Electrostatic Potential and Capacitance: Class 12 Physics · Chapter 2

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

Published 28 Sep 2026 · Updated 28 Sep 2026

170 min
2 simulations
20 practice questions
55 MCQs

What you will learn in Electrostatic Potential and Capacitance

Explain why the electrostatic force is conservative, and define electrostatic potential energy and potential with zero at infinity
Derive and use the potentials of a point charge, an electric dipole, a system of charges and a charged spherical shell
Describe equipotential surfaces and relate the electric field to the rate at which the potential changes
Find the potential energy of an assembly of charges, of a charge in an external potential and of a dipole in a uniform field
State and justify the electrostatic properties of conductors, including electrostatic shielding
Explain how polar and non-polar dielectrics polarise and how this weakens the field inside them
Derive the capacitance of a parallel plate capacitor with and without a dielectric, and combine capacitors in series and in parallel
Calculate the energy stored in a capacitor and the energy density of an electric field

Original companion text aligned to NCERT Physics Class 12, Chapter 2, pp. 45–80. Not affiliated with or endorsed by NCERT.

2.1 Introduction

Carry a bucket of water up the stairs to the tank on your roof and the effort is not wasted: it waits there, ready to come back as the water rushes down the pipe to the kitchen tap. In Class 11 you called this stored energy gravitational potential energy. Electric charges offer exactly the same bargain. Push one positive charge towards another and you must work against their repulsion; let go, and the charges fly apart, handing that work back as kinetic energy.

Why can the stored work be recovered at all? Gravity has a special property: the work it does on the bucket depends only on how high the bucket ends up, not on which staircase you climbed or how many times you paused. Forces with this property are called conservative; the force of a stretched spring is another. The force between two charges at rest belongs to the same family. Like the pull of gravity between two masses, it acts along the line joining the pair. Its size depends only on how far apart they are, and any force built that way is conservative, as the staircase argument of Section 2.2 will show. Everything else in this chapter rests on that one property.

On that foundation the chapter builds two ideas. The first is the electrostatic potential, which turns the energy bookkeeping of any arrangement of charges into a map of numbers. The second is the capacitor, a device that stores charge, and energy along with it, in the field between two conductors.

Here is the electrical version of lifting the bucket (Fig. 2.1). A positive charge $Q$ sits on an insulating stand at the origin. We carry a much smaller positive charge $q$ inward, starting at R and finishing at P, closer to $Q$ . We move it so gently that it never gains speed: at every instant our push $𝐅_{ext}$ is exactly as strong as the electric repulsion $𝐅_{E}$ and points the other way. The stand keeps $Q$ in place, and $q$ is so small that it hardly tugs on $Q$ anyway, so the arrangement we push against stays the same throughout.

2/3An external force $𝐅_{ext} = - 𝐅_{E}$ moves $q$ inward without speeding it up; the work piles up as potential energy $U$ .

Fig. 2.1Work done slowly against the repulsion is stored as potential energy; released, the charge turns it back into kinetic energy.NCERT Fig. 2.1 · p. 46

A large positive charge Q is held fixed on a stand at the left. A small positive test charge q starts far to the right; small arrows show the outward electric force and an equal inward external force. Two horizontal bars below track the potential energy U and the kinetic energy K. In the second step q has been pushed slowly much closer to Q: both force arrows are longer and the U bar has grown large while K is still zero. In the third step q has been released and is racing away to the right with a velocity arrow; the U bar has shrunk and the K bar has grown by the same amount.

DerivationWork stored as potential energy

With no net force on $q$ , its kinetic energy never changes, so every joule of work done by the external agent is stored. Adding up $𝐅_{ext} \cdot d 𝐫$ along the path from R to P gives the work

W_{R P} = \int_{R}^{P} 𝐅_{ext} \cdot d 𝐫 = - \int_{R}^{P} 𝐅_{E} \cdot d 𝐫

(2.1)

The electric force meanwhile does the equal and opposite amount, $- W_{R P}$ . We define the rise in the electrostatic potential energy of $q$ as this stored work:

Δ U = U_{P} - U_{R} = W_{R P}

(2.2)

Notice what Eq. (2.2) fixes and what it leaves open. It fixes only a change in $U$ , and that change depends on nothing but the end points R and P: the work is the same along every route between them (you will see why in Fig. 2.2), which is exactly what lets us attach one number to each point. What it leaves open is a constant: add $5 J$ , or any other amount, to $U$ at every point and every difference stays the same.

So we may pick the zero of $U$ wherever it is most convenient. For charges confined to a finite region, the obvious place is very far away, where they no longer affect each other at all. With $U_{\infty} = 0$ and R pushed out to infinity,

W_{\infty P} = U_{P} - U_{\infty} = U_{P}

(2.3)

In words: the potential energy of $q$ at P is the work an external agent must do to bring $q$ slowly in from infinity to P, with the other charges held where they are.

