Chapter 10: Electrostatics
In the 1750s, the American scientist Benjamin Franklin was convinced of something that most people of his time thought absurd: that the lightning splitting the sky during a storm was nothing but a giant spark — the same kind of spark that jumps from a rubbed rod to a fingertip, only unimaginably larger.
To test the idea, Franklin flew a kite into a stormy sky, with a metal key tied near the lower end of the wet string. Charge from the storm clouds travelled down the damp string, and small sparks jumped from the key to his knuckle. The frightening flash in the clouds and the tiny crackle from a plastic comb turned out to be the same physics, separated only by scale.
Franklin also gave us words we still use every day. He proposed that there are two kinds of electricity, and named them positive and negative. Out of that idea grew the lightning rod — a simple pointed metal conductor that quietly guides the enormous charge of a strike into the ground and saves buildings from destruction.
This chapter is about those charges: what they are, how they are separated, how strongly they push and pull one another, and how they fill the space around them with a field.
Diagram
Fig. 10.1 – Franklin’s kite in a storm cloud, with a metal key on the wet string and a small spark jumping from the key to a hand, alongside a modern building protected by a lightning rod.
Each of these questions is answered somewhere in this chapter. Keep them in mind as you read, and watch for the moment when the answer appears.
10.1 Introduction
Static Charges and Moving Charges
It is useful to be clear about the difference between electrostatics and current electricity right at the start. Current electricity, which powers almost everything in daily life, is produced by charges in motion. Electrostatics deals with charges that stay put. The same charges are involved in both; only their state of motion is different.
Charges are, in fact, present everywhere around us — inside every atom of every object. Yet we rarely notice them, because in ordinary matter the positive and negative charges are present in equal amounts and their effects cancel out. Their presence becomes noticeable only under special circumstances, when some of the charge gets separated.
Where We Actually See Electrostatics
Take the sweater example. When you remove a woollen sweater in winter on a dry day, you hear a crackling sound and the sweater seems to stick to your body. This happens because friction between your body (and the clothes beneath) and the sweater separates electric charges: one surface gains extra electrons and the other is left short of them. The two oppositely charged surfaces then attract each other, and the small sparks between them produce the crackling sound. On a humid day, the moisture in the air lets this separated charge leak away quickly, which is why the effect is far weaker in the monsoon than in dry winter.
So the subject begins with a simple question that the rest of this chapter will answer carefully: what is electric charge, how does it behave, and what does it do to the space around it?
10.2 Electric Charges
The Oldest Clue: Amber and Fur
The story of electricity begins with a puzzle that is more than two and a half thousand years old.
The Greeks could describe what happened, but not why. Today we can, because we know that electric charge is a basic property of the elementary particles out of which matter is made.
Charge Is a Property of Matter Itself
All ordinary matter is built from atoms, and atoms are built from three particles: the proton, the neutron and the electron.
A proton is taken to be positively charged, an electron negatively charged, and a neutron is electrically neutral — it carries no charge at all. The words “positive” and “negative” are simply labels for the two opposite kinds of charge that exist in nature.
Protons and neutrons together make up the nucleus of an atom. Since the nucleus contains the positive protons and the neutral neutrons, the nucleus as a whole is positively charged. The negatively charged electrons surround this nucleus.
In a normal atom, the number of electrons is exactly equal to the number of protons. Their charges therefore cancel, and the atom as a whole is electrically neutral. Because matter is made of such atoms, most matter around us is electrically neutral — which is exactly why we do not see electrical forces acting between everyday objects.
Charging by Conduction
If matter is neutral, how do we ever get a charged object? By separating the charge that is already there.
When two dissimilar substances — fur and amber, or a plastic comb and dry hair — are rubbed against each other, electrons are transferred from one substance to the other. Electrons are the particles that move; the heavy protons in the nuclei stay where they are.
The substance that receives the electrons develops a negative charge. The substance that loses them is left with an equal amount of positive charge. Notice that no charge has been created: it has only been moved from one body to the other.
Since charges are actually transferred from one body to another, this way of charging is called charging by conduction.
Rubbing is not the only way to separate charge. Charges can also be separated by chemical reactions (as in an electric cell), by convection (as in the churning air inside a thundercloud), and by diffusion (as across the membranes of living cells).
Charging by Induction
Something more subtle happens when a charged body is brought near an uncharged conductor without touching it.
The side of the conductor nearer the charged body develops a charge opposite to that on the charged body, while the far side develops a charge similar to it. This process is called induction, and it can be followed step by step in Fig. 10.2.
Diagram
Fig. 10.2 (a)–(d) – Four-step induction sequence on an insulated conductor: (a) insulated conductor with a positively charged rod brought near, showing induced negative charge on the near side and positive on the far side; (b) the conductor earthed, the positive charge neutralised by electrons drawn up from the Earth; (c) earthing removed, the negative charge still held on the conductor by the nearby positive rod; (d) rod removed, the negative charge spreading over the surface of the conductor.
The reason induction works is that the electrons in a conductor are free — they are not bound to individual atoms and can move easily through the material. When a charged body comes near, these free electrons respond at once. A charged body attracts or repels the electrons of the conductor depending on whether the charge on the body is positive or negative respectively.
As a result, the positive and negative charges of the conductor are redistributed and accumulate at the two ends — one end near the charged body and one end away from it. Notice in Fig. 10.2 (a) that the conductor as a whole is still neutral: it has simply been pulled apart internally.
Two conclusions follow from all of this. First, only two types of charge are found in nature: positive and negative. Second, in induction there is no transfer of charge between the charged body and the conductor. Nothing crosses the gap. So when the charged body is moved away, the charges in the conductor are free once more and simply spread back to their original arrangement.
Steps (b), (c) and (d) of Fig. 10.2 show how induction can be used to leave a conductor permanently charged. While the positive rod is held near, the conductor is earthed; electrons flow up from the Earth and neutralise the positive charge on the far side. The earthing is then removed, and the negative charge that remains is held in place by the attraction of the rod. Only when the rod is finally taken away does this extra negative charge distribute itself over the surface of the conductor. The conductor is now genuinely charged — and its charge is opposite to that of the rod.
Diagram
Fig. 10.2 (e) – Labelled gold leaf electroscope showing metal disc, insulation, metal stem, metal plate, gold leaf, earth connection and glass window, with an inset showing the leaf lifted away from the plate when charged.
10.3 Basic Properties of Electric Charge
Charge is not just a label we attach to particles. It behaves in definite, reliable ways: it adds up in a particular manner, it comes only in fixed packets, it is never destroyed, and it makes charged bodies push and pull on one another. These four properties — additivity, quantization, conservation and the force between charges — are the foundation of everything that follows in this chapter.
10.3.1 Additive Nature of Charge
Electric charge is additive, in much the same way that mass is additive. If you know the charge on every part of an object, you can find the charge on the whole object simply by adding them up.
The word algebraic is doing important work in that statement. While adding, the sign of each charge — positive or negative — must be taken into account. You are not adding sizes; you are adding signed quantities.
This has a consequence that seems strange at first. If two bodies carry equal and opposite charges, the net charge of the system of the two bodies together is zero. The charges have not vanished; they simply cancel in the sum.
This is exactly what happens inside an atom. The nucleus carries a positive charge, and this charge is equal in magnitude to the total negative charge of the electrons surrounding it. The algebraic sum is zero, which is why atoms — and therefore ordinary matter — are electrically neutral.
How Charge Differs from Mass
It is instructive to place the additive property of charge side by side with that of mass, because the comparison shows where the analogy breaks down.
The masses of the particles constituting an object are always positive, whereas the charges distributed on different parts of the object may be positive or negative.
The total mass of an object is always positive, whereas the total charge on the object may be positive, zero, or negative.
10.3.2 Quantization of Charge
You cannot buy sugar in any amount you like if the shop only sells sealed one-kilogram packets. Charge is like that: nature only supplies it in packets of one fixed size.
The minimum value of charge, as determined for the electron by Millikan’s oil drop experiment, is
\[e = 1.6 \times 10^{-19}\ \text{C}\]
This is called the elementary charge. Here, C stands for coulomb, which is the unit of charge in the SI system. (The coulomb is defined precisely in Article 10.4.3; for now, treat it simply as the SI unit of charge.)
Matter is built from protons (+ve) and electrons (−ve), and every one of them carries this same magnitude of charge. So the charge on any object must be an integral multiple of \(\pm e\):
\[q = \pm ne\]
where \(n\) is an integer (1, 2, 3, …) and \(e = 1.6 \times 10^{-19}\ \text{C}\) is the elementary charge. There is no way to place a charge of, say, \(0.5e\) or \(1.7e\) on a body, because there is no particle available carrying such a charge.
Further, the charge on an object can be increased or decreased only in multiples of \(e\). The reason is simply the mechanism of charging itself: during charging, a whole number of electrons is transferred from one body to the other. You can transfer one electron or a billion electrons, but never half an electron.
This property is known as quantization of charge, or the discrete nature of charge.
Why We Never Notice the Steps
If charge really comes in steps, why does charge seem to vary smoothly in every experiment we do?
The discrete nature of electric charge is usually not observable in practice, because the magnitude of the elementary charge \(e\) is extremely small. As a result, the number of elementary charges involved in charging an ordinary object becomes extremely large.
Take a concrete case. When a glass rod is rubbed with silk, a charge of the order of one microcoulomb (\(1\ \mu\text{C} = 10^{-6}\ \text{C}\)) appears on the glass rod (and an equal opposite charge on the silk). Since the elementary charge is \(e = 1.6 \times 10^{-19}\ \text{C}\), the number of elementary charges on the glass rod (or the silk) is
\[n = \frac{10^{-6}\ \text{C}}{1.6 \times 10^{-19}\ \text{C}} = 6.25 \times 10^{12}\]
Since this is a tremendously large number, adding or removing one electron changes the total by an utterly negligible fraction. The steps are far too fine to detect, and so the quantization of charge is not observed — one usually observes a continuous variation of charge instead.
The situation is like measuring sand by the truckload: the sand really does come in separate grains, but nobody notices the graininess when the quantity is that large.
10.3.3 Conservation of Charge
Return to the glass rod and silk. When a glass rod is rubbed with silk, the rod becomes positively charged and the silk becomes negatively charged. Careful measurement shows something remarkable: the amount of positive charge on the glass rod is found to be exactly the same as the amount of negative charge on the silk.
So if you take the glass rod and the silk together as one system, that system possesses zero net charge after rubbing — exactly as it did before rubbing. Charge was not created by the rubbing; it was only moved from one body to the other.
The result and conclusion of this experiment can be generalized. In any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant. Or, put the other way round: for an isolated system, total charge cannot be created nor destroyed.
