Electric Charges and Fields: Class 12 Physics · Chapter 1

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 26 Sep 2026 · Updated 28 Sep 2026

155 min
3 simulations
29 practice questions
45 MCQs

What you will learn in Electric Charges and Fields

Explain charging by rubbing as a transfer of electrons, and use the additivity, conservation and quantisation of charge
Distinguish conductors from insulators and explain why earthing removes charge from a conductor
Apply Coulomb's law in vector form and the superposition principle to find the force on a point charge
Calculate the electric field of point charges and systems of charges, and read field-line diagrams
Define electric flux and compute it for flat surfaces in a field
Derive the field of an electric dipole on its axis and equatorial plane, and the torque on a dipole in a uniform field
State Gauss's law and use it to find the fields of a long wire, an infinite plane sheet and a thin spherical shell

Original companion text aligned to NCERT Physics Class 12, Chapter 1, pp. 1–44. Not affiliated with or endorsed by NCERT.

1.1 Introduction

Pull a woollen sweater over your head on a dry December night and you may hear faint crackles, or even see tiny blue sparks in a dark room. Walk across a nylon carpet, touch a steel door handle and a small jolt nips your finger. A monsoon thunderstorm plays the same trick on a grand scale: a lightning stroke is a spark several kilometres long.

Fig. 1.1Inside a storm cloud, charge separates; when enough piles up, a giant spark jumps to the ground.NCERT p. 1

A cloud floats above flat ground. Plus signs fill the upper part of the cloud and minus signs line its base. The ground below carries a row of plus signs, attracted by the negative cloud base. A jagged lightning bolt joins the cloud base to the ground, and a small house stands to one side.

Each of these is charge that piled up on a surface, usually by rubbing, and then suddenly found a path to escape. The physics of charges at rest is called electrostatics: how such charges push and pull on one another, and how they change the space around them. In this chapter we meet electric charge and its properties, Coulomb's law, the electric field and field lines, flux, the electric dipole and, finally, Gauss's law, a shortcut that turns some hard field calculations into a few lines.

1.2 Electric Charge

The words electron and electricity both come from elektron, the Greek name for amber, a golden fossil resin. Rub a piece of amber with fur and it snatches up dry leaves and bits of straw; Thales of Miletus, around 600 BCE, is usually credited with noticing this. Later experimenters found the same behaviour in many other pairs of materials: glass rubbed with silk, plastic rubbed with fur, ebonite rubbed with wool.

When such rubbed objects are brought near one another, a clear pattern emerges. Two glass rods rubbed with silk push each other away. Two plastic rods rubbed with fur also push each other away. But a rubbed glass rod and a rubbed plastic rod pull towards each other. Every rubbed object in the world falls into one of these two families, so we say there are exactly two kinds of electric charge (विद्युत आवेश). Objects of the same family repel; objects of different families attract, as Fig. 1.2 shows.

Fig. 1.2Like charges repel; unlike charges attract.NCERT Fig. 1.1 · p. 2

Three panels hang pairs of small charged balls from a common bar. In the first panel both balls are positive and the threads spread apart; in the second both are negative and they also spread apart; in the third one ball is positive and the other negative, and the threads lean towards each other. Short arrows under each ball show the direction of the electric force on it.

Touch the rubbed glass rod back against its own silk and both lose their power to attract paper. The two kinds of charge undo each other, the way a withdrawal cancels a deposit. Benjamin Franklin captured this with the labels positive and negative, chosen so that equal amounts add to zero. Which kind gets the plus sign is pure convention: glass rubbed with silk is called positive, which makes the silk, and a plastic rod rubbed with fur, negative. A body whose charges cancel exactly is electrically neutral.

Definition

Electric charge

A basic property of matter that makes it exert and feel electric forces. It comes in two kinds, positive and negative: like charges repel and unlike charges attract.

ExampleA glass rod rubbed with silk becomes positively charged.

A simple detector of charge is the electroscope (Fig. 1.3). Its working part is one continuous piece of metal: a knob outside a glass jar, a rod through an insulating stopper, and a pair of very light foil leaves hanging inside, where draughts cannot disturb them. Because the metal is a single conductor, charge given to the knob is shared all the way down to the leaves. The two leaves then carry the same kind of charge and push each other apart, and a bigger charge opens them wider.

Fig. 1.3Charge from the rod spreads over the metal to both leaves, which then repel each other; more charge opens them wider.NCERT Fig. 1.2 · p. 3

Two identical electroscopes stand side by side. Each is a glass jar with a metal rod passing through an insulating stopper, a flat metal knob on top and two thin foil leaves hanging from the bottom of the rod. On the left, labelled uncharged, the leaves hang straight down, side by side. On the right, labelled charged, a negatively charged rod touches the knob; minus signs appear on the knob and beside both leaves, and the leaves have swung apart into an upside-down V.

Why does rubbing charge a body at all? Every atom holds positive protons in its nucleus and an equal number of negative electrons around it, so ordinary matter is crammed with charge that happens to balance. The outermost electrons of some atoms are only loosely held. When two different materials are pressed and dragged across each other, a few of these electrons are pulled across the boundary. One surface ends up with a surplus of electrons and becomes negative; the other is left short and becomes positive.

3/3The rod is now positive and the silk negative, by exactly the same amount.

Fig. 1.4Rubbing only moves electrons; it never creates charge.NCERT p. 3

A glass rod is drawn above a strip of silk cloth. Both hold five plus signs and five electrons and are labelled neutral. In the rubbing step the cloth presses against the rod and three electrons from the rod drop into the cloth, leaving dashed empty circles behind in the rod. In the last step the cloth is pulled away: the rod, three electrons short, is labelled net plus, and the silk, with three extra electrons, is labelled net minus.

Two points deserve emphasis. First, rubbing transfers charge; it never manufactures it. The glass becomes positive by exactly as much as the silk becomes negative. Second, only a minute fraction of the electrons ever move. A charged comb may gain billions of extra electrons, yet the comb itself contains something like $10^{24}$ of them.

Electric forces are not a curiosity of rubbed rods. The stiffness of a steel beam, the stretch of a rubber band, the grip of a gecko's toes on a wall, the friction that stops your slipper sliding and every chemical reaction in your body come down to charged particles pulling and pushing on one another. Apart from gravity, nearly every force you meet in daily life is electrical at heart.

Try at homeNCERT p. 2

Two kinds of charge from sticky tape

You’ll need

a roll of transparent sticky tape
a clean, dry table top
a pen

Predict first

Two B strips are peeled off the table the same way. When you bring them together they will…

attract each otherrepel each othernot affect each other

Stick two strips of tape, each about 10 cm long, side by side on a clean table. Fold a small tab at one end of each and write B (bottom) on both tabs.
Peel both strips off quickly. Hold them by the tabs and bring them close together. Note what happens.
Stick two fresh strips on the table and press a third strip, labelled T (top), onto each of them. Peel each pair off together, then pull the T strip off its B strip quickly.
Test every pairing: T with T, T with B, and each strip with your finger. Record your observations in the table.

Table 1.1Tape test observations

Pair brought close	Prediction	What you saw
B and B
T and T
T and B
Any strip and your finger

Your entries are saved automatically.

Can’t do it now? Try the simulation · Fig. 1.9

Timeline

Discovering electric charge

c. 600 BCE
Rubbed amber
Thales of Miletus records that rubbed amber attracts light objects.
1600
A new word
William Gilbert lists many materials that behave like amber and coins the Latin electricus.
1733
Two kinds
Charles du Fay shows there are two kinds of electrification that repel their own kind and attract the other.
1747
Plus and minus
Benjamin Franklin names the two kinds positive and negative.
1785
An inverse-square law
Charles-Augustin de Coulomb measures the force between charged spheres.
1897
The electron
J. J. Thomson identifies the electron, the carrier of negative charge in metals.
1909–1913
Charge is grainy
Robert Millikan's oil-drop experiments show that charge comes in whole-number multiples of $e$ .

1.3 Conductors and Insulators

Rub a plastic ruler on dry hair and it snaps up scraps of paper. Rub a steel ruler the same way, holding it in your hand, and nothing seems to happen. Is steel simply impossible to charge? The answer lies in how freely charge can move inside each material.

In a conductor (चालक) some charges are free to wander. Metals are full of loosely bound electrons that drift from atom to atom. Tap water, the human body and the earth also conduct, because they contain ions that can move. In an insulator (विद्युतरोधी) such as rubber, PVC, mica, dry paper or glass, every charge is tied to its own atom or molecule and cannot travel far. A third group, the semiconductors such as silicon, sits in between; you will meet them in Chapter 14.

Place the same small patch of extra electrons on a metal sphere and on a plastic rod (Fig. 1.5). On the metal the electrons repel one another and, being free to move, race apart until they cover the whole surface, all within a tiny fraction of a second. On the plastic they cannot move, so the patch stays exactly where it was put. Chapter 2 shows that on a conductor the charge finally settles entirely on the outer surface.

2/3On the metal the electrons repel one another and spread over the whole surface; on the plastic they cannot move.

Fig. 1.5Charge moves freely over a conductor but stays put on an insulator.NCERT p. 3

On the left a metal sphere stands on an insulating stand; on the right a plastic rod stands on another. In the first step a cluster of electrons is placed at one spot on each. In the second step the electrons on the metal have spread out evenly over its surface while those on the plastic stay bunched at one end. In the third step a wire connects the metal sphere to the earth and its electrons drain away, while the plastic rod keeps its cluster.

Now the steel ruler makes sense. It does pick up charge when rubbed, but steel, your hand, your body and the floor all conduct, so the charge escapes to the earth almost at once. Insulate the ruler from your hand, for example by holding it through a dry plastic bag, and it keeps its charge just like the plastic one. Giving a conductor a path to the earth so that its charge can drain away is called earthing or grounding. The third pin of a household plug earths the metal body of an appliance: if a live wire ever touches the body, current flows straight to the earth and trips the fuse or circuit breaker, instead of flowing through the person who touches it.

Try at homeNCERT p. 4

Can a steel ruler be charged?

You’ll need

a steel ruler
a plastic ruler
a dry plastic bag
dry hair or a woollen cloth
small scraps of paper

Predict first

Which will lift the paper bits?

the steel ruler held in the bare handthe steel ruler held through the plastic bagneither, since metals never charge

Tear a sheet of paper into scraps, each smaller than a grain of rice, and spread them on a table.
Grip the steel ruler in your bare hand, rub it briskly on dry hair or wool and hold it just above the scraps.
Slip a dry plastic bag over your hand, hold the ruler only through the bag, rub it again and bring it near the scraps.
Repeat with the plastic ruler held in your bare hand.