Definition

Electrostatic potential energy

The work done by an external agent in bringing a charge from infinity to a given point without accelerating it, with all the other charges held fixed. Only differences of it are physically meaningful.

ExampleIt is positive for a charge near a like charge and negative near an unlike charge.

2.2 Electrostatic Potential

The work in Eq. (2.2) grows in step with $q$ : double the test charge and every force $q 𝐄$ on it doubles, and so does the work. To describe the arrangement of source charges rather than the visitor, divide by $q$ . Height plays this role for gravity: the energy $m g h$ of a lifted object depends on its mass, but the height $h$ belongs to the place. The electrical counterpart of height is the electrostatic potential $V$ :

Potential difference

V_{P} - V_{R} = \frac{U_{P} - U_{R}}{q} = \frac{W_{R P}}{q}

(2.4)

Eq. (2.4) compares two points; it says nothing about either one on its own. To give each point a number of its own we need a reference level, as sea level is for heights. For charges confined to a finite region the natural reference is infinity, where their influence has died away. With $V_{\infty} = 0$ , the potential at a point is the work per unit positive charge needed to bring a test charge there from infinity.

Definition

Electrostatic potential

The work done by an external agent, per unit positive test charge, in bringing a test charge from infinity to a point without acceleration. Its SI unit is the volt: $1 V = 1 J / C$ .

ExampleA torch cell keeps its positive terminal about $1.5 V$ above its negative terminal, so every coulomb passing between them exchanges $1.5 J$ .

One technical point: a real test charge would tug the source charges slightly out of place. Using a vanishingly small test charge $δ q$ , and dividing the work $δ W$ it needs by $δ q$ , removes that worry without changing the answer.

Because the force is conservative, $V_{P} - V_{R}$ is the same for every route between the two points. Fig. 2.2 shows why for a single point charge. Any route can be imitated, as closely as you like, by a staircase of short radial steps and short arcs centred on the charge. Along an arc the force is at right angles to the motion and does no work, so only the radial steps count, and they always add up to the same trip inward.

3/3Break any path into arcs (no work: force at right angles to the motion) and radial steps, which always add up to the same total.

Fig. 2.2Whatever the route from R to P, the work per unit charge,

V_{P} - V_{R}

, is the same.NCERT Fig. 2.2 · p. 48

A positive charge Q sits at the lower left, surrounded by faint circular arcs centred on it. Point R lies far out to the right and point P closer in to the upper left. In the first step a straight path joins R to P, labelled with the work W. In the second step a winding orange path loops down and back up between the same points, labelled same work. In the third step a zigzag path hugs the straight line: it is made of short arcs along the circles, which need no work, and short radial steps towards Q, which carry all the work.

Potential difference is what a voltmeter reads and what the label on a cell or a socket states. A torch cell gives about $1.5 V$ , a car battery $12 V$ and a household socket in India $230 V$ (an alternating value, as you will see in Chapter 7). Between a thundercloud and the ground the potential difference can reach hundreds of millions of volts.

Quick check

1 of 3

Moving a

+ 2.0 μ C

charge slowly from point R to point P takes

6.0 m J

of work. What is

V_{P} - V_{R}

, in volts?

Show answer

Answer 3000 V

V_{P} - V_{R} = W / q = (6.0 \times 10^{- 3} J) / (2.0 \times 10^{- 6} C) = 3.0 \times 10^{3} V

2.3 Potential due to a Point Charge

Now for an actual calculation. Place a point charge $Q$ at the origin and ask for $V$ at a point P a distance $r$ away. The route does not matter, so choose the easiest: straight in along a radius, from infinity to P.

DerivationPotential of a point charge

At a distance $r^{'}$ from $Q$ , a unit positive test charge feels the Coulomb force

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

(2.5)

where ${\hat{𝐫}}^{'}$ points away from $Q$ . Move the test charge by a small radial step $Δ r^{'}$ . On the way in, $r^{'}$ shrinks, so $Δ r^{'}$ is negative. The external agent pushes with $- 𝐅$ , and its work over the step is

Δ W = - \frac{Q}{4 π ε_{0} r^{' 2}} Δ r^{'}

(2.6)

which is positive for $Q > 0$ , as it should be. Adding the steps from $r^{'} = \infty$ to $r^{'} = r$ turns the sum into an integral:

W = - \int_{\infty}^{r} \frac{Q}{4 π ε_{0} r^{' 2}} d r^{'} = {[\frac{Q}{4 π ε_{0} r^{'}}]}_{\infty}^{r} = \frac{Q}{4 π ε_{0} r}

(2.7)

This work per unit charge is, by definition, the potential at P:

Potential of a point charge

V (r) = \frac{1}{4 π ε_{0}} \frac{Q}{r}

(2.8)

Eq. (2.7) also has a neat graphical reading: $V (r)$ is the area under the graph of $E$ against distance, from $r$ all the way out to infinity.

