.
Two point charges of \(+9\ \mu\)C and \(-4\ \mu\)C are placed 20 cm apart. Find the point on the line joining them, between the two charges, at which the electric potential is zero.
Two parallel plates 8 mm apart are maintained at a potential difference of 320 V. Calculate the electric field intensity between them, and the force experienced by a charge of \(2\ \mu\)C placed in this region.
An oil drop carrying a charge of \(3.2\times10^{-19}\) C is held stationary between two horizontal plates 12 mm apart with a potential difference of 900 V across them. Calculate the mass of the drop. (Take \(g = 9.8\) m s⁻².)
Three point charges of \(+1\ \mu\)C, \(+2\ \mu\)C and \(-3\ \mu\)C are placed at the corners of an equilateral triangle of side 50 cm. Calculate the total electrostatic potential energy of the system.
An electric dipole of dipole moment \(5\times10^{-8}\) C m is placed in a uniform electric field of \(3\times10^{4}\) N C⁻¹. Calculate (a) the maximum torque on the dipole, (b) the work done in rotating it from the position of stable equilibrium through 90°, and (c) the work done in turning it completely through 180°.
A parallel plate capacitor has plates of area \(1.5\times10^{-2}\) m² separated by 3 mm of air. Calculate its capacitance, and the charge on each plate when it is connected across a 150 V supply.
Three capacitors of 4 μF, 6 μF and 12 μF are connected in series across a 100 V supply. Calculate the equivalent capacitance, the charge on each capacitor and the potential difference across the 6 μF capacitor.
Two capacitors of 5 μF and 15 μF are connected in parallel across a 40 V supply. Calculate the equivalent capacitance, the charge on each capacitor and the total energy stored in the combination.
A parallel plate capacitor of plate area \(4\times10^{-2}\) m² and plate separation 5 mm has a dielectric slab of thickness 3 mm and dielectric constant 5 introduced between its plates. Calculate the capacitance before and after the slab is introduced.
A capacitor of capacitance 8 μF is charged to a potential difference of 250 V. Calculate the charge stored and the energy stored. If it is then connected across an identical uncharged capacitor, calculate the common potential difference reached.
The electric field between the plates of an air-filled parallel plate capacitor of plate area \(2\times10^{-2}\) m² changes at a uniform rate of \(6\times10^{7}\) V m⁻¹ s⁻¹. Calculate the displacement current between the plates.
The spherical dome of a Van de Graaff generator has a radius of 40 cm. Calculate the charge it must carry for its potential to reach \(1.5\times10^{6}\) V, and the electric field intensity just outside its surface at that potential.
A dipole with its charges, \(-q\) and \(+q\), located at the points \((0, -b, 0)\) and \((0, +b, 0)\) is present in a uniform electric field E whose equipotential surfaces are planes parallel to the YZ planes. (a) What is the direction of the electric field E? (b) How much torque would the dipole experience in this field?
Three charges \(-q\), \(+Q\) and \(-q\) are placed at equal distance on a straight line. If the potential energy of the system of the three charges is zero, then what is the ratio of \(Q : q\)?
Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors \(C_1\) and \(C_2\) with their capacitances in the ratio 1:2, so that the energy stored in these two cases becomes the same.
Two charges of magnitudes \(-4Q\) and \(+2Q\) are located at points \((2a, 0)\) and \((5a, 0)\) respectively. What is the electric flux due to these charges through a sphere of radius \(4a\) with its centre at the origin?
A 6 μF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
One hundred twenty five small liquid drops, each carrying a charge of 0.5 μC and each of diameter 0.1 m, form a bigger drop. Calculate the potential at the surface of the bigger drop.
The dipole moment of a water molecule is \(6.3\times10^{-30}\) C m. A sample of water contains \(10^{21}\) molecules, whose dipole moments are all oriented in an electric field of strength \(2.5\times10^{5}\) N/C. Calculate the work to be done to rotate the dipoles from their initial orientation \(\theta_1 = 0\) to one in which all the dipoles are perpendicular to the field, \(\theta_2 = 90^\circ\).
A charge 6 μC is placed at the origin and another charge \(-5\) μC is placed on the y axis at a position A (0, 6.0) m.
- Calculate the total electric potential at the point P whose coordinates are (8.0, 0) m.
- Calculate the work done to bring a proton from infinity to the point P. What is the significance of the sign of the work done?
- In a parallel plate capacitor with air between the plates, each plate has an area of \(6\times10^{-3}\) m² and the separation between the plates is 2 mm.
- Calculate the capacitance of the capacitor.
- If this capacitor is connected to a 100 V supply, what would be the charge on each plate?
- How would the charge on the plates be affected if a 2 mm thick mica sheet of \(k = 6\) is inserted between the plates while the voltage supply remains connected?
- Find the equivalent capacitance between P and Q. Given, area of each plate = \(A\) and separation between plates = \(d\).
Diagram
Fig. 8.35 – The two multi-plate arrangements from the original exercise, each with terminals P and Q, plate area A and separation d marked.