The condition “isolated” is essential and is often tested in examinations. A system is isolated if no charge can enter or leave it. If a body is connected to the Earth, charge can flow in or out — and then its own charge is certainly not conserved, even though the charge of the larger system (body + Earth) still is.
10.3.4 Forces between Charges
Careful experiments with charged objects showed that they experience a force when brought close to each other — even when they are not touching. This force can be either attractive or repulsive, and which one it is depends entirely on the signs of the two charges.
Figure 10.3 describes this schematically: a rubber rod and a glass rod carrying opposite charges attract, while two rubber rods or two glass rods carrying like charges repel.
Diagram
Fig. 10.3 – Suspended charged rods showing attraction between a negatively charged rubber rod and a positively charged glass rod, and repulsion between two rubber rods and between two glass rods.
This simple rule is exactly what makes charging by induction work, as described in Section 10.2 and Fig. 10.2. The positive rod held near the conductor attracts the conductor’s free electrons to the near side and repels the like (positive) charge to the far side. Nothing crosses the gap; the rule “like repels, unlike attracts” does all the work at a distance.
The next question is the obvious one, and it is the question Coulomb set out to answer: how strong is this force? That is the subject of Section 10.4.
10.4 Coulomb’s Law
We now know that charges push and pull one another. The next step is to say how much. The electric interaction between two charged bodies can be expressed in terms of the forces they exert on each other, and this force can be measured.
For this study, Coulomb deliberately used point charges at rest. This is an idealisation worth stating carefully.
The idealisation matters because “the distance between two charges” is ambiguous for a large charged body — distance from which part of it? If the body is tiny compared with the separation, the ambiguity disappears.
Coulomb’s law is a fundamental law governing the interaction between charges at rest. Every result in the rest of this chapter can be traced back to it.
10.4.1 Scalar Form of Coulomb’s Law
Read the statement in three parts. “Directly proportional to the product of the magnitudes” means that doubling either charge doubles the force. “Inversely proportional to the square of the distance” means that doubling the separation cuts the force to one-fourth, not one-half. And “along the line joining the two charges” fixes the direction of the force — it never acts sideways.
Let \(q_1\) and \(q_2\) be two point charges at rest with respect to each other, separated by a distance \(r\). The magnitude \(F\) of the force between them is given by
\[F \propto \frac{q_1 q_2}{r^2}\]
\[F = K\, \frac{q_1 q_2}{r^2} \qquad \text{--- (10.1)}\]
Here \(F\) is the force in newton (N), \(q_1\) and \(q_2\) are the charges in coulomb (C), and \(r\) is the separation in metre (m). \(K\) is the constant of proportionality. Its magnitude depends on the units in which \(F\), \(q_1\), \(q_2\) and \(r\) are expressed, and also on the properties of the medium around the charges. That second dependence is important, and we take it up in the next subsection.
Equation (10.1) gives only the magnitude of the force. The sense of the force — pull or push — is decided by the signs of the charges: the force is attractive if the charges are unlike (one positive and one negative), and repulsive if the charges are similar (both positive or both negative). Notice in Fig. 10.4 how in every case the force arrows lie along the line joining the two charges.
Diagram
Fig. 10.4 – Three pairs of point charges separated by distance r: (−q₁, +q₂) with arrows pointing towards each other showing attraction; (+q, +q) with arrows pointing away showing repulsion; (−q, −q) with arrows pointing away showing repulsion.
10.4.2 Relative Permittivity or Dielectric Constant
While discussing Coulomb’s law it was assumed that the charges are held stationary in vacuum. But charges are not always in vacuum. When the charges are kept in a material medium, such as water, mica or paraffined paper, the medium affects the force between the charges.
The force between the two charges placed in a medium may be written as
\[F_{\text{med}} = \frac{1}{4\pi\varepsilon}\left(\frac{q_1 q_2}{r^2}\right) \qquad \text{--- (10.2)}\]
where \(\varepsilon\) is called the absolute permittivity of the medium. Permittivity is a property of the material: loosely speaking, it measures how much the medium “gets in the way” of the electric interaction.
The force between the same two charges placed in free space or vacuum at the same distance \(r\) is given by
\[F_{\text{vac}} = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1 q_2}{r^2}\right) \qquad \text{--- (10.3)}\]
where \(\varepsilon_0\) is the permittivity of free space.
Dividing Eq. (10.3) by Eq. (10.2), every quantity except the permittivities cancels:
\[\frac{F_{\text{vac}}}{F_{\text{med}}} = \frac{\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1 q_2}{r^2}\right)}{\dfrac{1}{4\pi\varepsilon}\left(\dfrac{q_1 q_2}{r^2}\right)} = \frac{\varepsilon}{\varepsilon_0}\]
The ratio \(\dfrac{\varepsilon}{\varepsilon_0}\) is the relative permittivity or dielectric constant of the medium, and is denoted by \(\varepsilon_r\) or \(K\).
\[K \text{ or } \varepsilon_r = \frac{\varepsilon}{\varepsilon_0} = \frac{F_{\text{vac}}}{F_{\text{med}}} \qquad \text{--- (10.4)}\]
Equation (10.4) gives us three ways of describing \(\varepsilon_r\):
\(\varepsilon_r\) is the ratio of the absolute permittivity of a medium to the permittivity of free space.
\(\varepsilon_r\) is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum, to the force between the same two point charges when placed at the same distance in the given medium. Since it is a ratio of two like quantities, \(\varepsilon_r\) is a dimensionless quantity — it has no unit.
\(\varepsilon_r\) is also called the specific inductive capacity of the medium.
Combining Eq. (10.4) with Eq. (10.2), the force between two point charges \(q_1\) and \(q_2\) placed at a distance \(r\) in a medium of relative permittivity \(\varepsilon_r\) is
\[F = \frac{1}{4\pi\varepsilon_0 \varepsilon_r}\, \frac{q_1 q_2}{r^2} \qquad \text{--- (10.5)}\]
Take water as an example. For water, \(\varepsilon_r = 80\). Then from Eq. (10.4),
\[\frac{F_{\text{vac}}}{F_{\text{water}}} = \varepsilon_r = 80\]
\[F_{\text{water}} = \frac{F_{\text{vac}}}{80}\]
This means that when two point charges are placed some distance apart in water, the force between them is reduced to \(\left(\dfrac{1}{80}\right)^{\text{th}}\) of the force between the same two charges placed at the same distance in vacuum.
10.4.3 Definition of Unit Charge from the Coulomb’s Law
We have been using the coulomb as the unit of charge without defining it. Coulomb’s law itself supplies the definition.
The force between two point charges \(q_1\) and \(q_2\), separated by a distance \(r\) in free space, is written using Eq. (10.2) in the form
\[F = \frac{1}{4\pi\varepsilon_0} \times \frac{q_1 q_2}{r^2} = 9 \times 10^{9} \times \frac{q_1 q_2}{r^2}\]
\[\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}\]
The number \(9 \times 10^{9}\) (in SI units, N m² C⁻²) is just the value of \(\dfrac{1}{4\pi\varepsilon_0}\), and it is worth committing to memory because it appears in almost every numerical you will solve.
Now put in the simplest possible values.
If \(q_1 = q_2 = 1\ \text{C}\) and \(r = 1.0\ \text{m}\),
then \(F = 9.0 \times 10^{9}\ \text{N}\).
From this, we define the coulomb (C), the unit of charge in SI units.
This force is a tremendously large force, not realisable in ordinary practical situations. In other words, one coulomb is an enormous amount of charge — you will never find a free coulomb sitting on a rod on your laboratory table. It is, therefore, necessary to express charge in smaller units for practical purposes. Subunits of the coulomb are used in electrostatics: the micro-coulomb (\(10^{-6}\ \text{C}\), \(\mu\)C), the nano-coulomb (\(10^{-9}\ \text{C}\), nC) and the pico-coulomb (\(10^{-12}\ \text{C}\), pC) are the normally used units.
10.4.4 Coulomb’s Law in Vector Form
The scalar form tells us how strong the force is. But force is a vector, and in problems with several charges we shall have to add these forces vectorially. So we need Coulomb’s law written with directions included.
As shown in Fig. 10.5, \(q_1\) and \(q_2\) are two similar point charges situated at points A and B. \(r_{12}\) is the distance of separation between them. \(\vec{F}_{21}\) denotes the force exerted on \(q_2\) by \(q_1\):
\[\vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \times \frac{q_1 q_2}{\left|\vec{r}_{21}\right|^2} \times \hat{r}_{21} \qquad \text{--- (10.6)}\]
Diagram
Fig. 10.5 – Charges +q₁ at A and +q₂ at B with position vectors r₁ and r₂ from origin O, separation r₁₂, and the repulsive forces F₁₂ on A and F₂₁ on B drawn outward along the line AB.
Here \(\hat{r}_{21}\) is the unit vector along \(\overline{AB}\), away from B. A unit vector is simply a vector of length 1 whose only job is to carry direction; the magnitude of the force is contained in the fraction in front of it.
Similarly, the force \(\vec{F}_{12}\) exerted on \(q_1\) by \(q_2\) is given by
\[\vec{F}_{12} = \frac{1}{4\pi\varepsilon_0} \times \frac{q_1 q_2}{\left|\vec{r}_{12}\right|^2} \times \hat{r}_{12} \qquad \text{--- (10.7)}\]
where \(\hat{r}_{12}\) is the unit vector along \(\overline{BA}\), away from A. \(\vec{F}_{12}\) acts on \(q_1\) at A and is directed along BA, away from A.
The two unit vectors point in exactly opposite directions, i.e.
\[\hat{r}_{12} = -\,\hat{r}_{21}\]
and hence
\[\vec{F}_{21} = -\,\vec{F}_{12}\]
Thus, the two charges experience forces of equal magnitude and opposite direction. These two forces form an action–reaction pair, exactly as Newton’s third law requires. Note carefully that they act on different bodies, so they never cancel each other.
10.5 Principle of Superposition
Coulomb’s law tells us the force between two point charges. But a real problem almost never has just two charges. What happens to the force on one charge when three, four or a hundred other charges are present around it? Does the presence of a third charge alter the force that the second one exerts?
The answer, established by experiment, is beautifully simple: it does not. Each pair of charges interacts as if the others were not there, and the forces then simply add as vectors.
Building the Resultant Force, One Pair at a Time
Consider a number of point charges \(q_1\), \(q_2\), \(q_3\), … kept at points \(A_1\), \(A_2\), \(A_3\), … as shown in Fig. 10.6. Suppose we want the total force on the charge \(q_1\).
Diagram
Fig. 10.6 – Charge q₁ at A₁ with charges q₂, q₃, q₄, q₅ around it at A₂, A₃, A₄, A₅, and the individual force vectors F₁₂, F₁₃, F₁₄, F₁₅ drawn on q₁.