What if $Q$ is negative, like the charge on a balloon rubbed on dry hair? Eq. (2.8) then gives $V < 0$ at every distance. Picture the unit test charge being pulled in by the balloon. To stop it rushing inward, your hand must hold it back all the way, pushing outward while the charge moves inward, and a force that opposes the motion does negative work. The closer the point, the more negative its potential.

Compare how $V$ and $E$ fade with distance in Fig. 2.3. The field falls as $1 / r^{2}$ , the potential only as $1 / r$ . Double the distance and $V$ halves, while $E$ drops to a quarter.

Fig. 2.3Both fade with distance, but the field fades faster: at twice the distance

V

is halved while

E

is quartered.NCERT Fig. 2.4 · p. 49

A positive point charge Q sits at the left end of a distance axis marked r0, 2r0, 3r0 and 4r0. At each distance a pair of bars compares the potential, a solid blue bar, with the field, a hatched orange bar, both scaled to 1 at r0, where a dashed line marks their common height. At 2r0 the potential bar is half as tall but the field bar only a quarter as tall. At 3r0 the bars are one third and one ninth, and at 4r0 one quarter and one sixteenth, so the field bars shrink much faster than the potential bars.

Example 2.1NCERT Example 2.1 · p. 49

A dust grain near a charged sphere

A small metal sphere on an insulating stand carries $+ 6.0 n C$ ; outside it, it acts like a point charge at its centre. A dust grain carrying $- 3.0 n C$ drifts in from far away and ends up at a point P, $12 c m$ from the centre. (a) What is the potential at P? (b) By how much does the grain's potential energy change on the way in? Does the field speed the grain up or slow it down? (c) The grain is then carried slowly once around a circle of radius $12 c m$ centred on the sphere, and finally moved out to $24 c m$ . How much work does the agent do in each of these two stages? Take $1 / 4 π ε_{0} = 9.0 \times 10^{9} N m^{2} C^{- 2}$ .

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

2.4 Potential due to an Electric Dipole

Recall the electric dipole from Chapter 1: a pair of charges $\pm q$ held a small distance $2 a$ apart. It is described by its dipole moment $𝐩$ , of size $2 a q$ , which runs from the negative charge to the positive one. Its field took some effort, because field vectors must be added head to tail. Potentials are plain numbers with signs, so superposition is just addition, and the dipole potential takes only a few lines.

2/3Seen from far away, $r_{2}$ exceeds $r_{1}$ by the projection of the dipole on OP: $r_{2} - r_{1} \approx 2 a \cos θ$ .

Fig. 2.4Far from a dipole, the two distances to P differ by

2 a \cos θ

; the potential is positive on the

+ q

side, negative on the

- q

side and zero on the equatorial plane.NCERT Fig. 2.5 · p. 50

A vertical dipole has +q at the top and -q at the bottom, with centre O and the dipole moment p pointing up. A point P lies up and to the right. In the first step lines r1 and r2 join P to the two charges, a dashed line r joins O to P, and the angle theta is marked between p and OP. In the second step perpendiculars dropped from the two charges onto line OP mark off a highlighted segment of length 2a cos theta: the extra distance to P from -q compared with +q. In the third step the upper half of the plane is tinted and labelled V greater than zero, the lower half V less than zero, and the horizontal equatorial line through O is labelled V equals zero.

DerivationPotential of a dipole far away

Measure positions from O, the midpoint of the dipole. Point P, at position $𝐫$ , is $r_{1}$ from $+ q$ and $r_{2}$ from $- q$ (Fig. 2.4). Adding the two point-charge potentials,

V = \frac{1}{4 π ε_{0}} (\frac{q}{r_{1}} - \frac{q}{r_{2}}) = \frac{q}{4 π ε_{0}} \frac{r_{2} - r_{1}}{r_{1} r_{2}}

(2.9)

The cosine rule gives the two distances exactly:

r_{1}^{2} = r^{2} + a^{2} - 2 a r \cos θ, r_{2}^{2} = r^{2} + a^{2} + 2 a r \cos θ

(2.10)

Far from the dipole ( $r ≫ a$ ) the lines from the two charges to P are almost parallel. Keeping only terms of first order in $a / r$ , each distance differs from $r$ by the projection $a \cos θ$ of half the dipole on the line OP:

r_{1} \approx r - a \cos θ, r_{2} \approx r + a \cos θ

(2.11)

(You can get the same result from Eq. (2.10) with the binomial approximation $\sqrt{1 - x} \approx 1 - x / 2$ for small $x$ .) The difference and the product that we need are therefore

r_{2} - r_{1} \approx 2 a \cos θ, r_{1} r_{2} \approx r^{2}

(2.12)