Compare

Conductors and insulators

Conductors

Contain charges that move freely (electrons in metals, ions in solutions)
Added charge spreads over the whole surface
Can be charged by rubbing only if held by an insulator
Examples: copper, aluminium, the human body, the earth

Insulators

Charges are bound to their atoms or molecules
Added charge stays where it was placed
Charge easily by rubbing
Examples: rubber, PVC, glass, mica, dry paper

Definition

Conductor

A material in which electric charges can move freely, so charge placed on it spreads over its surface.

ExampleCopper, aluminium, the human body.

1.4 Basic Properties of Electric Charge

We now know there are two kinds of charge and that they tend to cancel. Three more properties underpin every calculation in this book. First, a useful idealisation: a charged body counts as a point charge when its size is negligible compared with the distances that matter in the problem, and we then place its whole charge at a single point. A charged pollen grain a metre from another is an excellent point charge; two charged plates a millimetre apart are not.

1.4.1 Additivity of charges

Charge adds like money in a bank account: deposits count as positive, withdrawals as negative, and the balance is their sum. In the same way, add up every charge in a system, keeping its sign, and you have the system's total charge. Charge is a scalar, with no direction, which is what makes such simple addition possible. A system holding $+ 3$ , $- 5$ , $+ 2$ , $- 1$ and $+ 4 μ C$ has a total charge of $+ 3 μ C$ . A system holding equal amounts of both signs has zero net charge, however much charge it contains.

q_{total} = q_{1} + q_{2} + q_{3} + \dots + q_{n}

1.4.2 Charge is conserved

Rubbing gave us a first taste of a deeper rule: the silk's gain of charge exactly matched the glass's loss, and nothing appeared from nowhere. Experiments of every kind, from electrolysis to particle collisions, confirm the general law. The total charge of an isolated system never changes; charge can move from one body to another inside it, but the algebraic total stays fixed. This is the law of conservation of charge.

Conservation does not mean that charged particles are permanent. A free neutron decays into a proton, an electron and a neutral antineutrino: total charge $0$ before, $(+ e) + (- e) + 0 = 0$ after. A high-energy gamma ray passing a nucleus can turn into an electron and a positron, charges $- e$ and $+ e$ : again zero before and after. Particles come and go; the net charge never does.

1.4.3 Quantisation of charge

When Robert Millikan measured the charges on hundreds of tiny oil drops between 1909 and 1913, he never found an arbitrary value. Every charge was a whole-number multiple of one basic amount, now called $e$ : a proton carries $+ e$ and an electron $- e$ . Every free charge measured since obeys the same rule. This is the quantisation of charge:

Quantisation of charge

q = n e, n = 0, \pm 1, \pm 2, \pm 3, \dots

For ordinary matter the rule comes down to counting. A body with $n_{p}$ protons and $n_{e}$ electrons carries charge $(n_{p} - n_{e}) e$ , and the difference of two whole numbers is a whole number. Long before Millikan, Faraday's laws of electrolysis had hinted at the same thing: the charge needed to release one mole of any element is always a small whole-number multiple of the same amount, about $96 500 C$ .

Charge is measured in coulombs (C). SI defines the ampere first and the coulomb follows from it: a steady current of $1 A$ carries $1 C$ past any point each second, so $1 C = 1 A s$ . In coulombs, the elementary charge is

e = 1.602 \times 10^{- 19} C

where

$e$: the elementary charge, the magnitude of the charge on an electron or a proton

Turned around, one coulomb is about $6.24 \times 10^{18}$ elementary charges. A comb or a balloon charged by rubbing carries only a tiny fraction of that, usually somewhere between a few nanocoulombs and a microcoulomb: $1 μ C = 10^{- 6} C$ and $1 n C = 10^{- 9} C$ .

If charge is grainy, why does it seem to flow and vary smoothly? Because the grains are fantastically small. Even $1 μ C$ is more than six million million elementary charges, and adding or removing one of them changes the total by less than one part in $10^{12}$ . It is like a sack of rice: the sack is made of separate grains, yet a kilogram more or less looks like a smooth change (Fig. 1.6). For large-scale charges we can treat charge as continuous. For an oil drop carrying a dozen electrons, or an ion carrying two, quantisation cannot be ignored.

2/2Over trillions of electrons the steps are too small to see, so charge looks continuous.

Fig. 1.6Charge comes in steps of

e

, far too small to notice for everyday objects.NCERT p. 5

Two graphs plot the charge on a body against the number of electrons added. In the first step only the left graph is clear: close up, it is a staircase, and one riser is marked with a bracket of size e, because each added electron raises the charge by exactly one step. In the second step the right graph appears: over trillions of electrons the same relation looks like a smooth straight line, and a circular magnifier held over one small spot on the line reveals the tiny steps hidden inside.

Example 1.1NCERT Example 1.1 · p. 6

How big is a coulomb?

In a laboratory ion source, a steady beam delivers $2.0 \times 10^{11}$ singly charged positive ions per second to an insulated metal cup. How long must the beam run for the cup to collect (a) $1.0 C$ and (b) $1.0 μ C$ ? Take $e = 1.6 \times 10^{- 19} C$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 1.2NCERT Example 1.2 · p. 6

The charge hidden in an iron nail

An iron nail has a mass of $10 g$ . Iron has atomic number $26$ and molar mass $55.85 g / m o l$ . Find the total positive charge and the total negative charge in the nail. ( $N_{A} = 6.02 \times 10^{23} m o l^{- 1}$ , $e = 1.6 \times 10^{- 19} C$ )

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

How many electrons must be removed from a neutral object to give it a charge of

+ 8.0 n C

? (Enter the number, e.g. 3e10.)

Show answer

Answer 50000000000

n = q / e = (8.0 \times 10^{- 9} C) / (1.6 \times 10^{- 19} C) = 5.0 \times 10^{10}

electrons.

1.5 Coulomb's Law

So far we have said only that charges attract or repel. How strong is the push or pull, and how does it change with distance? In 1785 Charles-Augustin de Coulomb answered this with careful measurements on small charged spheres. His result, Coulomb's law (कूलॉम का नियम), is the foundation of everything that follows.

LawCoulomb's law

Two point charges at rest exert forces on each other along the straight line joining them. Each force is proportional to the product of the magnitudes of the charges and to the inverse square of their separation. Like charges repel; unlike charges attract.

Coulomb's law (magnitude)

F = k \frac{∣ q_{1} q_{2} ∣}{r^{2}}

(1.1)

where

$q_{1}, q_{2}$: the two point charges
$r$: the distance between them
$k$: a positive constant that depends on the choice of units

Coulomb's instrument, the torsion balance, turned a tiny force into a visible twist. A light horizontal rod hung from a fine wire, with a small charged ball at one end. A second charged ball brought near it pushed the rod round, twisting the wire until the wire's springiness balanced the electric force, and the angle of twist, read from a scale, measured that force. Charge was harder, because no instrument could measure it. Coulomb could, however, halve a charge without measuring it (Fig. 1.7). When a charged metal ball touches an identical uncharged one, symmetry demands that they end up with equal shares. Repeating the trick produces $q / 2$ , $q / 4$ , $q / 8$ , … from an unknown $q$ .

3/4Each sphere keeps $q / 2$ .

Fig. 1.7Touching an identical neutral sphere halves a charge without measuring it.NCERT p. 7

Two identical metal spheres, A and B, stand on insulating stands. In the first step A carries eight plus signs and is labelled q, while B is neutral. In the second step B slides over and touches A; each sphere now shows four plus signs, labelled q over 2. In the third step B moves back to its place, still holding q over 2. In the fourth step B has been discharged and touched to A again, so each sphere now holds q over 4.

With the charges left alone, doubling the separation cut the force to a quarter. With the separation fixed, halving one charge halved the force, and halving both cut it to a quarter. Together these observations give Eq. (1.1).

From historyNCERT p. 7

Charles-Augustin de Coulomb

Charles-Augustin de Coulomb (1736–1806) trained as a military engineer and spent about eight years in Martinique, supervising the building of a fort. His early research was thoroughly practical: friction, the strength of beams and the stability of earth walls, work still used by civil engineers. His torsion balance grew out of a prize-winning study of compass needles hung from fine threads. With it, in 1785, he measured the forces between small charged spheres, and he later showed that magnetic poles obey an inverse-square law too. Joseph Priestley had guessed the electric law in 1767 from a single clue: a charged hollow metal can exerts no force on charges inside it, the shell result of Section 1.14.3. Henry Cavendish confirmed it by experiment, but his results stayed unpublished for a century.

Coulomb never had a unit of charge; he only compared charges. Once the law is known, though, it can be used to define one. Pick any positive number for $k$ , and the unit of charge becomes whatever amount makes two such charges $1 m$ apart feel a force of $k$ newtons. SI does not work this way. Historically it fixed the ampere through the magnetic forces between current-carrying wires (today it fixes the value of $e$ ), and the coulomb is simply $1 A s$ . So $k$ is not ours to choose; it is measured:

k = \frac{1}{4 π ε_{0}} \approx 9 \times 10^{9} N m^{2} C^{- 2}, ε_{0} = 8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}

where

$ε_{0}$: the permittivity of free space (vacuum)

The factor $4 π$ is written in on purpose; it makes many later formulas, including Gauss's law, simpler. With it, Coulomb's law reads

F = \frac{1}{4 π ε_{0}} \frac{∣ q_{1} q_{2} ∣}{r^{2}}

(1.2)

Two charges of $1 C$ each, placed $1 m$ apart in vacuum, would repel with a force of about $9 \times 10^{9} N$ , roughly the weight of a million tonnes. Why is the coulomb so outsized? It was sized for the magnetic effects of currents. One ampere is a modest current, but magnetic forces between wires are far weaker than electric forces between charges, so the charge one ampere carries in a second is enormous by electrostatic standards. That is why electrostatics works in $μ$ C and nC.

Force is a vector, so we should say which way it points as well as how big it is. Fig. 1.8 sets up the geometry.

DerivationCoulomb's law in vector form

Let charges $q_{1}$ and $q_{2}$ sit at position vectors $𝐫_{1}$ and $𝐫_{2}$ from an origin O. The vector pointing from charge 1 to charge 2 is

𝐫_{21} = 𝐫_{2} - 𝐫_{1}, r_{21} = ∣ 𝐫_{21} ∣, {\hat{𝐫}}_{21} = \frac{𝐫_{21}}{r_{21}}

Similarly $𝐫_{12} = 𝐫_{1} - 𝐫_{2} = - 𝐫_{21}$ points from 2 to 1, and ${\hat{𝐫}}_{12} = - {\hat{𝐫}}_{21}$ . Write $𝐅_{21}$ for the force on $q_{2}$ due to $q_{1}$ . Its size is given by Eq. (1.2) and it lies along the line joining the charges, so

𝐅_{21} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{21}^{2}} {\hat{𝐫}}_{21}

(1.3)

Here the charges carry their signs, and one formula covers every case. If $q_{1}$ and $q_{2}$ have the same sign, $q_{1} q_{2} > 0$ and $𝐅_{21}$ points along ${\hat{𝐫}}_{21}$ , away from $q_{1}$ : repulsion. If the signs differ, $q_{1} q_{2} < 0$ and the force points back towards $q_{1}$ : attraction.