The force exerted on the charge \(q_1\) by \(q_2\) is \(\vec{F}_{12}\). The value of \(\vec{F}_{12}\) is calculated by ignoring the presence of all the other charges — that is, using Coulomb’s law for the pair \((q_1, q_2)\) alone. Similarly, we find \(\vec{F}_{13}\), \(\vec{F}_{14}\) and so on, one at a time, each time using Coulomb’s law and pretending the rest of the charges are absent.
Only after all of them have been found separately do we combine them. Notice in Fig. 10.6 that these individual forces point in different directions, so they must be added as vectors, not as numbers.
The total force \(\vec{F}_1\) on charge \(q_1\) is the vector sum of all such forces:
\[\vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} + \vec{F}_{14} + \dots\]
\[= \frac{1}{4\pi\varepsilon_0}\left[\frac{q_1 q_2}{\left|\vec{r}_{12}\right|^2}\,\hat{r}_{12} + \frac{q_1 q_3}{\left|\vec{r}_{13}\right|^2}\,\hat{r}_{13} + \dots \right]\]
where \(\hat{r}_{12}\), \(\hat{r}_{13}\), etc., are unit vectors directed to \(q_1\) from \(q_2\), \(q_3\), etc., and \(r_{12}\), \(r_{13}\), \(r_{14}\), etc., are the distances from \(q_1\) to \(q_2\), \(q_3\), etc., respectively. Each term is measured in newton (N), each charge in coulomb (C), each distance in metre (m).
The General Form for N Charges
Let there be \(N\) point charges \(q_1\), \(q_2\), \(q_3\), …, \(q_N\). The force \(\vec{F}\) exerted by these charges on a test charge \(q_0\) can be written using the summation notation \(\Sigma\) as follows:
\[\vec{F}_{\text{test}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \dots + \vec{F}_N \qquad \text{--- (10.9)}\]
\[= \sum_{n=1}^{N} \vec{F}_n = \frac{1}{4\pi\varepsilon_0} \sum_{n=1}^{N} \frac{q_0\, q_n}{r_{0n}^2}\, \hat{r}_{0n} \qquad \text{--- (10.10)}\]
where \(\hat{r}_{0n}\) is a unit vector directed from the \(n^{\text{th}}\) charge to the test charge \(q_0\), and \(r_{0n}\) is the separation between them, so that
\[\vec{r}_{0n} = r_{0n}\, \hat{r}_{0n}\]
Equation (10.10) looks formidable, but it says nothing more than the sentence we began with: compute each pairwise Coulomb force, then add the arrows.
In exam problems this is almost always applied in a fixed sequence — find the magnitude of each individual force from Coulomb’s law, mark its direction along the line joining that pair, resolve the forces into components (or use the parallelogram law for two forces), and only then combine. Skipping straight to a scalar sum is the commonest mistake made with superposition.
10.6 Electric Field
Coulomb’s law describes a force between two charges, but it leaves an uncomfortable question hanging: how does one charge “know” that the other is there, across empty space, with nothing in between?
The idea of the electric field answers this by splitting the problem into two halves. First, a charge \(Q\) modifies the space around itself. Second, any other charge placed in that modified space feels a force. The space around a charge \(Q\) gets modified in such a way that when a test charge is brought into this region, it experiences a Coulomb force.
Field as Force per Unit Charge
Mathematically, the electric field is defined as the force experienced per unit charge. This definition is what converts a vague “region of influence” into a measurable quantity.
Let \(Q\) and \(q\) be two charges separated by a distance \(r\). The Coulomb force between them is given by
\[\vec{F} = \frac{1}{4\pi\varepsilon_0}\, \frac{Qq}{r^2}\, \hat{r}\]
where \(\hat{r}\) is the unit vector along the line joining \(Q\) to \(q\).
Dividing this force by the charge \(q\) on which it acts removes \(q\) from the expression altogether. What is left depends only on the source charge \(Q\) and on the position. Therefore, the electric field due to charge \(Q\) is given by
\[\vec{E} = \frac{\vec{F}}{q} = \frac{Q}{4\pi\varepsilon_0 r^2}\, \hat{r} \qquad \text{--- (10.11)}\]
Here \(\vec{E}\) is the electric field (or electric intensity) at the point where \(q\) was placed, \(Q\) is the source charge in coulomb (C), \(r\) is the distance from \(Q\) in metre (m), and \(\varepsilon_0\) is the permittivity of free space.
Two statements follow, and both are conceptually important.
The Coulomb force acts across empty space (vacuum) and does not need any intervening medium for its transmission. And, crucially, the electric field exists around a charge irrespective of the presence of other charges. The field is not created by the test charge; the test charge merely reveals a field that was already there.
Since the Coulomb force is a vector, the electric field of a charge is also a vector, and it is directed along the direction of the Coulomb force experienced by a positive test charge.
The Field of a Point Charge
The magnitude of the electric field at a distance \(r\) from a point charge \(Q\) is the same at all points on the surface of a sphere of radius \(r\), as shown in Fig. 10.8. Its direction is along the radius of the sphere, pointing away from its centre if the charge is positive (and towards the centre if the charge is negative).
Diagram
Fig. 10.8 – A point charge +Q at the centre of a sphere of radius r, with the field vector E drawn along the radius, pointing outward, labelled E = Q/4πε₀r².
The SI unit of electric intensity is newton per coulomb (N C⁻¹). In practice, the electric field is more often expressed in volt per metre (V m⁻¹); the two units are identical, as will be shown in Article 10.6.2.
The dimensional formula of \(E\) is obtained from \(E = F/q_0\):
\[E = \frac{F}{q_0} = \frac{[\text{LMT}^{-2}]}{[\text{IT}]}\]
\[E = [\text{LMT}^{-3}\,\text{I}^{-1}]\]
10.6.1 Electric Field Intensity due to a Point Charge in a Material Medium
We saw in Section 10.4.2 that a medium weakens the force between charges. Naturally, it weakens the field too, and by exactly the same factor.
Consider a point charge \(q\) placed at point O in a medium of dielectric constant \(K\), as shown in Fig. 10.9.
Diagram
Fig. 10.9 – A point charge q at O inside a rectangular block of medium of dielectric constant k, with a test charge q₀ at a point P at distance r along OP.
Consider the point P in the electric field of the point charge \(q\), at a distance \(r\) from it. A test charge \(q_0\) placed at the point P will experience a force which is given by Coulomb’s law in a medium,
\[\vec{F} = \frac{1}{4\pi\varepsilon_0 K}\, \frac{q\, q_0}{r^2}\, \hat{r}\]
where \(\hat{r}\) is the unit vector in the direction of the force, i.e. along OP.
By the definition of electric field intensity (force per unit charge),
\[\vec{E} = \frac{\vec{F}}{q_0} = \frac{1}{4\pi\varepsilon_0 K}\, \frac{q}{r^2}\, \hat{r}\]
The direction of \(\vec{E}\) will be along OP when \(q\) is positive, and along PO when \(q\) is negative.
The magnitude of electric field intensity in a medium is therefore given by
\[E = \frac{1}{4\pi\varepsilon_0 K}\, \frac{q}{r^2} \qquad \text{--- (10.12)}\]
For air or vacuum, \(K = 1\), and Eq. (10.12) reduces to the familiar result
\[E = \frac{1}{4\pi\varepsilon_0}\, \frac{q}{r^2}\]
which is exactly the magnitude of Eq. (10.11), as it must be.
The Inverse Square Fall-Off
The Coulomb force between two charges and the electric field \(E\) of a charge both follow the inverse square law, \(F \propto 1/r^2\) and \(E \propto 1/r^2\). This means that as one moves away from the charge, both quantities fall steeply at first and then flatten out into a long tail, as the graph in Fig. 10.10 shows. The field never quite becomes zero — it only becomes negligibly small.
Diagram
Fig. 10.10 – Graph of E or F on the vertical axis against r on the horizontal axis, showing the steeply falling inverse-square curve labelled F ∝ 1/r², E ∝ 1/r².
Two Kinds of Electric Field
- Uniform electric field: A uniform electric field is a field whose magnitude and direction are the same at all points. For example, the field between two parallel plates, as in Fig. 10.11 (a). Here the field lines are parallel, equally spaced arrows.
Diagram
Fig. 10.11 (a) – Uniform electric field between two oppositely charged parallel plates, shown as evenly spaced parallel arrows from the positive plate to the negative plate.
- Non-uniform electric field: A field whose magnitude and direction are not the same at all points. For example, the field due to a point charge, as in Fig. 10.11 (b). In this case, the magnitude of the field is the same at a distance \(r\) from the point charge in any direction, but the direction of the field is not the same — it points radially outward, and so differs from point to point.
Diagram
Fig. 10.11 (b) – Non-uniform electric field of a point charge +q, shown as arrows radiating outward in all directions.
10.6.2 Practical Way of Calculating Electric Field
Equation (10.11) is fine for a point charge, but in the laboratory we rarely start with a known charge. What we actually control is a voltage. So we need a way of getting the field from a potential difference — and the parallel-plate arrangement gives it to us.
A pair of charged parallel plates is arranged as shown in Fig. 10.12. The electric field between them is uniform. A potential difference \(V\) is applied between the two parallel plates separated by a distance \(d\). The electric field between them is directed from plate A (positive) to plate B (negative).
Diagram
Fig. 10.12 – Two parallel plates A (positive) and B (negative) separated by distance d, with a charge +q between them and the uniform field directed from A to B.
A charge \(+q\) placed between the plates experiences a force \(F\) due to the electric field. If we have to move the charge against the direction of the field, i.e. towards the positive plate, we have to do some work on it. If we move the charge \(+q\) from the negative plate B to the positive plate A, the work done against the field is
\[W = Fd\]
where \(d\) is the separation between the plates. (The force is constant along the path because the field is uniform, so work is simply force times distance.)
The same work can be written using the potential difference \(V\) between the two plates, since potential difference is work done per unit charge:
\[W = Vq, \quad \text{but } W = Fd\]
\[\therefore Vq = Fd \quad \therefore \frac{F}{q} = \frac{V}{d} = E\]
But \(F/q\) is precisely the electric field. Therefore the electric field can be defined as
\[E = \frac{V}{d} \qquad \text{--- (10.13)}\]
where \(V\) is the potential difference in volt (V) and \(d\) is the separation in metre (m), giving \(E\) in volt per metre (V m⁻¹). This is the commonly used definition of electric field, and it is the one that makes N C⁻¹ and V m⁻¹ the same unit.
10.6.3 Electric Lines of Force
Numbers and formulae give the field at one point at a time. To see the whole field at once, we need a picture — and that picture is due to one of the great experimenters of physics.
See Fig. 10.14. If a test charge is placed in an electric field, it would be acted upon by a force at every point in the field, and will move along a path. The path along which the unit positive charge moves is called a line of force.