Substituting into Eq. (2.9) and writing $p = 2 q a$ ,

V = \frac{q}{4 π ε_{0}} \frac{2 a \cos θ}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{p \cos θ}{r^{2}}

(2.14)

Since $p \cos θ = 𝐩 \cdot \hat{𝐫}$ , with $\hat{𝐫}$ the unit vector from O towards P, the result takes a compact vector form:

Potential of a dipole

V = \frac{1}{4 π ε_{0}} \frac{𝐩 \cdot \hat{𝐫}}{r^{2}} (r ≫ a)

(2.15)

How good is the approximation? The terms we dropped are smaller than the one we kept by a factor of about $(a / r)^{2}$ . On the axis, for example, the exact result is $V = k p / (r^{2} - a^{2})$ , so at a distance of ten dipole lengths ( $r = 20 a$ ) the approximate formula is off by only a quarter of a per cent. For an ideal point dipole, shrunk to zero size with $p$ held fixed, it is exact.

Two directions deserve a special look. On the axis, $θ = 0$ or $π$ , so $\cos θ = \pm 1$ :

V = \pm \frac{1}{4 π ε_{0}} \frac{p}{r^{2}}

(2.16)

with the plus sign on the $+ q$ side and the minus sign on the $- q$ side. On the equatorial plane, $θ = π / 2$ and $V = 0$ : every point there is equally far from $+ q$ and $- q$ . Set beside a single charge, the dipole potential stands out in three ways:

It depends on direction. Walk around the dipole at a fixed distance and $V$ changes, from positive off the $+ q$ end, through zero at the side, to negative off the $- q$ end. Only the angle $θ$ from the axis matters, so the pattern is the same on every side of the axis, like the shape of a lampshade around its stand.
It fades faster. The dipole potential falls as $1 / r^{2}$ , compared with $1 / r$ for a single charge, because from far away $+ q$ and $- q$ almost cancel. (The field falls one power faster still, as $1 / r^{3}$ .)
Zero potential does not mean zero field. On the equatorial plane $V = 0$ , yet Chapter 1 gave a field of size $p / 4 π ε_{0} r^{3}$ there, pointing opposite to $𝐩$ . The field depends on how $V$ changes from point to point, not on its value (Section 2.6.1).

Example

Potential near a water molecule

A water molecule has a permanent dipole moment of about $6.2 \times 10^{- 30} C m$ . Treating it as a point dipole, find the potential it produces (a) $1.0 n m$ away along its axis, on the positive side; (b) at the same distance, but $60^{\circ}$ from the axis; (c) $2.0 n m$ away along the axis. Compare (a) and (c).

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

Quick check

1 of 3

At a point on the equatorial plane of an electric dipole, which is correct?

V = 0

and

E = 0

V \neq 0

and

E = 0

V \neq 0

and

E \neq 0

V = 0

but

E \neq 0

Show answer

Answer (D) $V = 0$ but $E \neq 0$

The point is equidistant from

+ q

and

- q

, so their potentials cancel. Their fields do not: both have components pointing from the

+ q

side towards the

- q

side, which add up.

2.5 Potential due to a System of Charges

Superposition makes the general case easy. Each charge in a collection contributes its own point-charge potential at P, and since potentials are signed numbers, the total is their algebraic sum. For charges $q_{1}, q_{2}, \dots, q_{n}$ at distances $r_{1 P}, r_{2 P}, \dots, r_{n P}$ from P,

V = V_{1} + V_{2} + \dots + V_{n}

(2.17)

Potential of a system of charges

V = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r_{1 P}} + \frac{q_{2}}{r_{2 P}} + \dots + \frac{q_{n}}{r_{n P}}) = \frac{1}{4 π ε_{0}} \sum_{i = 1}^{n} \frac{q_{i}}{r_{i P}}

(2.18)

No angles and no components: a positive charge adds to $V$ and a negative one subtracts, whatever direction it lies in. Finding $𝐄$ for the same charges would need every contribution resolved into components first, which is why it often pays to find $V$ first.

Charge that is spread out, through a volume or over a surface, is handled the same way. Split it into tiny pieces, each small enough to count as a point charge $d q$ at its own distance $r$ from P. Add up their contributions $d q / 4 π ε_{0} r$ ; for vanishingly small pieces the sum becomes an integral. Only distances enter, never directions, which is why the potential of a spread-out charge is usually far easier to find than its field.