Swapping the labels 1 and 2 gives $𝐅_{12}$ , the force that $q_{2}$ exerts on $q_{1}$ :

𝐅_{12} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{𝐫}}_{12} = - 𝐅_{21}

The two forces are equal in size and opposite in direction, exactly as Newton's third law demands.

2/3 $q_{1} q_{2} > 0$ : each force points away from the other charge.

Fig. 1.8The force acts along the line joining the charges, and

𝐅_{12} = - 𝐅_{21}

in every case.NCERT Fig. 1.3 · p. 8

From an origin O, position vectors r1 and r2 point to two charges q1 and q2, and a third arrow r21 runs from q1 to q2. In the second step both charges are positive and the forces point apart along the line joining them: F21 on q2 away from q1 and F12 on q1 away from q2. In the third step q2 is negative and both forces point towards each other, equal in size and opposite in direction.

One caution: Eq. (1.3) describes charges in vacuum, and air changes it very little. Put water, oil or glass between the charges and the molecules of that material rearrange themselves in response, weakening the force; Chapter 2 deals with this.

SimulationCoulomb's law lab

A stage one metre wide holds a fixed positive source charge of 2 microcoulombs and a small positive test charge you can drag anywhere. Arrows show the field around the source, and a readout shows the force on the test charge in newtons and its direction as you move it.

Fig. 1.9A fixed

+ 2 μ C

source charge and a movable

+ 1 μ C

test charge on a stage one metre wide. The readout gives the force in newtons.

Try this

Using the scale bar, put the test charge $0.20 m$ from the source and note the force. Move it to $0.40 m$ . Does the force fall to a quarter?
Carry the test charge in a circle around the source at a fixed distance. What changes in the readout, and what stays the same?
Predict the force at $0.10 m$ using $F = k q_{1} q_{2} / r^{2}$ , then check it on the stage.

Example 1.3NCERT Example 1.3 · p. 9

Electricity versus gravity in a hydrogen atom

In a hydrogen atom the electron and proton are, on average, $5.3 \times 10^{- 11} m$ apart. (a) Find the ratio of the electric force to the gravitational force between an electron and a proton, and between two electrons. (b) Find the electric force in the atom and the acceleration it would give each particle. Use $k = 8.99 \times 10^{9} N m^{2} C^{- 2}$ , $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$ , $e = 1.60 \times 10^{- 19} C$ , $m_{e} = 9.11 \times 10^{- 31} k g$ , $m_{p} = 1.67 \times 10^{- 27} k g$ .

Try it yourself first. Then reveal the solution one step at a time.

The inverse-square law gives a quick way to reason about changes without recomputing everything, as the next example shows.

Example 1.4NCERT Example 1.4 · p. 10

Touch, share, then repel

Two identical small metal spheres A and B on insulating stands carry charges of $+ 8.0 n C$ and $- 2.0 n C$ . With their centres $10 c m$ apart they attract each other. (a) Find this force. (b) The spheres are touched together and then put back in the same positions. What is the force now? (c) How close must the centres be brought for the new force to be as large as the original one? Treat the spheres as point charges when they are apart.

Try it yourself first. Then reveal the solution one step at a time.

1.6 Forces between Multiple Charges

A charge in a real device is seldom alone: it may have dozens of neighbours, each pulling or pushing on it. Coulomb's law deals with one pair at a time. To find the total force we need one more fact, and only experiment can supply it: the superposition principle (अध्यारोपण सिद्धांत).

PrinciplePrinciple of superposition

The force that one charge exerts on another is exactly what Coulomb's law gives for that pair, whatever other charges are nearby. The total force on any charge is the vector sum of the forces exerted on it by every other charge, each taken separately.

For three charges (Fig. 1.10), the force on $q_{1}$ is the Coulomb force from $q_{2}$ plus the Coulomb force from $q_{3}$ , added by the parallelogram (or triangle) rule:

𝐅_{1} = 𝐅_{12} + 𝐅_{13} = \frac{1}{4 π ε_{0}} [\frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{𝐫}}_{12} + \frac{q_{1} q_{3}}{r_{13}^{2}} {\hat{𝐫}}_{13}]

(1.4)

3/3The parallelogram rule gives the total force $𝐅_{1}$ .

Fig. 1.10Each pair interacts as if the others were absent; the forces then add as vectors.NCERT Fig. 1.5 · p. 11

Three charges are shown: q1 and q2 positive, q3 negative. In the second step two force arrows appear on q1: F12 pointing away from q2 and a shorter F13 pointing towards q3. In the third step dashed lines complete a parallelogram on these two arrows and a bold arrow along its diagonal shows the total force F1.

With $n$ charges we just keep adding. Here ${\hat{𝐫}}_{1 i}$ is the unit vector pointing from $q_{i}$ towards $q_{1}$ :

𝐅_{1} = 𝐅_{12} + 𝐅_{13} + \dots + 𝐅_{1 n} = \frac{q_{1}}{4 π ε_{0}} \sum_{i = 2}^{n} \frac{q_{i}}{r_{1 i}^{2}} {\hat{𝐫}}_{1 i}

(1.5)

These two ideas together, Coulomb's law for each pair and superposition for the crowd, carry the whole of electrostatics. Every result in this chapter and the next can be traced back to them.

Example 1.5NCERT Example 1.5 · p. 12

Symmetry, and the missing charge

Equal charges $q = + 2.0 μ C$ sit at the four corners of a square of side $a = 20 c m$ . A charge $Q = + 1.0 μ C$ is placed at the centre. (a) What is the net force on $Q$ ? (b) The charge at corner A is removed. What is the net force on $Q$ now?

Forces on the central charge

Q

.NCERT p. 12

Left: a square of side a with an equal positive charge at each corner and a smaller positive charge Q at the centre; four equal force arrows on Q point away from the four corners. Right: the same square with corner A empty; three force arrows remain on Q and a bold arrow points towards the empty corner A.

Try it yourself first. Then reveal the solution one step at a time.

Example 1.6NCERT Example 1.6 · p. 13

Three charges: forces that add to zero

Charges $q_{A} = + 2.0 μ C$ , $q_{B} = - 2.0 μ C$ and $q_{C} = - 2.0 μ C$ sit at the corners of a right-angled isosceles triangle, with the right angle at A and $A B = A C = a = 30 c m$ . Find the net force on each charge, and check that the three net forces add to zero.

Charges at the corners of a right-angled isosceles triangle.NCERT p. 13

A right-angled isosceles triangle is drawn with dashed sides. Corner A, bottom left, holds a positive charge and has a right-angle mark. Corner B, bottom right, and corner C, top left, each hold a negative charge. Sides AB and AC are labelled a and the slanting side BC is labelled a root 2. Small axes at the bottom right show x along AB and y along AC.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A charge of

+ 1.0 μ C

is at

x = 0

and a charge of

+ 4.0 μ C

is at

x = 0.30 m

. At what position

x

(in metres) between them does a third charge feel no net force?

Show answer

Answer 0.1 m

The two repulsions balance when

\frac{1}{x^{2}} = \frac{4}{(0.30 - x)^{2}}

, so

0.30 - x = 2 x

and

x = 0.10 m

. The point is closer to the smaller charge.

1.7 Electric Field

You already know one field, though you may not call it that. Near the Earth's surface every kilogram is pulled down with about $9.8 N$ , whether or not anything is there to be pulled: $g = 9.8 N / k g$ is a property of the place, not of the object. Electric forces can be described in the same way.

Instead of saying 'charge $Q$ pushes on charge $q$ ', we split the interaction in two. First, $Q$ sets up an electric field (विद्युत क्षेत्र): a vector $𝐄$ with a definite value at every point around it. Second, a charge placed at any point feels a force fixed by the field at that point, whatever produced it. The field is there even when no charge is present to feel it.

The field of a point charge $Q$ follows directly from Coulomb's law. At a point with position vector $𝐫$ measured from $Q$ ,

Field of a point charge

𝐄 (𝐫) = \frac{1}{4 π ε_{0}} \frac{Q}{r^{2}} \hat{𝐫}

(1.6)

where

$\hat{𝐫}$: the unit vector pointing from $Q$ towards the point
$r$: the distance from $Q$

A charge $q$ placed at that point then feels the Coulomb force

𝐅 = \frac{1}{4 π ε_{0}} \frac{Q q}{r^{2}} \hat{𝐫}

(1.7)

which is the charge times the field at its location:

𝐅 (𝐫) = q 𝐄 (𝐫)

(1.8)

Eq. (1.8) also tells us how to measure a field: place a known charge at the point, measure the force on it and divide. The probe is called a test charge $q_{0}$ ; the charges that create the field are the source charges. A thermometer poses the same problem. Dip a large, cold thermometer into a small cup of hot tea and you cool the very tea you wanted to measure. A large test charge would likewise push the source charges out of place, so the probe must be small enough not to disturb them. The field is defined in the limit

Definition of electric field

𝐄 = \lim_{q_{0} \to 0} \frac{𝐅}{q_{0}}

(1.9)

Because the force is proportional to $q_{0}$ , the ratio $𝐅 / q_{0}$ comes out the same whichever small test charge we use: the field belongs to the sources and the point, not to the probe. Put another way, $𝐄$ is the force per coulomb of positive charge placed at the point.

Definition

Electric field

The electric force per unit positive test charge at a point, $𝐄 = 𝐅 / q_{0}$ . It is a vector with SI unit N/C.

ExampleNear the ground on a fine day there is a downward field of about $100 N / C$ .

Eq. (1.8) also fixes the SI unit of field, N/C. (The next chapter gives an equivalent unit, V/m.) Three features of the point-charge field are worth fixing in mind, and Fig. 1.11 shows them:

Direction. For positive $Q$ the field points radially outward; for negative $Q$ , radially inward. A positive test charge is pushed along $𝐄$ ; a negative one is pushed the opposite way.
Size. It falls as $1 / r^{2}$ : twice as far out, the field is a quarter as strong.
Symmetry. The size depends only on $r$ , so it is the same everywhere on a sphere centred on the charge. The field is spherically symmetric.

1/3Outward arrows; at twice the distance the field is a quarter as strong.

Fig. 1.11The field of a point charge is radial and falls off as

1 / r^{2}

.NCERT Fig. 1.8 · p. 14

A point charge sits at the centre with field arrows at eight directions on two circles, the outer circle twice as far out. In the first step the charge is positive and every arrow points radially outward; arrows on the outer circle are a quarter as long, labelled E and E over 4. In the second step dashed circles mark spheres of radius r and 2r on which the field strength is the same everywhere. In the third step the charge is negative and every arrow points radially inward.