Diagram
Fig. 10.14 – A curved electric line of force with tangent vectors E_A at point A and E_B at point B drawn along the curve, showing the field direction at each point.
A line of force is thus defined as a curve such that the tangent at any point to this curve gives the direction of the electric field at that point. This is why the lines are drawn as curves rather than straight lines in general: the direction of the field usually changes from point to point, and the curve simply follows it.
Reading Strength from the Picture
The lines carry more information than direction alone. The density of field lines indicates the strength of the electric field at a given point in space, as illustrated in Fig. 10.15. Where the lines crowd together, the field is strong; where they spread apart, it is weak.
Diagram
Fig. 10.15 – Field lines converging from a widely spaced region at A (low field) to a closely spaced region at B (high field), showing how line density represents field strength.
Characteristics of Electric Lines of Force
The lines of force originate from a positively charged object and terminate on a negatively charged object.
The lines of force neither intersect nor meet each other, as it would mean that the electric field has two directions at a single point — which is impossible, since the force on a test charge there could only point one way.
The lines of force leave or terminate on a conductor normally, i.e. at right angles to its surface.
The lines of force do not pass through a conductor, i.e. the electric field inside a conductor is always zero; but they do pass through insulators.
The magnitude of the electric field intensity is proportional to the number of lines of force per unit area of the surface held perpendicular to the field.
Electric lines of force are crowded in a region where the electric intensity is large.
Electric lines of force are widely separated from each other in a region where the electric intensity is small.
The lines of force of a uniform electric field are parallel to each other and are equally spaced.
Figure 10.16 collects the standard pictures: the outward radial lines of a positive charge, the inward radial lines of a negative charge, the curved lines running from a positive to a negative charge, the mutual repulsion of the lines of two like charges, the lines meeting a conductor at right angles with \(E = 0\) inside it, and the crowding of lines where the field is strong.
Diagram
Fig. 10.16 (a)–(f) – (a) radial lines of force directed outward from +q; (b) radial lines directed inward towards −q; (c) curved lines from +q to −q for an unlike pair; (d) lines of force for two like charges, showing no lines between them; (e) lines terminating normally on a conductor with E = 0 inside; (f) lines converging so that more lines cross the same area at A than at B, so the intensity is more at A and less at B.
Finally, a caution that is easy to forget once the diagrams start to look convincing. The lines of force are purely a geometric construction which helps us visualise the nature of the electric field in a region. The lines of force have no physical existence — there are no threads in space; there is only the field, which has a definite magnitude and direction at every point.
10.7 Electric Flux
In the previous section we agreed on a convention: the number of lines of force per unit area is a measure of the intensity of the electric field \(\vec{E}\). Let us now turn that convention around and use it to define a new and very useful quantity.
If the field strength is “lines per unit area”, then
\[\therefore E = \frac{\text{Number of lines of force}}{\text{Area enclosing the lines of force}} \qquad \text{--- (10.14)}\]
and therefore, rearranging,
\[\text{Number of lines of force} = (E) \cdot (\text{Area})\]
This product — field times area — is what we call electric flux. Loosely, it counts how many field lines pass through a chosen surface. The word “flux” comes from the Latin for flow; nothing is actually flowing here, but the picture of lines streaming through a window is the right one to keep in mind.
When the Surface is Tilted
There is a complication. The number of lines that get through a window depends not only on how big the window is, but on how it is turned. Hold a hoop face-on to falling rain and much rain passes through; turn it edge-on and almost none does, even though the hoop has not changed size.
The same happens here. When the area is inclined at an angle \(\theta\) with the direction of the electric field, as shown in Fig. 10.17, the electric flux must be calculated as follows.
Diagram
Fig. 10.17 – A plane area S tilted in a uniform field: parallel field lines E strike the surface, the outward normal S makes an angle θ with E, and the angle θ is marked between the field direction and the normal.
To handle direction properly, a surface is represented by an area vector. The area vector \(d\vec{S}\) has a magnitude equal to the area of the small surface element, and its direction is along the normal (perpendicular) to that surface. Representing a surface by its normal, rather than by the surface itself, is what allows us to speak of an “angle between the field and the area”.
Let the angle between the electric field \(\vec{E}\) and the area vector \(d\vec{S}\) be \(\theta\). Then the electric flux passing through the small area \(dS\) is given by
\(d\phi\) = (component of \(dS\) along \(\vec{E}\)) · (area of \(d\vec{S}\))
\[d\phi = E\,(dS \cos\theta)\]
\[d\phi = E\, dS \cos\theta\]
\[d\phi = \vec{E} \cdot d\vec{S} \qquad \text{--- (10.15)}\]
Equation (10.15) is simply the dot product of the field vector and the area vector — which is exactly the compact way of writing “field times area times the cosine of the angle between them”.
Flux Through a Whole Surface
Equation (10.15) gives the flux through one small element of a surface. A real surface may be large, and may be curved, so that both \(\vec{E}\) and the direction of the normal change from place to place. To get the total flux, we add up the contributions of all the small elements — which, in the language of calculus, means integrating over the surface.
The total flux through the entire surface is
\[\Phi = \int d\phi = \int_{s} \vec{E} \cdot d\vec{S} = \vec{E} \cdot \vec{S} \qquad \text{--- (10.16)}\]
The last step, \(\Phi = \vec{E} \cdot \vec{S}\), is valid in the simple case where the field is uniform over a flat surface, so that \(\vec{E}\) can be taken out of the integral and \(\int d\vec{S} = \vec{S}\).
Note that flux is a scalar quantity, even though it is built out of two vectors. It can be positive, negative or zero: positive when the field lines emerge through the surface (angle with the normal less than 90°), negative when they enter it (angle greater than 90°), and zero when they merely graze it.
The SI unit of electric flux can be worked out from Eq. (10.16):
\[\Phi = \vec{E} \cdot \vec{S} = (\text{V/m})\ \text{m}^2 = \text{Vm}\]
So flux is measured in volt metre (V m) — which, since 1 V m⁻¹ = 1 N C⁻¹, is the same as N m² C⁻¹.
Electric flux may look like a mere bookkeeping device at this stage — a way of counting lines. Its real power appears in the next section, where Gauss showed that the total flux out of any closed surface depends on nothing at all except the charge locked inside it.
10.8 Gauss’ Law
Gauss’ law is analogous to Coulomb’s law in the sense that it too expresses a relationship between the electric field and electric charge. But it does so from a completely different angle, and this makes it an equivalent method for finding electric intensity — often a far easier one. It relates the values of the field on a closed surface to the total charge enclosed by that surface.
Setting Up the Proof
Consider a closed surface of any shape which encloses a number of positive electric charges, as shown in Fig. 10.18. Note the words “any shape” — the surface need not be a sphere, or even regular. To prove Gauss’ theorem, imagine a small charge \(+q\) present at a point O inside the closed surface, and imagine an infinitesimal area \(dS\) of this irregular closed surface.
Diagram
Fig. 10.18 – An irregular closed surface enclosing a charge +q at O; a small area element dS at point P, with OP = r, the outward normal to dS making an angle θ with the field E = q/4πε₀r², and the solid angle dω subtended at O.
The magnitude of the electric field intensity at the point P on \(dS\), due to the charge \(+q\) at the point O, is given by the point-charge result:
\[E = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r^2}\right)\]
The direction of \(\vec{E}\) is away from the point O, along OP. Let \(\theta\) be the angle subtended by the normal drawn to the area \(dS\) and the direction of \(\vec{E}\).
The electric flux \(d\phi\) passing through the area \(dS\) is then, from Eq. (10.15),
\[d\phi = E \cos\theta \; dS\]
\[= \frac{q}{4\pi\varepsilon_0 r^2}\cos\theta \; dS\]
\[= \left(\frac{q}{4\pi\varepsilon_0}\right)\left(\frac{dS \cos\theta}{r^2}\right)\]
\[d\phi = \left(\frac{q}{4\pi\varepsilon_0}\right) d\omega \qquad \text{--- (10.17)}\]
where
\[d\omega = \frac{dS \cos\theta}{r^2}\]
is the solid angle subtended by the area \(dS\) at the point O.
The step that turns the whole argument is this one. A solid angle is the three-dimensional counterpart of an ordinary angle: it measures how much of the “view” from O is taken up by the patch \(dS\). Its definition already contains the factor \(\cos\theta/r^2\) — precisely the two awkward factors in the flux. So the tilt of the patch and its distance both disappear into \(d\omega\), and the flux through the patch ends up depending on nothing but the charge and the solid angle.
Adding Up Over the Whole Surface
The total electric flux, \(\phi_E\), crossing the given closed surface can be obtained by integrating Eq. (10.17) over its area. Thus,
\[\Phi_E = \int d\phi = \int_{s} \vec{E} \cdot \vec{ds} = \int \frac{q}{4\pi\varepsilon_0}\, d\omega = \frac{q}{4\pi\varepsilon_0} \int d\omega\]
But the total solid angle subtended by the entire closed surface at the point O inside it is
\[\int d\omega = 4\pi\]
(just as the total ordinary angle around a point in a plane is \(2\pi\) radians, the total solid angle around a point in space is \(4\pi\) steradian).
Therefore,
\[\text{Total flux} = \frac{q}{4\pi\varepsilon_0}(4\pi)\]
\[\Phi_E = \int_{s} \vec{E} \cdot \vec{ds} = +\,q/\varepsilon_0 \qquad \text{--- (10.18)}\]
Look at what has vanished from the answer: the shape of the surface, the size of the surface, the position of the charge inside it, and the distance \(r\). Nothing survives except the charge itself and the constant \(\varepsilon_0\). This is true for every electric charge enclosed by a given closed surface.
So, total flux due to charge \(q_1\), over the given closed surface \(= +\,q_1/\varepsilon_0\);
total flux due to charge \(q_2\), over the given closed surface \(= +\,q_2/\varepsilon_0\);
total flux due to charge \(q_n\), over the given closed surface \(= +\,q_n/\varepsilon_0\).
What the Signs Mean
The positive sign in Eq. (10.18) indicates that the flux is directed outwards, away from the charge — as in Fig. 10.19 (a). If the charge is negative, the flux will be directed inwards, as shown in Fig. 10.19 (b), and the flux is then negative.
And if a charge lies outside the closed surface, the net flux through the surface will be zero, as in Fig. 10.19 (c). This makes sense from the line picture: every line of force that enters the closed surface on one side must leave it on the other, so the inward and outward contributions cancel exactly.
Diagram
Fig. 10.19 (a) – Outward flux lines emerging from a closed surface enclosing a positive charge +q.
Diagram
Fig. 10.19 (b) – Inward flux lines entering a closed surface enclosing a negative charge −q.
Diagram
Fig. 10.19 (c) – A charge q outside a closed surface, with field lines entering on one side and leaving on the other, so the net flux is zero.