Chapter 1 has already done the hard work for a thin spherical shell of radius $R$ carrying charge $q$ . Gauss's law showed that its field outside cannot be told apart from that of a point charge $q$ at the centre. Two fields that agree all the way from P out to infinity need the same work per unit charge, so outside the shell

V = \frac{1}{4 π ε_{0}} \frac{q}{r} (r \geq R)

(2.19(a))

Inside the shell the field is zero. A test charge can wander about in there without any work being done on it, so every inside point shares the value that $V$ has on the shell itself:

V = \frac{1}{4 π ε_{0}} \frac{q}{R} (r \leq R)

(2.19(b))

Fig. 2.5Across the surface of a charged shell the field jumps, but the potential is continuous.NCERT p. 52

Two side-by-side graphs share a shaded strip marking the inside of the shell, out to radius R. In the left graph the field is zero inside, jumps at R to kq over R squared and then falls as 1 over r squared. In the right graph the potential is flat at kq over R inside and falls smoothly as 1 over r outside, with no jump at R.

Example 2.2NCERT Example 2.2 · p. 52

Where is the potential zero?

A charge of $+ 4.0 n C$ sits at the origin and a charge of $- 1.0 n C$ sits on the $x$ -axis at $x = 30 c m$ (Fig. 2.6). Taking $V = 0$ at infinity, locate all the places on the $x$ -axis at which $V$ vanishes. Does the electric field vanish at those places too?

Fig. 2.6Along the line of the two charges,

V (x)

crosses zero twice: between them, and beyond the smaller charge.NCERT Fig. 2.7 · p. 52

A horizontal x-axis marked in centimetres carries a +4.0 nC charge at 0 and a -1.0 nC charge at 30 cm. A blue curve shows the potential V along the axis: large and positive near the positive charge, plunging to large negative values close to the negative charge, and rising again beyond it to a small positive hump far to the right. The curve crosses zero at two points, marked with circles at 24 cm and at 40 cm.

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

InteractivePotential probe

An interactive stage with two point charges on a horizontal line, 30 cm apart, whose values you can change in steps of 1 nC. A draggable probe shows the potential in volts and the field strength in newtons per coulomb at its position, with an arrow for the direction of the field and a highlighted curve through all points at the same potential as the probe. Below the stage, a graph plots the potential along the line through the charges and marks where it is zero. Preset buttons load the charges of Example 2.2, a dipole and two equal positive charges.

Fig. 2.7Drag the probe to read

V

and

E

. The highlighted curve is the equipotential through the probe; the graph shows

V

along the line of the charges.

Example 2.3NCERT Example 2.3 · p. 53

An electron and a proton near a charged sphere

A small metal sphere carrying $+ 5.0 n C$ is held fixed at O inside an evacuated chamber. Point N is $3.0 c m$ from O and point F is $9.0 c m$ from O. (a) Which point is at the higher potential, and by how much? (b) An electron is released from rest at F. Which way does it move, does the field do positive or negative work on it, and what is its kinetic energy as it passes N? (c) What happens to the electron's potential energy between F and N? (d) A proton is to be carried slowly from F to N. Must the agent push it or hold it back, and is the agent's work positive or negative? (e) If the proton is then let go at N, does its kinetic energy grow or shrink?

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

Quick check

1 of 3

Charges of

+ 2.0 n C

+ 2.0 n C

and

- 3.0 n C

each sit

10 c m

from a point P, in three different directions. What is the potential at P, in volts? (Take

k = 9.0 \times 10^{9} N m^{2} C^{- 2}

Show answer

Answer 90 V

Directions do not matter for

V

V = k (2.0 + 2.0 - 3.0) n C / 0.10 m = (9.0 \times 10^{9}) (1.0 \times 10^{- 9}) / 0.10 = 90 V

2.6 Equipotential Surfaces

Trekkers in the Western Ghats carry maps with contour lines, each joining points at the same height above sea level. Where the lines crowd together the slope is steep; where they spread apart the ground is gentle. Potential invites exactly the same kind of map.

Definition

Equipotential surface

A surface on which every point has the same electrostatic potential.

ExampleFor a single point charge, any sphere centred on the charge.

For a lone point charge the map is easy to draw. $V = Q / 4 π ε_{0} r$ depends on distance alone, so each equipotential surface is a sphere centred on the charge, and the family of them nests like the layers of an onion. The field lines are radii, so they pierce every layer squarely. That square crossing turns out to be a general law, not a coincidence of this simple case:

PropertyThe field is normal to equipotentials

Field lines always cross equipotential surfaces at right angles, and they run from higher to lower potential.

Suppose the field at some point S of an equipotential surface had a component $E_{∥}$ along the surface (Fig. 2.8). Move a small charge $q$ a short distance $δ l$ along the surface, in the direction of $E_{∥}$ . The field would then do work $q E_{∥} δ l$ on it. But for a conservative field this work must equal the fall in potential energy, $- q δ V$ , and $δ V = 0$ because both ends of the step lie on the same equipotential. So $E_{∥} = 0$ : the field has no component along the surface and is normal to it. It points the way a positive charge would lose potential energy, towards lower $V$ .

3/3So $𝐄$ meets every equipotential at a right angle, pointing towards lower potential.