1.7.1 Electric field due to a system of charges

For several source charges $q_{1}, q_{2}, \dots, q_{n}$ we lean on superposition once more. The field at a point P is the force per unit charge on a test charge at P, and that force is the vector sum of the separate Coulomb forces. So the fields simply add:

𝐄 (𝐫) = 𝐄_{1} + 𝐄_{2} + \dots + 𝐄_{n} = \frac{1}{4 π ε_{0}} \sum_{i = 1}^{n} \frac{q_{i}}{r_{i P}^{2}} {\hat{𝐫}}_{i P}

(1.10)

where

$r_{i P}$: the distance from $q_{i}$ to P
${\hat{𝐫}}_{i P}$: the unit vector pointing from $q_{i}$ towards P

The recipe is always the same: find the field of each charge at P, draw it as an arrow, then add the arrows as vectors, usually by resolving them into components. Example Example 1.8 works through it.

1.7.2 Physical significance of electric field

When you speak into a mobile phone, electrons surge back and forth in its antenna. Electrons in the antenna of a tower $3 k m$ away respond, but not at once: about ten microseconds later. What carried the influence during those microseconds? Not the electrons, which stayed where they were. It was the electromagnetic field, a disturbance travelling outward at the speed of light $c$ .

No influence travels faster than light. So when a charge $q_{1}$ is suddenly moved, a distant charge $q_{2}$ goes on feeling the old force until the news arrives (Fig. 1.12). Coulomb's law, which assumes instant action, cannot describe that gap; the field can. Fields store and carry energy (sunlight is field energy that left the Sun about eight minutes ago), carry momentum and change according to laws of their own, so physicists treat them as real things. Faraday was the first to think this way. For charges at rest the field is still a handy summary: work out $𝐄$ once and you know the force on any charge placed anywhere. In general, a field is any quantity with a value at every point; the temperature across a room is a scalar field, and $𝐄$ is a vector field.

3/3After a time $d / c$ the ripple reaches $q_{2}$ and its force changes direction.

Fig. 1.12Changes in a field spread out at the speed of light, so forces respond with a delay.NCERT p. 17

Two positive charges sit far apart on a dashed line; the right one feels a steady push to the right. In the second step the left charge jumps upward and curved ripples start spreading out from it, while the force on the right charge is unchanged. In the third step the ripples have reached the right charge and only now does its force arrow tilt to point away from the left charge's new position; a label reads delay equals d over c.

Example 1.7NCERT Example 1.7 · p. 17

Falling through an electric field

Two horizontal metal plates in an evacuated tube are $2.0 c m$ apart, with a uniform field of $5.0 \times 10^{3} N / C$ between them. (a) With the field pointing up, an electron is released from rest just below the top plate. How long does it take to reach the bottom plate? (b) The connections to the plates are swapped, reversing the field, and a proton is released from rest in the same place. How long does it take? (c) Compare both times with that for a small stone falling $2.0 c m$ , and check that gravity can be ignored for the particles. ( $e = 1.6 \times 10^{- 19} C$ , $m_{e} = 9.11 \times 10^{- 31} k g$ , $m_{p} = 1.67 \times 10^{- 27} k g$ )

Try it yourself first. Then reveal the solution one step at a time.

Example 1.8NCERT Example 1.8 · p. 18

Adding fields from two unequal charges

Point charges $q_{1} = + 4.0 \times 10^{- 8} C$ and $q_{2} = - 1.0 \times 10^{- 8} C$ are fixed $20 c m$ apart. Find (a) the electric field at P, midway between them, and (b) the field at R, $10 c m$ from P along the perpendicular bisector. (c) Where on the line through the two charges is the field zero? Use $k = 9 \times 10^{9} N m^{2} C^{- 2}$ .

Charges and points for Example 1.8.NCERT p. 18

A positive charge q1 and a negative charge q2 lie 20 cm apart on a dashed horizontal line that extends beyond both charges and is labelled line through the charges. Point P is midway between them, 10 cm from each. Point R lies 10 cm directly above P, and dashed lines join R to both charges.

Try it yourself first. Then reveal the solution one step at a time.

1.8 Electric Field Lines

A field assigns an arrow to every point in space, far too many to draw. Fig. 1.13 shows how to tame them for a positive point charge. Draw arrows at a few points along several directions: all of them point away from the charge, and they shrink with distance. Then run a smooth curve through each chain of arrows. These curves are electric field lines (विद्युत क्षेत्र रेखाएँ). At every point, the tangent to the line through it gives the direction of $𝐄$ , and an arrowhead shows which way along the tangent.

Joining up the arrows throws away their lengths, yet the picture still encodes strength. Lines are packed tightly where the field is strong and spread out where it is weak. Only the relative spacing carries meaning: drawing twice as many lines everywhere changes nothing physical. In three dimensions, 'packing' means the number of lines passing through a small area held perpendicular to them.

3/3The same lines cross a patch four times larger at twice the distance.

Fig. 1.13Where the field is strong the lines bunch together; where it is weak they thin out.NCERT Fig. 1.12 · p. 19

A positive charge sits at the centre. In the first step short arrows point outward along eight directions, getting shorter further from the charge. In the second step the arrows are replaced by sixteen continuous field lines radiating outward with arrowheads. In the third step a narrow cone is shaded; two arcs, one near and one twice as far, are each crossed by the same three lines, and labels show the far patch has four times the area of the near patch.

DerivationWhy line density follows the inverse-square law

First we need the solid angle, the three-dimensional cousin of an ordinary angle. A narrow cone with its tip at a point cuts a small area $Δ S$ out of a sphere of radius $r$ centred on that point. The cone's solid angle is this area divided by $r^{2}$ , measured in steradians (a whole sphere is $4 π$ sr). Since the area cut out grows as $r^{2}$ , the same cone gives the same answer on every sphere:

Δ Ω = \frac{Δ S}{r^{2}}

Consider a thin cone of solid angle $Δ Ω$ with its tip on a point charge. The lines of a point charge are straight and radial, so every line inside the cone stays inside it. Let $n$ lines lie in the cone. At distances $r_{1}$ and $r_{2}$ the cone's cross-section has areas $r_{1}^{2} Δ Ω$ and $r_{2}^{2} Δ Ω$ , and the same $n$ lines cross both. The number of lines per unit area is therefore

\frac{n}{r_{1}^{2} Δ Ω} at r_{1}, \frac{n}{r_{2}^{2} Δ Ω} at r_{2}

Since $n$ and $Δ Ω$ are fixed, the density of lines falls as $1 / r^{2}$ , exactly like the field of a point charge. Drawing lines that begin on the charge and never end in empty space automatically gets the strength right.

Michael Faraday, who had little formal mathematics, pictured fields in this way and spoke of lines of force. We prefer the name field lines, because a magnetic field line does not show the direction of the magnetic force on a moving charge, which is perpendicular to it. Every arrangement of charges has its own pattern. Two equal positive charges produce lines that bulge away from the gap between them, as if the charges were being shoved apart. A positive and a negative charge produce lines that arc from one to the other, as if tying them together. Scroll through the simulation to compare these patterns.

Scroll to explore

Field-line patterns

Point charges on a stage with their field lines drawn. As you scroll, the stage switches between a single positive charge, a single negative charge, two equal positive charges and a pair of equal and opposite charges; you can drag the charges to reshape the pattern.

A single positive charge. The lines point radially outward and run off to infinity. Their spacing grows with distance, showing the weakening field.

A single negative charge. The same pattern with every arrow reversed: lines come in from infinity and end on the charge.

Two equal positive charges. The lines swerve away from the region between the charges. Midway between them the field is zero, and the lines seem to shove the charges apart.

Equal and opposite charges. Lines leave $+ q$ and end on $- q$ , bunching up in the gap. This pair, an electric dipole, is the subject of Section 1.10. Drag a charge and watch the pattern follow.

All these pictures share a few general properties, which you can check against any field-line diagram:

Every electrostatic field line begins on a positive charge and ends on a negative one. A lone charge sends its lines off to infinity (if it is positive) or gathers them in from infinity (if negative).
Through a region containing no charge, a line never breaks off and never starts from nothing.
No two lines ever meet or cross: at a crossing point the field would need two directions at once.
An electrostatic field line never closes on itself. Chapter 2 traces this to a special property of fields due to charges at rest: they are conservative.

1.9 Electric Flux

Think of a solar panel on a rooftop in Jaipur. Facing the Sun squarely, it intercepts the most light. Tilted away, it intercepts less, because it presents a smaller area across the beam. Turned edge-on, it catches almost none. The amount intercepted depends on the size of the panel, the strength of the light and the angle between the beam and the panel.

Electric flux (विद्युत फ्लक्स) measures, in the same way, how much electric field passes through a surface. Nothing actually flows, but the idea matches the field-line picture closely: flux counts how many field lines cross the surface.

A patch of surface has both a size and a facing, so we describe it by a vector, $Δ 𝐒 = Δ S \hat{𝐧}$ . Its length is the area $Δ S$ , and its direction is the unit normal $\hat{𝐧}$ , sticking straight out of the patch. There are two possible normals, one on each side. For a single open patch we simply pick one and stay consistent; for a patch that is part of a closed surface the rule is always the outward normal. A curved surface is cut into patches small enough to count as flat, each with its own normal.

Now put a small patch in a field $𝐄$ , with its normal at angle $θ$ to $𝐄$ (Fig. 1.15). The lines that cross it are those passing through its projection across the field, which has area $Δ S \cos θ$ . So we define the flux through the patch as

Electric flux through a small patch

Δ ϕ = 𝐄 \cdot Δ 𝐒 = E Δ S \cos θ

(1.11)

2/3 $θ = 60^{\circ}$ : the flux halves, since $\cos 60^{\circ} = 0.5$ .

Fig. 1.15Flux depends on the area presented across the field,

Δ S \cos θ

.NCERT Fig. 1.15 · p. 22

Eleven horizontal field lines run left to right, with a thin flat patch seen edge-on in the middle and its normal vector n. In the first step the patch faces the field, theta 0 degrees, and eight lines cross it. In the second step it is tilted so its normal is at 60 degrees to the field; only four lines cross, and a dashed segment shows the projected area delta S cos theta. In the third step the patch lies along the field, theta 90 degrees, and no line crosses it.

The factor $\cos θ$ can be attached to either quantity. Attached to the area, $Δ S \cos θ$ is the patch's shadow on a plane perpendicular to the field. Attached to the field, $E \cos θ$ is the part of the field that pokes straight through the patch. Both give the same flux. Flux is a scalar. It is positive when the field crosses the patch in the direction of $\hat{𝐧}$ ( $θ < 90^{\circ}$ ) and negative when it crosses the other way ( $θ > 90^{\circ}$ ), and its SI unit is $N m^{2} / C$ .