Statement of Gauss’ Law
According to the principle of superposition, the total flux \(\phi\) due to all the charges enclosed within the given closed surface is the sum of the individual contributions:
\[\Phi_E = \frac{q_1}{\varepsilon_0} + \frac{q_2}{\varepsilon_0} + \frac{q_3}{\varepsilon_0} + \dots + \frac{q_n}{\varepsilon_0} = \sum_{i=1}^{i=n} \frac{q_i}{\varepsilon_0} = \frac{Q}{\varepsilon_0}\]
\[\int \vec{E} \cdot d\vec{S} = \frac{Q}{\varepsilon_0}\]
where \(Q\) is the total charge within the surface.
Gauss’ law is applicable to any closed surface of regular or irregular shape. That freedom is the whole point of the law, and it is what makes it a practical tool: we are allowed to choose a surface convenient for calculation.
10.9 Electric Dipole
So far we have dealt with single charges. But in nature, charges very often occur in pairs — equal amounts of positive and negative charge, sitting a small distance apart. A water molecule is such a pair. So is a molecule of ammonia. This arrangement is common enough, and important enough, to deserve a name of its own.
An electric dipole is shown in Fig. 10.21, with charge \(-q\) at A and \(+q\) at B.
Diagram
Fig. 10.21 – Electric dipole with −q at A and +q at B separated by 2l; the axial line x–y drawn through both charges, and the equatorial line P–Q drawn through the centre perpendicular to the axis; the dipole moment p marked from −q to +q.
Three lines matter in every dipole problem, so learn them now.
The line joining the two charges is called the dipole axis. A line passing through the dipole axis is called the axial line. A line passing through the centre of the dipole and perpendicular to the axial line is called the equatorial line, as shown in Fig. 10.21. (The names are borrowed from geography: the axis runs “pole to pole”, and the equator cuts across the middle.)
Dipole Moment
The strength of a dipole is measured in terms of a quantity called the dipole moment.
Let \(q\) be the magnitude of each charge and \(2\vec{l}\) be the distance from the negative charge to the positive charge. Then the product \(q\,(2\vec{l})\) is called the dipole moment \(\vec{p}\).
\[\vec{p} = q\,(2\vec{l}) \qquad \text{--- (10.19)}\]
Notice that the dipole moment depends on both the size of the charges and how far apart they are. Two tiny charges far apart can have the same dipole moment as two large charges close together.
A dipole moment is a vector whose magnitude is \(q\,(2l)\) and whose direction is from the negative to the positive charge. (In chemistry the opposite convention is sometimes used; in physics, always take it from \(-q\) to \(+q\).)
The unit of dipole moment is the coulomb-metre (Cm), or the debye (D), where
\[1\ \text{D} = 3.33 \times 10^{-30}\ \text{Cm}\]
To get a feel for the size: if two charges \(+e\) and \(-e\) are separated by \(1.0\ \text{Å}\) (one angstrom, \(10^{-10}\) m — roughly the size of an atom), the dipole moment is \(1.6 \times 10^{-29}\) Cm, or 4.8 D. For example, a water molecule has a permanent dipole moment of \(6.172 \times 10^{-30}\) Cm or 1.85 D. Its direction is from oxygen to hydrogen. (See the note on the natural dipole below.)
Diagram
Fig. 10.22 – A water molecule with the two H–O bonds at 105°, the effective +q on the hydrogen side and −q on the oxygen side, showing the permanent dipole moment.
10.9.1 Couple Acting on an Electric Dipole in a Uniform Electric Field
What happens when a dipole is placed in a uniform electric field? The positive charge is pushed one way and the negative charge the other — so the dipole does not move off as a whole, but it does turn.
Consider an electric dipole placed in a uniform electric field \(E\). The axis of the electric dipole makes an angle \(\theta\) with the direction of the electric field, as shown in Fig. 10.23 (a).
Diagram
Fig. 10.23 (a) – A dipole of length 2l with −q and +q placed in a uniform field E, its axis making angle θ with E, and the dipole moment p = q(2l) marked along the axis.
Figure 10.23 (b) shows the couple acting on an electric dipole in a uniform electric field.
Diagram
Fig. 10.23 (b) – Couple acting on the dipole: force +qE on the charge +q along E, force −qE on the charge −q opposite to E, forming a couple that rotates the dipole.
The force acting on the charge \(-q\) at A is
\[\vec{F}_A = -\,q\vec{E}\]
in the direction opposite to that of \(\vec{E}\), and the force acting on the charge \(+q\) at B is
\[\vec{F}_B = +\,q\vec{E}\]
in the direction of \(\vec{E}\).
Since \(\vec{F}_A = -\,\vec{F}_B\), the two forces are equal in magnitude and opposite in direction. Their vector sum is zero, so there is no net force on the dipole and its centre does not accelerate. But they do not act along the same line: they are separated by a distance, and two equal and opposite forces separated by a distance form a couple.
The moment of the couple is called torque, and is defined by
\[\vec{\tau} = \vec{d} \times \vec{F}\]
where \(d\) is the perpendicular distance between the two equal and opposite forces.
∴ Magnitude of Torque = Magnitude of force × Perpendicular distance
∴ Torque on the dipole \(= \vec{\tau} = \overline{BP} \times q\vec{E}\)
Working out the perpendicular distance from the geometry of Fig. 10.23 (b), it is \(2l \sin\theta\), and so
\[\therefore \tau = qE\,2l\sin\theta \qquad \text{--- (10.20)}\]
But \(\vec{p} = q \times 2\vec{l}\), i.e. \(p = q(2l)\). Substituting,
\[\therefore \tau = pE \sin\theta \qquad \text{--- (10.21)}\]
In vector form,
\[\vec{\tau} = \vec{p} \times \vec{E} \qquad \text{--- (10.22)}\]
where \(\tau\) is in newton metre (N m), \(p\) in coulomb metre (C m), and \(E\) in newton per coulomb (N C⁻¹) or volt per metre (V m⁻¹).
Reading the Two Extreme Cases
Equation (10.21) contains the whole physics of a dipole in a field, if you read its two limits.
If \(\theta = 90^\circ\), then \(\sin\theta = 1\) and \(\tau = pE\). When the axis of the electric dipole is perpendicular to the uniform electric field, the torque of the couple acting on the electric dipole is maximum.
If \(\theta = 0\), then \(\tau = 0\). This is the minimum torque on the dipole.
Put these together and the behaviour is clear: torque tends to align the dipole axis along the direction of the electric field. The turning effect is strongest when the dipole is broadside to the field, and it dies to nothing exactly when the dipole has swung into line with the field. That is the position the dipole settles into.
10.9.2 Electric Intensity at a Point due to an Electric Dipole
A dipole has zero net charge, but it is not electrically silent. We now calculate the field it produces — first at a point on its axis, then at a point on its equator. These are the two standard positions, and they are the two that examinations ask about.
Case 1: At a Point on the Axis of a Dipole
Consider an electric dipole consisting of two charges \(-q\) and \(+q\) separated by a distance \(2l\), as shown in Fig. 10.25. Let P be a point at a distance \(r\) from the centre C of the dipole, lying on the axial line. The electric intensity \(\vec{E}_a\) at P due to the dipole is the vector sum of the field due to the charge \(-q\) at A and the field due to \(+q\) at B.
Diagram
Fig. 10.25 – Dipole with −q at A and +q at B, centre C, and a point P on the axial line at distance r from C; the fields E_A and E_B at P drawn along the axis in opposite senses, with |E_B| > |E_A|.
The electric field intensity at P due to the charge \(-q\) at A is
\[\vec{E}_A = \frac{1}{4\pi\varepsilon_0}\, \frac{(-q)}{(r + l)^2}\ \widehat{PD}\]
where \(\widehat{PD}\) is the unit vector directed along \(\overline{PD}\). Note the distance: the far charge A is at \((r + l)\) from P.
The electric intensity at P due to the charge \(+q\) at B is
\[\vec{E}_B = \frac{1}{4\pi\varepsilon_0}\, \frac{q}{(r - l)^2}\ \widehat{PQ}\]
where \(\widehat{PQ}\) is a unit vector directed along \(\overline{PQ}\). The near charge B is at only \((r - l)\) from P.
The magnitude of \(\vec{E}_B\) is greater than that of \(\vec{E}_A\) (because BP < AP). This is the crucial asymmetry: the two fields point in opposite directions, but they do not cancel, because the nearer charge always wins.
The resultant field \(\vec{E}_a\) at P on the axis, due to the dipole, is
\[\vec{E}_a = \vec{E}_B + \vec{E}_A\]
The magnitude of \(\vec{E}_a\) is given by taking the difference of the two magnitudes:
\[\left|\vec{E}_a\right| = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{(r-l)^2} - \frac{q}{(r+l)^2}\right]\]
Taking the LCM \((r^2 - l^2)^2\) and expanding the squares in the numerator:
\[\left|\vec{E}_a\right| = \frac{q}{4\pi\varepsilon_0}\left[\frac{r^2 + l^2 + 2lr - r^2 + 2lr - l^2}{(r^2 - l^2)^2}\right]\]
The \(r^2\) and \(l^2\) terms cancel, leaving only \(4lr\):
\[\left|\vec{E}_a\right| = \frac{2(2lq)r}{4\pi\varepsilon_0 (r^2 - l^2)^2}\]
But \(2lq = p\), the dipole moment. Therefore,
\[\left|\vec{E}_a\right| = \frac{1}{4\pi\varepsilon_0}\, \frac{2pr}{(r^2 - l^2)^2} \qquad \text{--- (10.23)}\]
\(\vec{E}_a\) is directed along PQ, which is the direction of the dipole moment \(\vec{p}\), i.e. from the negative to the positive charge, parallel to the axis.
The far-field approximation. In most practical situations the point P is very far from the dipole compared with the dipole’s own size. If \(r \gg l\), then \(l^2\) can be neglected compared to \(r^2\), so \((r^2 - l^2)^2 \approx r^4\), and Eq. (10.23) becomes
\[\left|\vec{E}_a\right| = \frac{1}{4\pi\varepsilon_0}\, \frac{2p}{r^3} \qquad \text{--- (10.24)}\]
The field will be along the direction of the dipole moment \(\vec{p}\).
Case 2: At a Point on the Equatorial Line
Now let P be a point on the equatorial line, at a distance \(r\) from the centre C of the dipole, as shown in Fig. 10.26 (a).
Diagram
Fig. 10.26 (a) – Dipole with −q at A and +q at B, centre C; point P on the equatorial line at distance r from C; fields E_A (directed from P towards A) and E_B (directed away from B) drawn at P, each making angle θ with the equatorial direction.