Fig. 2.8No work is needed to move a charge along an equipotential, so the field can have no component along it.NCERT Fig. 2.9 · p. 54

Three dashed equipotential curves run from top to bottom, labelled 60 V, 40 V and 20 V from left to right, with a small positive test charge at point S on the 40 V curve. In the first step an arrow along the curve shows the charge sliding along it, labelled W equals zero. In the second step a dashed arrow for a hypothetical tilted field at S is split into a part across the curve and a part along it; the part along it would do work. In the third step the field arrow at S crosses the curve at a right angle, marked with a small square, and several other field arrows cross all three curves at right angles, pointing from 60 V towards 20 V.

A uniform field gives the plainest pattern of all. The potential falls steadily in the direction of $𝐄$ and does not change sideways, so its equipotentials are flat, parallel sheets standing across the field, equally spaced for equal steps of $V$ . You will meet exactly this pattern between the plates of a capacitor (Section 2.12). Close to each charge of a dipole, or of a pair of like charges, the equipotentials are small near-spheres; further out they merge into larger closed surfaces. The simulation below shows several of these patterns with their field lines.

Scroll to explore

Equipotentials of some arrangements

Point charges on a stage with their field lines and dashed equipotential curves, drawn at a fixed set of potential values: green for positive potential, orange for negative and grey for zero. As you scroll, the stage shows one positive charge, one negative charge, two rows of opposite charges, a dipole and two equal positive charges. You can drag the charges and switch the field lines, field vectors and equipotentials on or off.

One positive charge. The dashed equipotentials are circles, spheres in three dimensions, centred on the charge; they are green because $V > 0$ . The field lines cross every circle at right angles. Close to the charge the circles crowd together, where the field is strongest.

One negative charge. The same circles, now orange because the potential is negative everywhere. The field lines point inwards, from higher (less negative) to lower (more negative) potential.

Two rows of opposite charges. Midway between the rows the equipotentials straighten into nearly flat, parallel lines, and the field lines run straight across them. This is a rough model of a charged capacitor, whose field between the plates is uniform (Section 2.12).

A dipole. The grey dashed line midway between $+ q$ and $- q$ is the zero-potential surface, the equatorial plane. Positive equipotentials wrap around the $+ q$ end and negative ones around the $- q$ end.

Two equal positive charges. Close to each charge the equipotentials are small circles; further out they merge into one peanut-shaped curve enclosing both. Midway between the charges the two fields cancel, so no field line passes through that point, yet the potential there is positive: two positive contributions add.

Let's explore

Map equipotentials in a tray of salt water

You’ll need

a shallow transparent plastic tray on a sheet of graph paper
salt water
two strips of aluminium foil and a coin as electrodes
a 9 V battery with crocodile-clip leads
a digital multimeter set to DC volts

Predict first

With two long strip electrodes at opposite ends of the tray, what shape will the lines of equal voltage take?

circles around each stripstraight lines joining the two stripsstraight lines parallel to the stripsno regular pattern

Put the graph paper under the transparent tray and pour in salt water about $1 c m$ deep. Stand a foil strip upright along each short end of the tray and connect the strips to the battery.
Clip the multimeter's black lead to the negative strip. Dip the tip of the red lead into the water at any point and note the reading.
Find five points that all read $3.0 V$ , mark them on the graph paper and join them with a smooth line. Repeat for $4.5 V$ and $6.0 V$ .
Replace one strip with a coin standing on its edge (a small, almost point-like electrode) and map the same three voltages again.

Table 2.1Your equipotential map

Equipotential	Shape with two strips	Shape with a strip and a coin
3.0 V
4.5 V
6.0 V

Your entries are saved automatically.

With two strips the equal-voltage lines come out straight, parallel to the electrodes and evenly spaced: the field between them is nearly uniform. With the coin, the lines curve around it and crowd together close to it, where the field is strongest. At every point, the direction in which the reading changes fastest is at right angles to your lines, just as $𝐄$ is at right angles to equipotentials. (The meter drives a tiny current through the water, but the pattern of potentials matches the electrostatic one.)

Can’t do it now? Try the simulation · Fig. 2.11

2.6.1 Relation between field and potential

The contour-map picture already suggests the link between $𝐄$ and $V$ : steep ground where contours crowd, strong field where equipotentials crowd. Fig. 2.10 makes this quantitative.

3/3At A a $10 V$ drop happens over $1 c m$ , so $E \approx 1000 V / m$ . At B it takes $4 c m$ , so $E \approx 250 V / m$ .