For a large or curved surface we divide it into small patches, find the flux through each and add:

ϕ \approx \sum 𝐄 \cdot Δ 𝐒

(1.12)

The sum is only approximate because each patch is treated as having a single value of $𝐄$ . Making the patches smaller and smaller removes the error; in that limit the sum becomes a surface integral.

Definition

Electric flux

For a small flat patch, $Δ ϕ = 𝐄 \cdot Δ 𝐒$ ; for a large surface, the sum of this over all its patches. It measures how much field passes through the surface.

ExampleA $0.10 m^{2}$ patch facing a $200 N / C$ field has a flux of $20 N m^{2} / C$ .

ExampleNCERT p. 22

Turning a frame in a field

A flat rectangular frame, $30 c m \times 20 c m$ , is placed in a uniform electric field of $2.0 \times 10^{3} N / C$ . Find the flux through it when its normal makes an angle of (a) $0^{\circ}$ , (b) $60^{\circ}$ , (c) $90^{\circ}$ and (d) $180^{\circ}$ with the field.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A uniform field of

400 N / C

makes an angle of

30^{\circ}

with the normal to a flat surface of area

0.050 m^{2}

. Find the flux through it, in

N m^{2} / C

N m²/C

Show answer

Answer 17.3 N m²/C

ϕ = E A \cos θ = 400 \times 0.050 \times \cos 30^{\circ} \approx 17.3 N m^{2} / C

1.10 Electric Dipole

Most objects around you are electrically neutral, yet a neutral object can still produce a field if its positive and negative charges sit in different places. The simplest example is an electric dipole (विद्युत द्विध्रुव): charges $+ q$ and $- q$ held a fixed distance $2 a$ apart. The midpoint between them is the dipole's centre and the line through them is its axis. We give the dipole a direction along the axis, pointing from $- q$ to $+ q$ .

What kind of field should we expect? Close to the dipole you are much nearer one charge than the other, and that charge dominates. Far away, both charges are at almost the same distance, so their fields nearly cancel, though not completely. A reasonable guess, then: a dipole's field should die away with distance faster than the $1 / r^{2}$ field of a single charge. The calculation below confirms it.

1.10.1 The field of an electric dipole

Rather than tackle a general point, we work out the field where the geometry is simplest: along the axis, and in the equatorial plane, the plane through the centre at right angles to the axis. Fig. 1.16 shows both cases. Throughout, $\hat{𝐩}$ is the unit vector along the axis from $- q$ to $+ q$ .

3/3Equal fields; the sideways parts cancel and the net field at Q points opposite to $𝐩$ .

Fig. 1.16On the axis the net field points along

𝐩

; on the equatorial line it points against

𝐩

.NCERT Fig. 1.17 · p. 24

A negative charge and a positive charge sit a distance 2a apart on a horizontal axis, with the dipole moment p pointing from negative to positive. In the second step a point P on the axis shows a long arrow E plus q pointing away from the positive charge and a shorter opposite arrow E minus q; their sum, a bold arrow E, points along p. In the third step a point Q on the perpendicular bisector shows two equal field arrows, one away from the positive charge and one towards the negative charge; the parallelogram of these gives a bold resultant pointing opposite to p.

DerivationField at a point on the axis

Put the point P on the axis at distance $r$ from the centre O, beyond $+ q$ . Then P is $r - a$ from $+ q$ and $r + a$ from $- q$ . The field of $- q$ points towards $- q$ , that is along $- \hat{𝐩}$ ; the field of $+ q$ points away from $+ q$ , along $+ \hat{𝐩}$ :

𝐄_{- q} = - \frac{q}{4 π ε_{0} (r + a)^{2}} \hat{𝐩}

(1.13(a))

𝐄_{+ q} = \frac{q}{4 π ε_{0} (r - a)^{2}} \hat{𝐩}

(1.13(b))

Add them and put the bracket over a common denominator, using $(r + a)^{2} - (r - a)^{2} = 4 a r$ :

𝐄 = \frac{q}{4 π ε_{0}} [\frac{1}{(r - a)^{2}} - \frac{1}{(r + a)^{2}}] \hat{𝐩} = \frac{q}{4 π ε_{0}} \frac{4 a r}{(r^{2} - a^{2})^{2}} \hat{𝐩}

(1.14)

Far away ( $r ≫ a$ ) we can drop $a^{2}$ beside $r^{2}$ in the denominator:

𝐄 = \frac{4 q a}{4 π ε_{0} r^{3}} \hat{𝐩} (r ≫ a)

(1.15)

DerivationField at a point on the equatorial plane

Now let the point be on the equatorial plane at distance $r$ from the centre. Both charges are the same distance $\sqrt{r^{2} + a^{2}}$ away, so their fields are equal in size:

E_{+ q} = \frac{q}{4 π ε_{0}} \frac{1}{r^{2} + a^{2}}

(1.16(a))

E_{- q} = \frac{q}{4 π ε_{0}} \frac{1}{r^{2} + a^{2}}

(1.16(b))

$𝐄_{+ q}$ points away from $+ q$ and $𝐄_{- q}$ points towards $- q$ . Their components perpendicular to the axis are equal and opposite, so they cancel. Their components parallel to the axis both point from $+ q$ towards $- q$ , opposite to $\hat{𝐩}$ , and they add. If $θ$ is the angle each field makes with the axis, $\cos θ = a / \sqrt{r^{2} + a^{2}}$ , and

𝐄 = - (E_{+ q} + E_{- q}) \cos θ \hat{𝐩} = - \frac{2 q a}{4 π ε_{0} (r^{2} + a^{2})^{3 / 2}} \hat{𝐩}

(1.17)

For $r ≫ a$ :

𝐄 = - \frac{2 q a}{4 π ε_{0} r^{3}} \hat{𝐩} (r ≫ a)

(1.18)

Compare Eqs. (1.15) and (1.18). In both, the charge and the separation enter only through the combination $2 q a$ : a dipole with half the charge and twice the separation has exactly the same far field. That combination deserves a name and a direction. It is the electric dipole moment (विद्युत द्विध्रुव आघूर्ण):

Dipole moment

𝐩 = q \times 2 a \hat{𝐩}

(1.19)

where

$p = 2 q a$: its magnitude, in $C m$
$\hat{𝐩}$: direction: along the axis from $- q$ to $+ q$

In terms of $𝐩$ the far fields take a compact form:

𝐄_{axis} = \frac{2 𝐩}{4 π ε_{0} r^{3}} (r ≫ a)

(1.20)

𝐄_{equator} = - \frac{𝐩}{4 π ε_{0} r^{3}} (r ≫ a)

(1.21)

Definition

Electric dipole moment

For charges $\pm q$ separated by $2 a$ , the vector $𝐩 = 2 q a \hat{𝐩}$ , pointing from $- q$ to $+ q$ . SI unit: $C m$ .

Three features stand out. First, the field falls off as $1 / r^{3}$ , one power faster than a point charge's, confirming our guess. Second, at equal distances the axial field is twice the equatorial one, and the two point in opposite senses relative to $𝐩$ . Third, unlike a point charge's field, a dipole's field is not the same in all directions: it depends on the angle between $𝐫$ and $𝐩$ as well as on $r$ .

The exact expressions differ noticeably from Eqs. (1.20) and (1.21) only when $r$ is not much larger than $a$ . Physicists therefore often use an idealised point dipole: let the separation shrink towards zero while the charge grows, keeping $p = 2 q a$ fixed. For a point dipole the two equations hold at every distance, and a small molecule seen from a few molecular diameters away behaves very nearly like one.

Compare

A point charge and a dipole

Point charge $q$

Total charge $q \neq 0$
Field falls as $1 / r^{2}$
Same strength in every direction at a given $r$
Field lines run out to infinity

Dipole $𝐩$

Total charge zero
Field falls as $1 / r^{3}$
Strength depends on direction: axis twice the equatorial value
Field lines run from $+ q$ to $- q$

Example 1.9NCERT Example 1.9 · p. 25

Exact and approximate dipole fields

A small dipole is made of charges $\pm 2.0 μ C$ held $4.0 m m$ apart. (a) Find the field at two points $20 c m$ from its centre: A on the axis, beyond the positive charge, and B on the equatorial plane. Work each one exactly and then with the far-field formula. (b) Along the axis, how close to the centre can you go before the far-field formula comes out $1 %$ too small? ( $k = 9 \times 10^{9} N m^{2} C^{- 2}$ )

Try it yourself first. Then reveal the solution one step at a time.

1.10.2 Physical significance of dipoles

Dipoles matter far beyond pairs of charged beads, because many molecules are dipoles. In a water molecule the oxygen atom pulls the shared electrons towards itself, and the molecule's bent shape keeps both hydrogen atoms on the same side. The oxygen end is slightly negative and the hydrogen end slightly positive, so the molecule has a dipole moment even with no field applied (Fig. 1.17). Such molecules are called polar molecules. Water's polarity helps it dissolve salt, and it is why a microwave oven, which twists water dipoles back and forth billions of times a second, heats food.

Carbon dioxide ( $C O_{2}$ ) and methane ( $C H_{4}$ ) tell the opposite story. Each of their bonds is slightly polar, but the molecules are so symmetric that the bond dipoles cancel: the centre of the positive charges coincides with the centre of the negative charges, and there is no permanent dipole moment. An external field can still stretch such a molecule a little, pulling the two centres apart and giving it an induced dipole moment. (The centre of the positive charges is found like a centre of mass, $𝐫_{c} = \sum q_{i} 𝐫_{i} / \sum q_{i}$ , summing over the positive charges only; the centre of the negative charges is found the same way.)

Fig. 1.17Shape decides polarity: a symmetric molecule cancels its bond dipoles, a bent one does not.NCERT p. 25

On the left a carbon dioxide molecule lies in a straight line, oxygen, carbon, oxygen, with partial charges delta minus on the oxygens and delta plus on carbon; two short arrows below point in opposite directions and cancel, giving p equals zero. On the right a water molecule is bent, oxygen on top and two hydrogens below, with delta minus on oxygen and delta plus on each hydrogen; a bold arrow p points from the oxygen towards the hydrogens.

Definition

Polar molecule

A molecule whose centres of positive and negative charge do not coincide, so it has a permanent dipole moment.

ExampleWater, $H_{2} O$ .

1.11 Dipole in a Uniform External Field

Take a rigid dipole, such as a water molecule, whose moment $𝐩$ does not depend on any applied field. Place it in a uniform field $𝐄$ , at an angle $θ$ to the field (Fig. 1.18). Will it drift, turn, or both?