The electric field at point P due to the charge \(-q\) at A is
\[\vec{E}_A = \frac{1}{4\pi\varepsilon_0}\, \frac{(-q)}{(\text{AP})^2}\ \widehat{PA}\]
where \(\widehat{PA}\) is the unit vector in the direction along \(\overline{PA}\).
Similarly, the electric field at P due to the charge \(+q\) at B is
\[\vec{E}_B = \frac{1}{4\pi\varepsilon_0}\, \frac{(+q)}{(\text{BP})^2}\ \widehat{PQ}\]
where \(\widehat{PQ}\) is the unit vector directed along \(\overline{PQ}\) or \(\overline{BP}\).
The electric field at P is the sum of \(\vec{E}_A\) and \(\vec{E}_B\):
\[\therefore \vec{E}_{eq} = \vec{E}_A + \vec{E}_B\]
Finding the distances. Consider the triangle ACP. Since the equatorial line is perpendicular to the axis, ACP is right-angled at C, and by Pythagoras’ theorem
\[(\text{AP})^2 = (\text{PC})^2 + (\text{AC})^2 = r^2 + l^2 = (\text{BP})^2\]
So P is equally distant from both charges. Hence
\[\therefore \left|\vec{E}_A\right| = \frac{1}{4\pi\varepsilon_0}\, \frac{q}{(r^2 + l^2)} \qquad \text{--- (10.25)}\]
\[\left|\vec{E}_B\right| = \frac{1}{4\pi\varepsilon_0}\, \frac{q}{(r^2 + l^2)} \qquad \text{--- (10.26)}\]
\[\left|\vec{E}_A\right| = \left|\vec{E}_B\right|\]
This is the essential difference from the axial case: on the equator, the two fields are exactly equal in magnitude. They do not point in the same direction, though — so we must resolve them.
Resolving the two fields. The resultant of the fields \(\vec{E}_A\) and \(\vec{E}_B\) acting at point P can be calculated by resolving these vectors along the equatorial line and along a direction perpendicular to it.
Diagram
Fig. 10.26 (b) – Components of the fields at P: E_A and E_B resolved into E_A sinθ and E_B sinθ along the y-axis (equal and opposite, cancelling) and E_A cosθ and E_B cosθ along the x-axis (equal and adding).
Consider Fig. 10.26 (b). Let the \(y\)-axis coincide with the equator of the dipole; the \(x\)-axis will then be parallel to the dipole axis, as shown. The origin is at the point P.
The \(y\)-components of \(E_A\) and \(E_B\) are \(E_A \sin\theta\) and \(E_B \sin\theta\) respectively. They are equal in magnitude but opposite in direction, so they cancel each other. There is no contribution from them towards the resultant.
The \(x\)-components of \(E_A\) and \(E_B\) are \(E_A \cos\theta\) and \(E_B \cos\theta\) respectively. They are of equal magnitude and are in the same direction, so they add:
\[\therefore \left|\vec{E}_{eq}\right| = E_A \cos\theta + E_B \cos\theta \qquad \text{--- (10.27)}\]
By using Eq. (10.25) and (10.26), and putting \(\cos\theta = \dfrac{l}{\sqrt{r^2 + l^2}}\) from the geometry of Fig. 10.26 (a):
\[\left|\vec{E}_{eq}\right| = 2 E_A \cos\theta\]
\[= 2\left(\frac{q}{4\pi\varepsilon_0 (r^2 + l^2)}\right) \frac{l}{\sqrt{r^2 + l^2}}\]
\[= \frac{2ql}{4\pi\varepsilon_0 (r^2 + l^2)^{3/2}}\]
The far-field approximation. If \(r \gg l\), then \(l^2\) is very small compared to \(r^2\), and, recognising \(2ql = p\),
\[\left|\vec{E}_{eq}\right| = \frac{1}{4\pi\varepsilon_0}\, \frac{p}{(r^2)^{3/2}} = \frac{1}{4\pi\varepsilon_0}\, \frac{p}{r^3} \qquad \text{--- (10.28)}\]
The direction of this field is along \(-\vec{p}\) (anti-parallel to \(\vec{p}\)), as shown in Fig. 10.26 (c). On the equator, the field points opposite to the dipole moment — the reverse of the axial case.
Diagram
Fig. 10.26 (c) – Point P on the equatorial line with the resultant field E_eq drawn anti-parallel to the dipole moment p.
Axial Versus Equatorial: The Factor of Two
Comparing Eq. (10.28) with Eq. (10.24), we find that the electric intensity at an axial point is twice that at a point on the equatorial position, lying at the same distance from the centre of the dipole.
\[E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}, \qquad E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}\]
Both fall off as \(1/r^3\); only the constant differs, and the directions are opposite. Remembering these two results together — “axial is twice equatorial, and points along \(\vec{p}\); equatorial points against \(\vec{p}\)” — will save you a great deal of time in objective questions.
10.10 Continuous Charge Distribution
Everything we have done so far has been built out of point charges. But a charged rod, a charged sheet of plastic or a charged sphere does not contain three or four charges — it contains something like \(10^{13}\) of them, packed so closely that they form, for all practical purposes, a smooth smear of charge.
Counting them one by one and adding \(10^{13}\) Coulomb forces is out of the question. Instead we treat the charge as continuous.
The condition in that definition matters. “Continuous” is not a property of the charge itself — charge is always quantized, as we saw in Section 10.3.2 — it is a statement about the scale on which we are looking. From a metre away, the individual electrons on a charged rod are indistinguishable from a smooth line of charge. From a nanometre away, they most certainly are not.
The charge distribution is continuous in the sense that a system of closely spaced charges is equivalent to a total charge which is continuously distributed along a line, or over a surface, or throughout a volume.
To find the electric field due to a continuous charge distribution, we define the following terms for different types of charge distribution. Each is simply “how much charge per unit of something”.
(a) Linear Charge Density (λ)
As shown in Fig. 10.27, a charge \(q\) is uniformly distributed along a linear conductor of length \(l\). The linear charge density \(\lambda\) is defined as
\[\lambda = \frac{q}{l} \qquad \text{--- (10.29)}\]
The SI unit of \(\lambda\) is coulomb per metre (C/m).
Diagram
Fig. 10.27 – A thin straight rod with + signs spread evenly along its length, representing a uniform linear charge distribution.
For example, charge distributed uniformly on a straight thin rod or a thin nylon thread. If the charge is not distributed uniformly over the length of the thin conductor, then \(\lambda\) varies from place to place, and the charge \(dq\) on a small element of length \(dl\) can be written as
\[dq = \lambda\, dl\]
This last step is the one that makes the calculation possible: a tiny element of the rod carries a tiny charge \(dq\), which is small enough to be treated as a point charge. We then find its field and integrate over the whole rod.
(b) Surface Charge Density (σ)
Suppose a charge \(q\) is uniformly distributed over a surface of area \(A\), as shown in Fig. 10.28. Then the surface charge density \(\sigma\) is defined as
\[\sigma = \frac{q}{A} \qquad \text{--- (10.30)}\]
The SI unit of \(\sigma\) is coulomb per square metre (C/m²).
Diagram
Fig. 10.28 – A flat irregular surface with + signs spread over it, representing a uniform surface charge distribution.
For example, charge distributed uniformly on a thin disc or on a synthetic cloth. If the charge is not distributed uniformly over the surface of a conductor, then the charge \(dq\) on a small area element \(dA\) can be written as
\[dq = \sigma\, dA\]
(c) Volume Charge Density (ρ)
Suppose a charge \(q\) is uniformly distributed throughout a volume \(V\). Then the volume charge density \(\rho\) is defined as the charge per unit volume:
\[\rho = \frac{q}{V} \qquad \text{--- (10.31)}\]
The SI unit of \(\rho\) is coulomb per cubic metre (C/m³).
Diagram
Fig. 10.29 – A cylindrical solid with + signs distributed throughout its interior, representing a uniform volume charge distribution.
For example, charge on a solid plastic sphere or a solid plastic cube. If the charge is not distributed uniformly over the volume of a material, then the charge \(dq\) over a small volume element \(dV\) can be written as
\[dq = \rho\, dV\]
Finding the Field of a Continuous Distribution
Once the charge density is known, the recipe is always the same, and it is the principle of superposition (Section 10.5) applied an infinite number of times.
Divide the distribution into infinitesimally small elements, each carrying a charge \(dq\) (\(= \lambda\,dl\), or \(\sigma\,dA\), or \(\rho\,dV\), depending on the geometry). Each such element is small enough to be treated as a point charge, so the field it produces at the observation point is given by the point-charge formula. The electric field due to a continuous charge distribution can then be calculated by adding the electric fields due to all these small charges — a vector sum that, in the limit of infinitely many infinitesimal elements, becomes an integral.
In practice, when the distribution is highly symmetric — a long straight wire, a large plane sheet, a uniformly charged sphere — Gauss’ law (Section 10.8) with a well-chosen Gaussian surface gives the same answer far more quickly than the integration does. That is precisely the convenience Gauss’ law was introduced for.
10.11 Summary
Electric Charge
Electrostatics is the study of static electric charges, the forces between them, and the effects they produce in the form of electric fields and electric potentials. Current electricity, in contrast, is produced by moving charges.
Electric charge is a basic property of the elementary particles of matter. A proton is positively charged, an electron is negatively charged, and a neutron carries no charge. An atom has equal numbers of protons and electrons, so most matter around us is electrically neutral.
Charging by conduction: electrons are actually transferred from one body to another (for example, by rubbing two dissimilar substances). The body receiving electrons becomes negative; the other is left equally positive.
Charging by induction: a charged body brought near a conductor, without touching it, causes the free electrons of the conductor to redistribute. The near side develops the opposite charge and the far side a similar charge. No charge is transferred across the gap.
Charge can also be separated by chemical reactions (cells), convection (clouds) and diffusion (living cells).
Basic Properties of Charge
Additive nature: the total charge on an object is the algebraic sum of the charges on its parts (signs must be included). Unlike mass, the total charge may be positive, zero or negative.
Quantization: charge exists only as an integral multiple of the elementary charge,
\[q = \pm ne, \qquad e = 1.6 \times 10^{-19}\ \text{C}\]
where \(n\) is an integer. The elementary charge was determined by Millikan’s oil drop experiment. Because \(e\) is so small, the steps are not noticed in practice — e.g. a charge of \(10^{-6}\) C on a glass rod corresponds to \(n = 6.25 \times 10^{12}\) elementary charges.
Conservation: for an isolated system, total charge can neither be created nor destroyed; it may only be transferred from one part of the system to another.
Force between charges: like charges repel, unlike charges attract.
Coulomb’s Law
A point charge is a charge whose dimensions are negligibly small compared to its distance from other bodies.