Fig. 2.10The field is strongest where equipotentials are closest:

E \approx ∣ Δ V ∣ / Δ l

, measured at right angles to the surfaces.NCERT Fig. 2.12 · p. 55

Curved equipotential lines labelled 50, 40, 30, 20 and 10 volts run from top to bottom, close together on the left and increasingly far apart towards the right; a scale bar shows 1 cm. In the second step blue field arrows cross the lines at right angles, pointing from 50 V towards 10 V; they are longest on the left where the lines crowd and shortest on the right. In the third step two measurement bars mark point A, between the 50 V and 40 V lines 1 cm apart, and point B, between the 20 V and 10 V lines 4 cm apart; a panel reads E at A about 1000 V/m and E at B about 250 V/m.

DerivationField from the spacing of equipotentials

Carry a unit positive charge a short distance $δ l$ in some direction that makes an angle $θ$ with $𝐄$ . The field does work $E δ l \cos θ$ on it, so its potential changes by $δ V = - E δ l \cos θ$ . Two directions stand out.

If the step lies along an equipotential, $δ V = 0$ , so $\cos θ = 0$ : the field is at right angles to the surface, as we found above. If instead the step is taken straight along the field ( $θ = 0$ ), across from one equipotential to the next, the potential falls as fast as it can, and

δ V = - ∣ 𝐄 ∣ δ l \Rightarrow ∣ 𝐄 ∣ = - \frac{δ V}{δ l}

(2.20)

Here $δ l$ is the perpendicular distance between two neighbouring equipotentials and $δ V$ , negative in the direction of the field, is the difference between their potentials. The size of the field is therefore

Field from potential

E = \frac{∣ δ V ∣}{δ l}

(2.21)

Which way: downhill. $𝐄$ points along the steepest descent of $V$ , which always runs straight across the equipotentials.
How strong: as strong as the slope is steep. $E$ is the fall in $V$ per metre, measured straight across the equipotentials, so where equal steps of $V$ crowd together the field is strong.

Along any straight line, say the $x$ -axis, the same reasoning gives the component of the field along that line: $E_{x} = - d V / d x$ . The field component is the downhill slope of the potential. This also explains the unit $V / m$ , which you will see as often as $N / C$ : they are the same, since $1 V / m = 1 J C^{- 1} m^{- 1} = 1 N / C$ .

SimulationField and equipotentials

A dipole with its dashed equipotential curves drawn at a fixed set of potential values: green for positive, orange for negative and grey for zero potential. A small positive test charge can be dragged anywhere; an arrow shows the force on it, and readouts give the field strength and the direction of the force. Buttons switch to a single charge or two like charges, show field lines or field vectors, and release the test charge so that it moves.

Fig. 2.11Dashed curves are equipotentials (green positive, orange negative, grey zero). Drag the test charge and watch the force arrow.NCERT p. 55

Try this

Drag the test charge slowly along one dashed curve. The force arrow stays at right angles to the curve all the way.
Put the test charge on the grey dashed line midway between the charges, where $V = 0$ . Does the field readout drop to zero there?
Drag the test charge a little to the right of $+ q$ and press Release test charge. Which way does it run? Does it cross the dashed curves from green through grey to orange, towards lower potential, or the other way?

Example

Field from equipotential spacing

In a region of uniform field, the equipotential planes at $120 V$ and $70 V$ are $2.0 c m$ apart. (a) Find the size and direction of the field. (b) How much work does the field do on a $+ 5.0 μ C$ charge that moves from the $120 V$ plane to the $70 V$ plane? (c) What is the potential on a plane $0.80 c m$ beyond the $70 V$ plane?

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

2.7 Potential Energy of a System of Charges

So far a fixed set of source charges produced the potential, and we asked about the energy of a visitor. Now turn the question round: how much energy is locked up in an arrangement of charges itself? The answer is the total work needed to build it, bringing each charge in from infinity while those already placed are held still (Fig. 2.12).

3/3 $q_{3}$ feels both $q_{1}$ and $q_{2}$ . The ledger total is the potential energy $U$ of the arrangement.

Fig. 2.12Assembling charges one at a time: each newcomer pays for the potential of those already in place.NCERT Fig. 2.14 · p. 56

On the left, charges arrive one by one at the corners of a triangle; on the right, a ledger records the work for each arrival. In the first step only q1 is present and the ledger reads W1 equals 0. In the second step q2 has arrived at distance r12 from q1 and the ledger adds W2 equals k q1 q2 over r12. In the third step q3 has arrived at the top, at distances r13 and r23, the ledger adds W3 equals k q3 times the sum of q1 over r13 and q2 over r23, and a final line gives U as the sum of the three.