DerivationTorque on a dipole

The charge $+ q$ feels a force $q 𝐄$ and the charge $- q$ feels $- q 𝐄$ . The field is the same at both charges, so these forces cancel exactly: the net force is zero, and the dipole's centre does not accelerate.

The two forces do not, however, act along the same line, so together they form a couple that turns the dipole. Take torques about the centre O. The charge $+ q$ sits at displacement $𝐚$ from O and $- q$ at $- 𝐚$ , where $𝐚$ has length $a$ and points from $- q$ to $+ q$ . Adding $𝐫 \times 𝐅$ for the two charges gives

𝝉 = 𝐚 \times (q 𝐄) + (- 𝐚) \times (- q 𝐄) = 2 q 𝐚 \times 𝐄

The vector $2 q 𝐚$ is exactly the dipole moment $𝐩$ , so

Torque on a dipole

𝝉 = 𝐩 \times 𝐄

(1.22)

Its size is $τ = p E \sin θ$ , which is also one force, $q E$ , times the perpendicular distance $2 a \sin θ$ between the two lines of action. Because the net force is zero, the torque is the same about every point, not just about O. For the dipole in Fig. 1.18, $𝐩 \times 𝐄$ points into the page, so the dipole turns clockwise, towards $𝐄$ .

1/3Net force zero, but the forces act on different lines: a couple of arm $2 a \sin θ$ .

Fig. 1.18Equal and opposite forces form a couple:

τ = p E \sin θ

, largest at

90^{\circ}

and zero when aligned.NCERT Fig. 1.19 · p. 27

Horizontal field lines point right. A dipole with its positive charge at the upper right makes an angle theta of 50 degrees with the field. In the first step the positive charge feels a force qE to the right and the negative charge a force minus qE to the left; dashed lines of action are separated by 2a sin theta, and a curved arrow shows the dipole turning clockwise. In the second step the dipole has turned to 25 degrees and the arm between the forces is smaller. In the third step the dipole lies along the field, the two forces act along one line and the torque is zero.

The torque always tries to turn $𝐩$ into line with $𝐄$ . It is largest when the dipole is at right angles to the field ( $\sin 90^{\circ} = 1$ ) and vanishes when $𝐩$ is parallel to $𝐄$ , the stable position. It also vanishes when $𝐩$ points exactly against $𝐄$ , but that balance is unstable: the slightest nudge and the dipole swings round.

ExampleNCERT p. 27

Torque on a small dipole

A dipole consists of charges $\pm 2.0 n C$ held $3.0 m m$ apart. It is placed in a uniform field of $4.0 \times 10^{5} N / C$ with its axis at $30^{\circ}$ to the field. Find (a) the net force, (b) the torque, and (c) the largest torque this field could exert on it.

Try it yourself first. Then reveal the solution one step at a time.

If the field is not uniform, the forces on $+ q$ and $- q$ differ in size, and the dipole feels a net force as well as, in general, a torque. Take the simplest case: $𝐩$ along $𝐄$ , with the field pointing along $x$ and growing stronger with $x$ . Then $+ q$ sits where the field is stronger than at $- q$ , and the net force $q [E (x + a) - E (x - a)]$ points towards the stronger field; for a small dipole it is about $p d E / d x$ . Turn the dipole round, so that $𝐩$ points against $𝐄$ , and the net force reverses, pushing it towards the weaker field.

This is why a charged body attracts neutral ones: a balloon rubbed on your hair sticks to a wall, and a charged comb lifts scraps of paper. The charged body's field nudges the charges in the neutral object slightly apart, creating an induced dipole that lines up along the field. The field is strongest close to the charged body, so the aligned dipole is drawn towards it. The sign of the charged body does not matter, because the induced dipole always lines up with the field.

Try at homeNCERT p. 28

Bend water without touching it

You’ll need

a plastic comb or a balloon
dry hair or a woollen cloth
a tap with a steady thin stream
a silk scarf or polythene bag

Predict first

When the charged comb comes near the stream, the water will…

bend towards the combbend away from the combnot move, because water is neutral

Open a tap so that a thin, smooth stream of water falls (about as thick as a pencil lead).
Charge a plastic comb or an inflated balloon by rubbing it briskly on dry hair or wool.
Bring it close to the stream, a few centimetres below the tap, without touching the water. Watch the stream.
Discharge the comb by running your fingers over it, then charge it on a different material (a silk scarf or a polythene bag) and repeat.

The stream bends towards the comb, and it still bends towards it when the comb carries charge of the other sign. The comb's field polarises the water: charge of the opposite sign gathers on the side of the stream nearer the comb (water molecules are polar, and tap water also conducts a little). The field is strongest close to the comb, so the near side is pulled harder than the far side is pushed, and the stream swings towards the comb.

Can’t do it now? Try the simulation · Fig. 1.19

SimulationDipole in a non-uniform field

A large positive charge is fixed near the left edge of the stage. A small dipole sits to its right, a negative charge nearer the large charge and a positive charge a little further away. Field lines spread out from the large charge and arrows show the force on each charge. You can drag the two small charges to turn the dipole round, or to move it nearer or further away, and watch the forces on its two ends change.

Fig. 1.19A small dipole (charges

\pm 0.25 q

) beside a large fixed charge

+ 4 q

, whose field is strong close to it and weak far away. Drag either end of the dipole.

Try this

Turn on Forces if it is off. Which end of the dipole is pulled harder, and which way is the net force on the dipole?
Swap the two small charges so that the positive end faces the large charge. Which way is the dipole pushed now?
Move both small charges far to the right, one at a time, keeping them the same distance apart. What happens to the difference between the forces on the two ends, and why?

Quick check

1 of 3

A dipole of moment

3.0 \times 10^{- 9} C m

is at

90^{\circ}

to a uniform field of

2.0 \times 10^{4} N / C

. What torque acts on it, in N m? (e.g. 6e-5)

N m

Show answer

Answer 0.00006 N m

τ = p E \sin 90^{\circ} = (3.0 \times 10^{- 9}) (2.0 \times 10^{4}) = 6.0 \times 10^{- 5} N m

1.12 Continuous Charge Distribution

A census reports how many people live per square kilometre of a district. Nobody gets that figure by listing where each person stands; the population of a large area is counted and divided by the area. Charge on large objects is described the same way. A charged metal sheet may carry $10^{12}$ extra electrons, far too many to track one by one, so we describe its charge by a density instead.

Take a small patch of a charged surface, of area $Δ S$ , carrying charge $Δ Q$ . 'Small' has a special meaning here: tiny compared with the object, so that the density can vary from place to place, yet huge compared with atoms, so that it holds millions of charges. The surface charge density at that place is

Surface charge density

σ = \frac{Δ Q}{Δ S}

(1.23)

where

$σ$: surface charge density, in $C / m^{2}$

Repeating this all over the surface gives $σ$ as a smooth function of position. It is an average: it deliberately blurs out the individual electrons and the empty space between them, just as a population density ignores the gaps between houses. Charge spread along a thin wire, or through a volume, is described in the same way (Fig. 1.20):

Linear charge density

λ = \frac{Δ Q}{Δ l}

(1.24)

where

$λ$: linear charge density, in $C / m$

Volume charge density

ρ = \frac{Δ Q}{Δ V}

(1.25)

where

$ρ$: volume charge density, in $C / m^{3}$

Fig. 1.20Each element is tiny on our scale but still holds a vast number of elementary charges.NCERT Fig. 1.21 · p. 28

Three panels. The first shows a curved charged wire with a short highlighted piece delta l, and the formula lambda equals delta Q over delta l. The second shows a charged flat sheet with a small patch delta S, and sigma equals delta Q over delta S. The third shows a charged solid blob with a tiny cube delta V inside, and rho equals delta Q over delta V.

Definition

Charge density

Charge per unit length ( $λ$ , in C/m), per unit area ( $σ$ , in $C / m^{2}$ ) or per unit volume ( $ρ$ , in $C / m^{3}$ ) of a continuous distribution.

To find the field of such a distribution, we go back to point charges. Chop the charged region into small volume elements $Δ V$ , each carrying charge $ρ Δ V$ and small enough to count as a point charge. The element at distance $r^{'}$ from P contributes

Δ 𝐄 = \frac{1}{4 π ε_{0}} \frac{ρ Δ V}{r^{' 2}} {\hat{𝐫}}^{'}

(1.26)

where

${\hat{𝐫}}^{'}$: the unit vector from the element towards P

and superposition adds up all the contributions:

𝐄 \approx \frac{1}{4 π ε_{0}} \sum_{all Δ V} \frac{ρ Δ V}{r^{' 2}} {\hat{𝐫}}^{'}

(1.27)

Here $ρ$ , $r^{'}$ and ${\hat{𝐫}}^{'}$ differ from one element to the next, so in general this is a long sum; with calculus it becomes an integral, which we do not attempt here. The message is that Coulomb's law and superposition give the field of any charge distribution, discrete, continuous or a mixture. For symmetric distributions, Gauss's law will give the answer with almost no summing at all.

Quick check

1 of 3

A thin plastic rod

1.5 m

long carries a charge of

6.0 μ C

spread uniformly along it. What is its linear charge density, in

μ

C/m?

μC/m

Show answer

Answer 4 μC/m

λ = Q / l = 6.0 μ C / 1.5 m = 4.0 μ C / m

1.13 Gauss's Law

We now combine flux and Coulomb's law. Start with the simplest case: a point charge $q$ at the centre of an imaginary sphere of radius $r$ . What is the total flux out through the sphere?

DerivationFlux of a point charge through a sphere

Divide the sphere into small patches. For a patch $Δ 𝐒$ the field there is given by Eq. (1.6), so

Δ ϕ = 𝐄 \cdot Δ 𝐒 = \frac{q}{4 π ε_{0} r^{2}} \hat{𝐫} \cdot Δ 𝐒

(1.28)

On a sphere the outward normal at every point lies along the radius, so $Δ 𝐒$ is parallel to $\hat{𝐫}$ and $\hat{𝐫} \cdot Δ 𝐒 = Δ S$ :

Δ ϕ = \frac{q}{4 π ε_{0} r^{2}} Δ S

(1.29)

Every patch is the same distance $r$ from the charge, so the factor in front is the same for all patches. Adding up, the patches together make the whole area $4 π r^{2}$ :

ϕ = \frac{q}{4 π ε_{0} r^{2}} \sum Δ S = \frac{q}{4 π ε_{0} r^{2}} \times 4 π r^{2} = \frac{q}{ε_{0}}

(1.30)

The radius has cancelled. A bigger sphere has more area, but the field on it is weaker in exactly the same proportion, so the flux is the same for every sphere. This is where the $4 π$ in $k = 1 / 4 π ε_{0}$ pays off.

The result is far more general than a sphere around a single charge. It holds for any closed surface around any collection of charges, and it is called Gauss's law.