The force between two point charges at rest is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, and acts along the line joining them:
\[F = K\,\frac{q_1 q_2}{r^2} \qquad \text{--- (10.1)}\]
- In vacuum,
\[F_{\text{vac}} = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1 q_2}{r^2}\right), \qquad \frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}\]
\[\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}\]
- In a medium of absolute permittivity \(\varepsilon\),
\[F_{\text{med}} = \frac{1}{4\pi\varepsilon}\left(\frac{q_1 q_2}{r^2}\right)\]
- Relative permittivity (dielectric constant):
\[K \text{ or } \varepsilon_r = \frac{\varepsilon}{\varepsilon_0} = \frac{F_{\text{vac}}}{F_{\text{med}}} \qquad \text{--- (10.4)}\]
It is dimensionless, and is also called specific inductive capacity. A material medium reduces the force between charges by the factor \(\varepsilon_r\). For water, \(\varepsilon_r = 80\), so the force falls to \(1/80\) of its vacuum value. The medium is assumed homogeneous, isotropic and infinitely large.
\[F = \frac{1}{4\pi\varepsilon_0 \varepsilon_r}\, \frac{q_1 q_2}{r^2} \qquad \text{--- (10.5)}\]
Unit of charge: one coulomb is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of \(9.0 \times 10^{9}\) N. Subunits used in practice: µC (\(10^{-6}\) C), nC (\(10^{-9}\) C), pC (\(10^{-12}\) C). A charge of 1 C corresponds to \(6.25 \times 10^{18}\) electrons.
Vector form:
\[\vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \times \frac{q_1 q_2}{|\vec{r}_{21}|^2} \times \hat{r}_{21}, \qquad \vec{F}_{12} = \frac{1}{4\pi\varepsilon_0} \times \frac{q_1 q_2}{|\vec{r}_{12}|^2} \times \hat{r}_{12}\]
with \(\hat{r}_{12} = -\hat{r}_{21}\) and hence \(\vec{F}_{21} = -\vec{F}_{12}\): the two forces form an action–reaction pair. Since they act along the line joining the charges, the electrostatic force is a central force.
- The electrostatic force between two protons is about \(1.23 \times 10^{36}\) times (about 36 orders of magnitude) stronger than the gravitational force between them. Both forces obey the inverse square law and both are central; but gravity is always attractive, while the electrostatic force may be attractive or repulsive.
Principle of Superposition
- When a number of charges are interacting, the resultant force on a particular charge is the vector sum of the forces exerted on it by the individual charges, each calculated as if the others were absent:
\[\vec{F}_{\text{test}} = \sum_{n=1}^{N} \vec{F}_n = \frac{1}{4\pi\varepsilon_0} \sum_{n=1}^{N} \frac{q_0 q_n}{r_{0n}^2}\, \hat{r}_{0n} \qquad \text{--- (10.10)}\]
Electric Field
- The region around a charged object in which a Coulomb force is experienced by another charge is called the electric field. It is defined as force per unit charge, and exists around a charge irrespective of the presence of other charges.
\[\vec{E} = \frac{\vec{F}}{q} = \frac{Q}{4\pi\varepsilon_0 r^2}\,\hat{r} \qquad \text{--- (10.11)}, \qquad E = \lim_{q \to 0} \frac{\vec{F}}{q}\]
SI unit: N C⁻¹, equivalently V m⁻¹. Dimensions: \([\text{LMT}^{-3}\text{I}^{-1}]\).
In a medium of dielectric constant \(K\),
\[E = \frac{1}{4\pi\varepsilon_0 K}\, \frac{q}{r^2} \qquad \text{--- (10.12)}\]
For air or vacuum, \(K = 1\).
Both \(F\) and \(E\) obey the inverse square law: \(F \propto 1/r^2\), \(E \propto 1/r^2\).
A uniform field has the same magnitude and direction at all points (e.g. between two parallel plates); a non-uniform field does not (e.g. the field of a point charge).
For two parallel plates separated by a distance \(d\) with a potential difference \(V\) across them,
\[E = \frac{V}{d} \qquad \text{--- (10.13)}\]
This is the commonly used definition of electric field. A small voltage across a small gap can therefore give a large field (e.g. a spark plug: 20 V across 1.25 mm gives \(1.6 \times 10^{4}\) V m⁻¹).
- Knowing the field at a point lets us find the force on any charge placed there, from \(F = qE\).
Electric Lines of Force
An electric line of force is an imaginary curve drawn such that the tangent at any point on it gives the direction of the electric field at that point. It is the path along which a unit positive charge would move.
Key characteristics: lines start on positive charge and end on negative charge; they never intersect; they meet a conductor normally; they do not pass through a conductor (the field inside a conductor is zero) but do pass through insulators; the field magnitude is proportional to the number of lines per unit area held perpendicular to the field; lines crowd where the field is strong and spread out where it is weak; the lines of a uniform field are parallel and equally spaced.
The lines are a geometric construction only — they have no physical existence.
Electric Flux
- Electric flux through a small area element:
\[d\phi = E\, dS \cos\theta = \vec{E} \cdot d\vec{S} \qquad \text{--- (10.15)}\]
where \(\theta\) is the angle between \(\vec{E}\) and the normal to the surface.
- Total flux through a surface:
\[\Phi = \int d\phi = \int_{s} \vec{E} \cdot d\vec{S} = \vec{E} \cdot \vec{S} \qquad \text{--- (10.16)}\]
- Flux is a scalar. SI unit: volt metre (V m), i.e. N m² C⁻¹.
Gauss’ Law
- The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by \(\varepsilon_0\):
\[\Phi_E = \oint \vec{E} \cdot d\vec{S} = \frac{Q}{\varepsilon_0} \qquad \text{--- (10.18)}\]
Derivation uses the solid angle \(d\omega = \dfrac{dS\cos\theta}{r^2}\), with \(\int d\omega = 4\pi\) for a point enclosed by the surface.
Positive flux is directed outwards (positive enclosed charge); negative flux is directed inwards (negative enclosed charge); a charge outside the closed surface contributes zero net flux.
The law holds for a closed surface of any shape, regular or irregular. The total flux depends only on the enclosed charge — not on the size, shape, or radius of the surface.
A Gaussian surface is an imaginary closed surface enclosing the charge distribution and symmetric about it (a sphere for a point charge, a cylinder for a linear distribution), chosen to make \(\int \vec{E} \cdot d\vec{s}\) easy to evaluate.
Electric Dipole
- A pair of equal and opposite charges separated by a finite distance is an electric dipole. Its dipole moment is
\[\vec{p} = q\,(2\vec{l}) \qquad \text{--- (10.19)}\]
a vector directed from the negative to the positive charge. Unit: coulomb-metre (C m) or debye (D), with \(1\ \text{D} = 3.33 \times 10^{-30}\) C m. Water has a permanent dipole moment of \(6.172 \times 10^{-30}\) C m (1.85 D), directed from oxygen to hydrogen.
Polar molecules (H₂O, NH₃, SO₂, NaCl) have naturally separated centres of positive and negative charge; non-polar molecules (H₂, Cl₂, CO₂, CH₄) do not, but a dipole can be induced in them by an external field.
In a uniform field, the two forces \(+q\vec{E}\) and \(-q\vec{E}\) form a couple: the net force is zero, but there is a torque
\[\tau = qE\,2l\sin\theta = pE\sin\theta, \qquad \vec{\tau} = \vec{p} \times \vec{E} \qquad \text{--- (10.20), (10.21), (10.22)}\]
The torque is maximum (\(\tau = pE\)) when \(\theta = 90^\circ\) and zero when \(\theta = 0\). Torque tends to align the dipole axis along the field.
- Field at an axial point (distance \(r\) from the centre):
\[\left|\vec{E}_a\right| = \frac{1}{4\pi\varepsilon_0}\, \frac{2pr}{(r^2 - l^2)^2} \qquad \text{--- (10.23)}\]
and for \(r \gg l\),
\[\left|\vec{E}_a\right| = \frac{1}{4\pi\varepsilon_0}\, \frac{2p}{r^3} \qquad \text{--- (10.24)}\]
directed along \(\vec{p}\).
- Field at an equatorial point:
\[\left|\vec{E}_{eq}\right| = \frac{2ql}{4\pi\varepsilon_0 (r^2 + l^2)^{3/2}}\]
and for \(r \gg l\),
\[\left|\vec{E}_{eq}\right| = \frac{1}{4\pi\varepsilon_0}\, \frac{p}{r^3} \qquad \text{--- (10.28)}\]
directed anti-parallel to \(\vec{p}\).
- Hence the field at an axial point is twice that at an equatorial point lying at the same distance from the centre of the dipole.
Continuous Charge Distribution
A system of closely spaced charges is treated as a continuous distribution when the charges lie very close together compared with their distance from the point where the field is to be found.
Linear charge density: \(\lambda = \dfrac{q}{l}\) (C/m), with \(dq = \lambda\,dl\) — (10.29)
Surface charge density: \(\sigma = \dfrac{q}{A}\) (C/m²), with \(dq = \sigma\,dA\) — (10.30)
Volume charge density: \(\rho = \dfrac{q}{V}\) (C/m³), with \(dq = \rho\,dV\) — (10.31)
The field due to a continuous distribution is found by adding (integrating) the fields due to all the small elements of charge.
10.12 Check Your Concepts
A glass rod rubbed with silk becomes positively charged. Does the rod gain protons, or lose electrons? Explain your answer using the structure of the atom, and state what happens to the silk.
Why is charging by induction possible only with a conductor, and not with a good insulator like dry wood?
A charged rod attracts small bits of uncharged paper. Explain how a body carrying no net charge can experience a net attractive force. Would the effect change if the rod carried a negative charge instead of a positive one?
A body is said to carry a charge of \(0.8 \times 10^{-19}\) C. Why is such a claim impossible? Which property of charge does it violate?
Charge is quantized, yet in the laboratory we always observe charge to vary smoothly and continuously. Explain why the “steps” are never noticed.
A charged conductor is touched to the Earth and its charge apparently disappears. Has the law of conservation of charge been violated? Justify your answer, paying attention to the word “isolated” in the statement of the law.
Two charges are separated by a fixed distance. What happens to the force between them if (a) one charge is tripled, (b) the distance is tripled, (c) both changes are made together? Answer using Coulomb’s law, without any calculation.
Coulomb’s force and gravitational force both obey the inverse square law, and the electrostatic force is about 10³⁶ times the stronger of the two. Why, then, is the motion of planets governed by gravity and not by electrostatics?
A \(+1\ \mu\text{C}\) charge and a \(+16\ \mu\text{C}\) charge are held near each other. Which charge experiences the greater force? Justify your answer using the vector form of Coulomb’s law.
State the principle of superposition. Explain why the force on a charge due to a second charge is unaffected by the presence of a third charge.