LawGauss's law

The net electric flux out of any closed surface equals the total charge inside that surface divided by $ε_{0}$ .

Gauss's law

ϕ = \oint 𝐄 \cdot d 𝐒 = \frac{q_{enc}}{ε_{0}}

(1.31)

where

$q_{enc}$: the algebraic sum of all charges inside the closed surface
$\oint$: the sum of $𝐄 \cdot Δ 𝐒$ over the whole closed surface

3/3Lines entering also leave; the net flux is zero.

Fig. 1.21Only the charge inside decides the net flux through a closed surface.NCERT Fig. 1.22 · p. 29

Twenty-four field lines spread out from a positive point charge. In the first step a dashed sphere surrounds the charge and every line crosses it once; a label gives the flux as q over epsilon zero. In the second step the sphere is replaced by a lumpy closed surface that still encloses the charge and is still crossed by every line once, so the flux is unchanged. In the third step a small dashed sphere sits off to the right, enclosing no charge: dots mark where lines enter and leave it, the same number each way, so the flux is zero.

A surface that encloses no charge must have zero net flux. Try it on a closed bowl: a hemisphere of radius $R$ capped by a flat lid, standing in a uniform field $𝐄$ that points straight up through the lid. Every line that enters through the curved underside leaves through the lid. The lid faces the field squarely, so its flux is $+ E π R^{2}$ . The curved part presents the same shadow, a disc of area $π R^{2}$ , but the field enters it, so its flux is $- E π R^{2}$ . The total is zero, as Gauss's law requires.

Keep these points in mind whenever you use the law:

Shape does not matter. The result holds for every closed surface, whatever its shape or size.
Only enclosed charge counts, signs included. $q_{enc}$ is the algebraic sum of the charges inside; where they sit inside makes no difference.
But the field is everybody's. The $𝐄$ in the flux is the total field. A charge just outside the surface changes $𝐄$ all over it; it simply sends as much flux in as out.
You choose the surface. A closed surface drawn for this purpose is a Gaussian surface (गाउसीय पृष्ठ). It is imaginary, so draw it wherever symmetry makes $E$ simple. Keep it off point charges, where $E$ grows without limit; cutting through smeared-out charge is fine.
Symmetry is what makes it useful. The law hands you $E$ directly only when you can find a surface on which $E$ is constant, and along the normal, wherever flux crosses.
It is a test of Coulomb's law. If the force went as $1 / r^{2.01}$ , the flux through a sphere around a charge would change slightly with the sphere's radius. Precise checks of Gauss's law are therefore precise checks of the inverse-square law.

Definition

Gaussian surface

An imaginary closed surface chosen to apply Gauss's law, usually shaped to match the symmetry of the charges.

Example 1.10NCERT Example 1.10 · p. 31

Flux through a cube in a growing field

In a certain region the electric field is $𝐄 = β x^{2} \hat{𝐢}$ , with $β = 500 N C^{- 1} m^{- 2}$ . A cube of side $0.20 m$ has its faces parallel to the coordinate planes, with its left face at $x = 0.10 m$ and its right face at $x = 0.30 m$ . Find (a) the net flux out of the cube and (b) the charge inside it.

The field points along

x

and grows as

x^{2}

.NCERT p. 31

A cube with edges 0.2 m long sits on the x-axis with its left face at x equals 0.1 m and its right face at x equals 0.3 m. A short field arrow E L points into the left face and a much longer arrow E R points out of the right face, showing that the field grows with x.

Try it yourself first. Then reveal the solution one step at a time.

Example 1.11NCERT Example 1.11 · p. 32

Where is the charge?

In a region the electric field is $𝐄 = + 350 \hat{𝐢} N / C$ for $x > 0$ and $𝐄 = - 350 \hat{𝐢} N / C$ for $x < 0$ . A closed box with a cross-section of $8.0 c m \times 6.0 c m$ lies along the $x$ -axis, with its end faces at $x = - 10 c m$ and $x = + 20 c m$ . Working out the flux through each face, find the charge inside the box. Would the answer change if the box were slid along the $x$ -axis until both of its ends lay at positive $x$ ?

Try it yourself first. Then reveal the solution one step at a time.

Notice that the ends need not be symmetric about $x = 0$ : the field has the same size everywhere, so only the fact that the box straddles the plane $x = 0$ matters. If the box lay wholly at $x > 0$ , the field would enter the left face ( $- 1.68$ ) and leave the right face ( $+ 1.68$ ), giving zero flux and zero charge. The charge must therefore sit in the plane $x = 0$ itself, as a sheet with $σ = q / A$ .
$σ = \frac{3.0 \times 10^{- 11} C}{4.8 \times 10^{- 3} m^{2}} \approx 6.2 \times 10^{- 9} C / m^{2}$

Quick check

1 of 3

A closed surface encloses charges of

+ 3.0 μ C

and

- 1.0 μ C

. A charge of

+ 5.0 μ C

sits just outside. What is the net flux out of the surface, in

N m^{2} / C

? (e.g. 1.5e5)

N m²/C

Show answer

Answer 226000 N m²/C

Only enclosed charge counts:

q_{enc} = 2.0 μ C

, so

ϕ = q_{enc} / ε_{0} = (2.0 \times 10^{- 6}) / (8.854 \times 10^{- 12}) \approx 2.26 \times 10^{5} N m^{2} / C

. The outside charge changes

𝐄

on the surface but not the net flux.

1.14 Applications of Gauss's Law

For a general charge distribution the sum in Eq. (1.27) is hard work. When the distribution is highly symmetric, Gauss's law does the job in a few lines. The recipe is always the same: (1) use symmetry to decide the direction of $𝐄$ and what its size can depend on; (2) choose a Gaussian surface on which $E$ is constant wherever the flux is not zero; (3) equate flux to $q_{enc} / ε_{0}$ and solve for $E$ .

1.14.1 Field due to an infinitely long straight uniformly charged wire

Take a very long, thin straight wire carrying charge $λ$ per unit length. Symmetry first. Rotate the whole arrangement about the wire, or slide it along the wire: it looks exactly the same, so the field can depend only on the perpendicular distance $r$ from the wire. For direction, pair each small piece of the wire with an equal piece the same distance on the other side of the foot of the perpendicular from P. Their fields at P have equal components along the wire in opposite senses, which cancel, leaving only the component pointing straight away from the wire. So $𝐄$ is radial: outward if $λ > 0$ , inward if $λ < 0$ .

DerivationField of a long charged wire

Choose a Gaussian surface that fits this symmetry: a cylinder of radius $r$ and length $l$ with the wire along its axis (Fig. 1.22).

Fig. 1.22By symmetry the field is radial, so only the curved surface carries flux.NCERT Fig. 1.26 · p. 33

A long straight wire lined with plus signs runs across the figure. A dashed Gaussian cylinder of radius r and length l surrounds a section of it. Field arrows point straight outward from the curved surface, perpendicular to the wire, while no arrows cross the flat end faces.

The field is radial, so it runs parallel to the flat end faces and gives them no flux. On the curved surface $𝐄$ is along the outward normal and has the same size everywhere (every point is at distance $r$ ). The flux is therefore $E$ times the curved area:

ϕ = E \times 2 π r l

The cylinder encloses a length $l$ of wire, carrying charge $λ l$ . Gauss's law gives

E \times 2 π r l = \frac{λ l}{ε_{0}} \Rightarrow E = \frac{λ}{2 π ε_{0} r}

In vector form, with $\hat{𝐧}$ the unit vector pointing radially away from the wire in the plane through P perpendicular to it,

Long charged wire

𝐄 = \frac{λ}{2 π ε_{0} r} \hat{𝐧}

(1.32)

The field of a line charge falls as $1 / r$ : double the distance and the field halves, a gentler fall than a point charge's $1 / r^{2}$ . Two cautions. The scalar $E$ in $𝐄 = E \hat{𝐧}$ carries the sign of $λ$ : for a negatively charged wire $E < 0$ and the field points inward, against $\hat{𝐧}$ , while its size is $∣ 𝐄 ∣ = ∣ λ ∣ / 2 π ε_{0} r$ . And the argument needed an infinite wire, because only then does symmetry forbid a field component along the wire. For a real wire of length $L$ , Eq. (1.32) works well near its middle at distances much smaller than $L$ , and fails near the ends.

1.14.2 Field due to a uniformly charged infinite plane sheet

Now take an infinite flat sheet with uniform surface charge density $σ$ , and let the $x$ -axis be normal to it. Sliding the sheet in its own plane, or turning it about the $x$ -axis, changes nothing. So $𝐄$ cannot depend on $y$ or $z$ , and it cannot have a component along the sheet. For $σ > 0$ it points straight away from the sheet on both sides.

DerivationField of an infinite charged sheet

Use a Gaussian 'pillbox': a short cylinder that pierces the sheet at right angles, with flat end faces of area $A$ on either side (Fig. 1.23).

Fig. 1.23Flux leaves only through the two end faces.NCERT Fig. 1.27 · p. 34

A thin vertical sheet covered in plus signs is seen edge-on. A dashed cylinder pierces it at right angles, with flat end faces of area A on either side, labelled face 1 and face 2. Field arrows point away from the sheet on both sides, passing straight out through the two end faces.

The field runs parallel to the curved side, so that side carries no flux. On face 1 the outward normal points along $- x$ and so does $𝐄$ ; on face 2 both point along $+ x$ . Each end face contributes $+ E A$ , so the total flux is $2 E A$ . The charge enclosed is the patch of sheet inside the cylinder, $σ A$ :

2 E A = \frac{σ A}{ε_{0}} \Rightarrow E = \frac{σ}{2 ε_{0}}

In vector form, with $\hat{𝐧}$ the unit normal pointing away from the sheet on each side,

Infinite charged sheet

𝐄 = \frac{σ}{2 ε_{0}} \hat{𝐧}

(1.33)

Surprisingly, $E$ does not depend on the distance from the sheet. One way to see why: look at the sheet from twice as far away. Each patch of it is now four times weaker, but a cone of fixed solid angle from your eye takes in four times as much of the sheet, so nothing changes. The field of an infinite sheet is therefore uniform on each side: away from the sheet if $σ > 0$ , towards it if $σ < 0$ . Crossing the sheet, the $x$ -component of $𝐄$ jumps from $- σ / 2 ε_{0}$ to $+ σ / 2 ε_{0}$ . A surface charge makes the field discontinuous; a volume charge (as in Example Example 1.12) does not. A real, finite sheet behaves like this only at distances small compared with its size, away from its edges.

1.14.3 Field due to a uniformly charged thin spherical shell

Finally, take a thin spherical shell of radius $R$ carrying charge $q$ spread uniformly, so $σ = q / 4 π R^{2}$ . The arrangement looks the same from every direction about its centre O, so at any point P the field must be radial and its size can depend only on the distance $r = O P$ . The natural Gaussian surface is a sphere of radius $r$ centred on O (Fig. 1.24).