Three equal charges are fixed at the vertices of an equilateral triangle. Without calculating, state the net force on an identical charge placed at the centroid, and justify the answer using symmetry.
Why is a test charge required to be positive and vanishingly small in the precise definition \(E = \lim_{q \to 0} \vec{F}/q\)? What would go wrong if a large test charge were used?
Does an electric field exist at a point where no charge is placed to detect it? Explain.
Why can a spark plug operating at only about 20 V produce a field strong enough to ionize a fuel–air mixture, while a 230 V mains supply across a room does nothing of the kind?
Why can two electric lines of force never intersect each other? What physically impossible situation would an intersection imply?
Lines of force have no physical existence. What, then, is the point of drawing them? List two pieces of information that can be read directly from a field-line diagram.
A flat sheet is placed in a uniform electric field. In what orientation is the flux through it maximum, and in what orientation is it zero? Explain with reference to the angle used in \(d\phi = E\,dS\cos\theta\).
A closed surface encloses a small charge, while a very large charge sits just outside it. Which charge determines the net flux through the surface, and which charges determine the value of \(\vec{E}\) at a point on the surface? Explain the difference.
A point charge is placed at the centre of a sphere. The radius of the sphere is doubled. What happens to (a) the electric field at the surface and (b) the total flux through the surface? Explain why the two answers differ.
What is a Gaussian surface? Why is a sphere chosen for a point charge and a cylinder for a long line of charge, when Gauss’ law is valid for a surface of any shape?
An electric dipole is placed in a uniform electric field. Show that it experiences no net force, but does experience a torque. In which orientation is the torque maximum, and in which is it zero?
Distinguish between polar and non-polar molecules, with one example of each. How can a dipole be produced in a non-polar molecule?
The field of a point charge falls off as \(1/r^2\), while the field of a dipole falls off as \(1/r^3\). Explain physically why the dipole’s field dies away faster.
At the same distance from the centre of a dipole, how does the field at an axial point compare with the field at an equatorial point, both in magnitude and in direction relative to \(\vec{p}\)?
A metal sphere and a plastic sphere are each given the same total charge. On which sphere would you describe the charge by a surface charge density, and on which by a volume charge density? Give the reason.
Why is a thick metal chain allowed to hang from a moving petrol tanker so that it touches the road? What would be the danger if it were removed?
What is the magnitude of charge on an electron?
State the law of conservation of charge.
Define a unit charge.
What is a uniform electric field?
State the units of linear charge density.
What is the unit of dipole moment?
What is relative permittivity?
10.13 Practice with Numericals
A body acquires a charge of \(-4.8\ \mu\text{C}\). Calculate the number of electrons transferred to it. Take \(e = 1.6 \times 10^{-19}\) C.
A metal sphere loses \(5.0 \times 10^{13}\) electrons. Calculate the magnitude and sign of the charge acquired by the sphere.
Two point charges of \(+3.0\ \mu\text{C}\) and \(+5.0\ \mu\text{C}\) are placed 30 cm apart in air. Calculate the magnitude of the force between them and state whether it is attractive or repulsive.
Two point charges of \(+2.0\ \mu\text{C}\) and \(-3.0\ \mu\text{C}\) are placed 15 cm apart in air. (a) Find the force between them. (b) Find the force if the same charges, at the same separation, are placed in a medium of dielectric constant 4.
Two identical charges repel each other with a force of \(9.0 \times 10^{-2}\) N when placed 10 cm apart in vacuum. Calculate the magnitude of each charge.
Two charges of \(+4.0\ \mu\text{C}\) and \(+9.0\ \mu\text{C}\) are fixed 50 cm apart. Find the position on the line joining them where a third charge would experience zero net force.
Charges of \(+2.0\ \mu\text{C}\), \(-2.0\ \mu\text{C}\) and \(+2.0\ \mu\text{C}\) are placed at the vertices A, B and C of an equilateral triangle of side 20 cm. Calculate the magnitude of the resultant force on the charge at A.
Calculate the magnitude and direction of the electric field at a distance of 20 cm from a point charge of \(-5.0\ \mu\text{C}\) placed in air.
At what distance from a point charge of \(+8.0\ \mu\text{C}\) is the magnitude of the electric field equal to \(2.0 \times 10^{5}\) N C⁻¹?
An electric field of magnitude \(3.0 \times 10^{4}\) N C⁻¹ acts at a point. Calculate the magnitude of the force experienced by (a) a proton and (b) an electron placed at that point, given that the magnitude of the charge on each is \(1.6 \times 10^{-19}\) C. State the direction of each force relative to the field.
The electric field due to a point charge in a certain medium of dielectric constant 5 is \(1.8 \times 10^{4}\) N C⁻¹ at a distance of 10 cm from the charge. Calculate the magnitude of the charge.
Two parallel plates are 2.0 mm apart and a potential difference of 400 V is applied across them. Calculate (a) the electric field between the plates and (b) the magnitude of the force on a charge of \(2.0\ \text{nC}\) placed between them.
A uniform electric field of \(4.0 \times 10^{3}\) N C⁻¹ passes through a flat circular surface of radius 10 cm. Calculate the electric flux through the surface when (a) the normal to the surface is parallel to the field, (b) the normal makes an angle of 60° with the field, and (c) the plane of the surface is parallel to the field.
A charge of \(+4.0\ \mu\text{C}\) is placed at the centre of a cube of side 20 cm. Calculate (a) the total flux through the cube and (b) the flux through one face of the cube.
A closed surface encloses charges of \(+6.0\ \mu\text{C}\), \(-2.0\ \mu\text{C}\) and \(+3.0\ \mu\text{C}\). A charge of \(+10\ \mu\text{C}\) lies just outside the surface. Calculate the net electric flux through the closed surface.
A point charge of \(+2.0\ \mu\text{C}\) is placed at the centre of a spherical surface of radius 50 cm. Calculate (a) the electric field at the surface and (b) the total flux through the sphere. (c) State how each answer changes if the radius is trebled.
An electric dipole consists of charges of \(\pm 5.0\ \text{nC}\) separated by 2.0 mm. Calculate its dipole moment, in C m and in debye. (Take \(1\ \text{D} = 3.33 \times 10^{-30}\) C m.)
A dipole of moment \(3.0 \times 10^{-8}\) C m is placed in a uniform electric field of \(5.0 \times 10^{4}\) N C⁻¹. Calculate the torque acting on it when its axis makes an angle of (a) 0°, (b) 30° and (c) 90° with the field.
An electric dipole of length 4.0 cm experiences a maximum torque of \(8.0 \times 10^{-3}\) N m when placed in a uniform electric field of \(2.0 \times 10^{4}\) N C⁻¹. Calculate the magnitude of the charge on the dipole.
A dipole of moment \(6.0 \times 10^{-9}\) C m is placed in air. Using the \(r \gg l\) approximation, calculate the electric field at a point 15 cm from its centre (a) on its axial line and (b) on its equatorial line. State the direction of the field in each case with respect to \(\vec{p}\).
A thin rod of length 40 cm carries a uniformly distributed charge of \(+3.2\ \mu\text{C}\). Calculate (a) its linear charge density and (b) the charge on a 5.0 cm length of the rod.
A charge of \(+1.5\ \mu\text{C}\) is spread uniformly over a thin disc of radius 10 cm. Calculate the surface charge density of the disc.
A solid plastic sphere of radius 5.0 cm carries a charge of \(+4.0\ \mu\text{C}\) distributed uniformly through its volume. Calculate its volume charge density.
Compare the electrostatic force and the gravitational force between an electron and a proton separated by \(5.3 \times 10^{-11}\) m. Take the mass of the electron as \(9.1 \times 10^{-31}\) kg, the mass of the proton as \(1.67 \times 10^{-27}\) kg and \(G = 6.674 \times 10^{-11}\ \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\).
Two parallel plates have a potential difference of 10 V between them. If the plates are 0.5 mm apart, what will be the strength of the electric field?
Two small spheres 18 cm apart have equal negative charges and repel each other with a force of 6 × 10⁻³ N. Find the total charge on both spheres.
A charge +q exerts a force of magnitude −0.2 N on another charge −2q. If they are separated by 25.0 cm, determine the value of q.
Four charges of +6 × 10⁻⁸ C each are placed at the corners of a square whose sides a are 3 cm each. Calculate the resultant force on each charge and show its direction on a diagram drawn to scale.
The electric field in a region is given by \(\vec{E}\) = 5.0 kN/C. Calculate the electric flux through a square of side 10.0 cm in the following cases:
- the square is along the XY plane
- the square is along the XZ plane
- the normal to the square makes an angle of 45° with the Z axis.
Three equal charges of 10 × 10⁻⁸ C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm and 25 cm respectively. Find the force exerted on the charge located at the 90° angle.
A potential difference of 5000 volt is applied between two parallel plates 5 cm apart. A small oil drop having a charge of 9.6 × 10⁻¹⁹ C falls between the plates. Find (a) the electric field intensity between the plates and (b) the force on the oil drop.
Calculate the electric field due to a charge of −8.0 × 10⁻⁸ C at a distance of 5.0 cm from it.
10.14 Objective Questions
- A positively charged glass rod is brought close to a metallic rod isolated from ground. The charge on the side of the metallic rod away from the glass rod will be
- same as that on the glass rod and equal in quantity
- opposite to that on the glass rod and equal in quantity
- same as that on the glass rod but lesser in quantity
- same as that on the glass rod but more in quantity
- An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will
- be attracted to the +ve plate
- be attracted to the −ve plate
- remain stationary
- move parallel to the plates
- A charge of +7 µC is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be
- 1 : 4
- 1 : 2
- 1 : 1
- 1 : 16
- Two charges of 1.0 C each are placed one metre apart in free space. The force between them will be
- 1.0 N
- 9 × 10⁹ N
- 9 × 10⁻⁹ N
- 10 N
- Two point charges of +5 µC are so placed that they experience a force of 80 × 10⁻³ N. They are then moved apart, so that the force is now 2.0 × 10⁻³ N. The distance between them is now
- 1/4 the previous distance
- double the previous distance
- four times the previous distance
- half the previous distance
- A metallic sphere A isolated from ground is charged to +50 µC. This sphere is brought in contact with another isolated metallic sphere B of half the radius of sphere A. The charge on the two spheres will now be in the ratio
- 1 : 2
- 2 : 1
- 4 : 1
- 1 : 1
- Which of the following produces a uniform electric field?
- point charge
- linear charge
- two parallel plates
- charge distributed on a circular any
- Two point charges of A = +5.0 µC and B = −5.0 µC are separated by 5.0 cm. A point charge C = 1.0 µC is placed at 3.0 cm away from the centre on the perpendicular bisector of the line joining the two point charges. The charge at C will experience a force directed towards
- point A
- point B
- a direction parallel to line AB
- a direction along the perpendicular bisector