DerivationField of a charged spherical shell

Outside the shell ( $r > R$ ). On the Gaussian sphere through P the field has the same size $E$ everywhere and points along the outward normal, so the flux is $E \times 4 π r^{2}$ . The sphere encloses the whole charge, $q = σ \times 4 π R^{2}$ :

E \times 4 π r^{2} = \frac{q}{ε_{0}} \Rightarrow E = \frac{q}{4 π ε_{0} r^{2}} = \frac{σ R^{2}}{ε_{0} r^{2}}

Outside a charged shell

𝐄 = \frac{q}{4 π ε_{0} r^{2}} \hat{𝐫} (r \geq R)

(1.34)

Inside the shell ( $r < R$ ). The flux through the Gaussian sphere is again $E \times 4 π r^{2}$ , but now the sphere encloses no charge at all:

Inside a charged shell

E \times 4 π r^{2} = 0 \Rightarrow E = 0 (r < R)

(1.35)

1/2The Gaussian sphere encloses all of $q$ : the field is that of a point charge at O.

Fig. 1.24Outside, the shell acts like a point charge at its centre; inside, the field vanishes.NCERT Fig. 1.28 · p. 35

A thin spherical shell of radius R carries plus signs evenly around it. In the first step a larger dashed Gaussian sphere passes through an outside point P; field arrows point radially outward and the label gives E equals q over four pi epsilon zero r squared. In the second step a smaller dashed Gaussian sphere lies inside the shell through an inside point P; it encloses no charge, the outside arrows fade, and the label reads E equals zero.

So, from outside, a uniformly charged shell looks exactly like a point charge at its centre; Newton proved the same about the gravity of a spherical planet. Inside, the field vanishes at every point, not just at the centre. Because this conclusion rests on the inverse-square law, experiments that hunt for a field inside a charged hollow conductor are among the most sensitive tests that the exponent in Coulomb's law is exactly $2$ .

Example 1.12NCERT Example 1.12 · p. 36

Inside and outside a charged ball

A ball of insulating material, radius $R = 5.0 c m$ , carries a charge $Q = + 2.0 μ C$ spread uniformly through its volume. Find an expression for the field at distance $r$ from its centre, both inside and outside, and evaluate it at $r = 2.0 c m$ , $5.0 c m$ and $10 c m$ . ( $k = 9 \times 10^{9} N m^{2} C^{- 2}$ )

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

A long straight wire carries

λ = 2.0 μ C / m

. What is the field

30 c m

from it, in N/C? (e.g. 4.5e4)

N/C

Show answer

Answer 120000 N/C

E = λ / (2 π ε_{0} r) = 2 k λ / r = 2 (9 \times 10^{9}) (2.0 \times 10^{- 6}) / 0.30 = 1.2 \times 10^{5} N / C

Table 1.2Quantities met in this chapterNCERT p. 40

Quantity	Symbol	SI unit	Dimensions
Electric field	$𝐄$	$N / C$ (or $V / m$ )	$[M L T^{- 3} A^{- 1}]$
Electric flux	$ϕ$	$N m^{2} / C$ (or $V m$ )	$[M L^{3} T^{- 3} A^{- 1}]$
Dipole moment	$𝐩$	$C m$	$[L T A]$
Linear charge density	$λ$	$C / m$	$[L^{- 1} T A]$
Surface charge density	$σ$	$C / m^{2}$	$[L^{- 2} T A]$
Volume charge density	$ρ$	$C / m^{3}$	$[L^{- 3} T A]$
Permittivity of free space	$ε_{0}$	$C^{2} N^{- 1} m^{- 2}$	$[M^{- 1} L^{- 3} T^{4} A^{2}]$
Vector area element	$Δ 𝐒$	$m^{2}$	$[L^{2}]$

Flash cards · 12

Formula flashcards

1 / 12

Recap

Electric Charges and Fields: key points and summary

Charge comes in two kinds, positive and negative: like charges repel, unlike charges attract. Rubbing charges bodies by transferring electrons; glass rubbed with silk is positive by convention (Section 1.2).
Conductors let charge move freely; insulators hold it in place. Earthing drains charge from a conductor (Section 1.3).
Charge is additive (a scalar with a sign), conserved in every isolated system and quantised: $q = n e$ with $e = 1.602 \times 10^{- 19} C$ (Section 1.4).
Coulomb's law: $𝐅_{21} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{21}^{2}} {\hat{𝐫}}_{21}$ , with $k = 1 / 4 π ε_{0} \approx 9 \times 10^{9} N m^{2} C^{- 2}$ . The forces on the two charges are equal and opposite (Section 1.5).
Between an electron and a proton, electrical attraction beats gravitational attraction by a factor of about $10^{39}$ (Example 1.3).
Superposition: each pair force is unaffected by other charges, and the total force is the vector sum of the pair forces (Section 1.6).
The electric field is the force per unit positive test charge, $𝐄 = 𝐅 / q_{0}$ . A point charge's field is radial and falls as $1 / r^{2}$ ; fields of several charges add as vectors (Section 1.7).
Field lines run from positive to negative charges, never cross, never form closed loops in electrostatics, and crowd where the field is strong (Section 1.8).
Flux through a small patch is $Δ ϕ = 𝐄 \cdot Δ 𝐒$ ; for closed surfaces the normal points outward (Section 1.9).
A dipole ( $𝐩 = 2 q a$ , from $- q$ to $+ q$ ) has a far field falling as $1 / r^{3}$ : $2 𝐩 / (4 π ε_{0} r^{3})$ on the axis and $- 𝐩 / (4 π ε_{0} r^{3})$ on the equatorial plane (Section 1.10).
In a uniform field a dipole feels no net force but a torque $𝝉 = 𝐩 \times 𝐄$ ; in a non-uniform field it also feels a net force, which is why charged objects attract neutral ones (Section 1.11).
Continuous charge is described by densities $λ$ , $σ$ and $ρ$ ; Coulomb's law and superposition still give the field (Section 1.12).
Gauss's law: the net flux out of any closed surface is $q_{enc} / ε_{0}$ (Section 1.13).
With symmetry, Gauss's law gives $E = λ / 2 π ε_{0} r$ for a long wire, $E = σ / 2 ε_{0}$ for an infinite sheet, and $q / 4 π ε_{0} r^{2}$ outside but zero inside a charged shell (Section 1.14).

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Quick check: MCQs on Electric Charges and Fields

1.2 Electric Charge4 questions
1.3 Conductors and Insulators3 questions
1.4.3 Quantisation of charge4 questions
1.5 Coulomb's Law4 questions
1.6 Forces between Multiple Charges3 questions
1.7.2 Physical significance of electric field4 questions
1.8 Electric Field Lines3 questions
1.9 Electric Flux3 questions
1.10.2 Physical significance of dipoles4 questions
1.11 Dipole in a Uniform External Field3 questions
1.12 Continuous Charge Distribution3 questions
1.13 Gauss's Law3 questions
1.14.3 Field due to a uniformly charged thin spherical shell4 questions

Glossary · 15

Key terms (English–Hindi)

charge densityआवेश घनत्वaavesh ghanatva: Charge per unit length ( $λ$ ), area ( $σ$ ) or volume ( $ρ$ ).First seen in the chapter →
conductorचालकchaalak: A material in which charge moves freely, such as a metal.First seen in the chapter →
conservation of chargeआवेश संरक्षणaavesh sanrakshan: The total charge of an isolated system never changes.First seen in the chapter →
Coulomb's lawकूलॉम का नियमkooloam ka niyam: The force between two point charges is proportional to $q_{1} q_{2}$ and to $1 / r^{2}$ , along the line joining them.First seen in the chapter →

dipole momentद्विध्रुव आघूर्णdvidhruv aaghoorn: $𝐩 = 2 q a$ , directed from $- q$ to $+ q$ ; SI unit C m.First seen in the chapter →

electric chargeविद्युत आवेशvidyut aavesh: A property of matter that makes it exert and feel electric forces; it is positive or negative, and like charges repel.First seen in the chapter →
electric dipoleविद्युत द्विध्रुवvidyut dvidhruv: Two equal and opposite charges $\pm q$ a small distance $2 a$ apart.First seen in the chapter →
electric fieldविद्युत क्षेत्रvidyut kshetra: The force per unit positive test charge at a point, $𝐄 = 𝐅 / q_{0}$ , in N/C.First seen in the chapter →
electric field lineविद्युत क्षेत्र रेखाvidyut kshetra rekha: A curve whose tangent gives the field direction; crowded lines mean a strong field.First seen in the chapter →
electric fluxविद्युत फ्लक्सvidyut flux: $\sum 𝐄 \cdot Δ 𝐒$ over a surface; a measure of the field lines crossing it.First seen in the chapter →

Gaussian surfaceगाउसीय पृष्ठgaussiya prishth: An imaginary closed surface used to apply Gauss's law.First seen in the chapter →

insulatorविद्युतरोधीvidyutrodhi: A material in which charges cannot move freely, such as glass or plastic.First seen in the chapter →

polar moleculeध्रुवीय अणुdhruveey anu: A molecule with a permanent dipole moment, such as water.First seen in the chapter →

quantisation of chargeआवेश का क्वांटमीकरणaavesh ka kvaantameekaran: Every free charge is a whole-number multiple of $e = 1.6 \times 10^{- 19} C$ : $q = n e$ .First seen in the chapter →

superposition principleअध्यारोपण सिद्धांतadhyaropan siddhaant: The total force (or field) from several charges is the vector sum of the separate contributions.First seen in the chapter →

Electric Charges and Fields: Class 12 Physics · Chapter 1

What you will learn in Electric Charges and Fields

1.1 Introduction

1.2 Electric Charge

1.3 Conductors and Insulators

1.4 Basic Properties of Electric Charge

1.4.1 Additivity of charges

1.4.2 Charge is conserved

1.4.3 Quantisation of charge

1.5 Coulomb's Law

1.6 Forces between Multiple Charges

1.7 Electric Field

1.7.1 Electric field due to a system of charges

1.7.2 Physical significance of electric field

1.8 Electric Field Lines

1.9 Electric Flux

1.10 Electric Dipole

1.10.1 The field of an electric dipole

1.10.2 Physical significance of dipoles

1.11 Dipole in a Uniform External Field

1.12 Continuous Charge Distribution

1.13 Gauss's Law

1.14 Applications of Gauss's Law

1.14.1 Field due to an infinitely long straight uniformly charged wire

1.14.2 Field due to a uniformly charged infinite plane sheet

1.14.3 Field due to a uniformly charged thin spherical shell

Electric Charges and Fields: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Electric Charges and Fields

Key terms (English–Hindi)

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