DerivationEnergy of an assembly of point charges

Two charges. Keep a ledger of the work, as in Fig. 2.12. The first charge, $q_{1}$ , arrives in empty space: nothing pushes back, so its entry in the ledger is zero. Its arrival changes the landscape, though, because every point now has the potential $q_{1} / 4 π ε_{0} r$ . The second charge pays its way into that landscape: bringing $q_{2}$ to a point $r_{12}$ from $q_{1}$ costs $q_{2}$ times the potential there. The ledger total is the energy of the pair:

Energy of two point charges

U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}

(2.22)

Three charges. The first two arrivals cost what they did before. Write $𝐫_{1}, 𝐫_{2}, 𝐫_{3}$ for the three positions:

W_{2} = q_{2} V_{1} (𝐫_{2}) = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}

(2.23)

Together, $q_{1}$ and $q_{2}$ now produce at any point P the potential

V_{1, 2} (P) = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r_{1 P}} + \frac{q_{2}}{r_{2 P}})

(2.24)

and the third charge must be pushed in against both:

W_{3} = q_{3} V_{1, 2} (𝐫_{3}) = \frac{1}{4 π ε_{0}} (\frac{q_{1} q_{3}}{r_{13}} + \frac{q_{2} q_{3}}{r_{23}})

(2.25)

Adding the three steps:

Energy of three point charges

U = \frac{1}{4 π ε_{0}} (\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{1} q_{3}}{r_{13}} + \frac{q_{2} q_{3}}{r_{23}})

(2.26)

The pattern is clear: one term for every pair of charges. Four charges give six pair terms, and $n$ charges give $n (n - 1) / 2$ .

Reverse the order, bringing $q_{3}$ first and $q_{1}$ last, and you collect exactly the same three pair terms. For a conservative force this had to happen. $U$ is a property of the finished arrangement, rather like the balance in a bank account, which does not record the order in which the deposits were made.

Each pair term carries the sign of $q_{i} q_{j}$ , and the sign has a meaning. Take a proton and an electron $5.3 \times 10^{- 11} m$ apart, roughly their separation in a hydrogen atom: $U = - e^{2} / 4 π ε_{0} r \approx - 4.3 \times 10^{- 18} J$ . A negative $U$ marks a bound pair: pulling the two apart to infinity would take that much work. (In a real atom the electron's motion also enters the bookkeeping, as you will see in Chapter 12.) Replace the electron by a second proton and the same number comes out positive: now energy is stored, ready to be released as the pair flies apart.

Example 2.4NCERT Example 2.4 · p. 57

Energy of a triangle of charges

Charges of $+ 3.0 n C$ at A, $+ 3.0 n C$ at B and $- 3.0 n C$ at C sit at the corners of an equilateral triangle of side $10 c m$ (Fig. 2.13). (a) How much work was needed to assemble them? (b) With A and B held fixed, the charge at C is moved to M, the midpoint of AB. How much work does this take? Check the answer in two ways.

Fig. 2.13The arrangement for Example 2.4: the charge at C is moved to M.NCERT p. 57

An equilateral triangle of side 10 cm has a +3.0 nC charge at the lower left corner A, a +3.0 nC charge at the lower right corner B and a -3.0 nC charge at the top corner C. The midpoint M of the base AB is marked, and a dashed arrow runs from C straight down to M.

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

2.8 Potential Energy in an External Field

So far the charges whose energy we counted were also the ones making the field. Often the two roles are split. Think of an electron in the beam of an electron microscope, or a smoke particle drifting between the plates of an electrostatic precipitator in a chimney. Each moves through a field set up by electrodes that we never need to describe in detail: all we are given is the potential $V (𝐫)$ they produce. The one assumption is that those sources stay as they are while our charge moves among them.

2.8.1 Potential energy of a single charge

Since $V (𝐫)$ already measures the work per unit charge for the trip in from infinity, a charge $q$ placed at $𝐫$ simply needs $q$ times as much. Its potential energy in the external field is

Energy of a charge in an external potential

U = q V (𝐫)

(2.27)

Eq. (2.27) leads to a handy unit of energy for atomic and nuclear physics. An electron or a proton that moves through a potential difference of $1 V$ changes its energy by $e \times 1 V = 1.6 \times 10^{- 19} J$ . This amount is one electron volt.

Definition

Electron volt (eV)

The energy gained or lost by a particle of charge $e$ when it moves through a potential difference of one volt: $1 e V = 1.6 \times 10^{- 19} J$ .

ExampleA photon of visible light carries about 2 to 3 eV.

Table 2.2Multiples of the electron volt

Unit	In eV	In joules	Where you meet it
keV	$10^{3}$	$1.6 \times 10^{- 16}$	dental and medical X-rays (tens of keV)
MeV	$10^{6}$	$1.6 \times 10^{- 13}$	energy released in single nuclear reactions
GeV	$10^{9}$	$1.6 \times 10^{- 10}$	the rest energy of a proton (about 0.94 GeV)
TeV	$10^{12}$	$1.6 \times 10^{- 7}$	proton beams in the Large Hadron Collider

Example

An electron gun

In an electron gun, such as the one in an electron microscope, an electron starts from rest and is accelerated through a potential difference of $2.5 k V$ . Find its kinetic energy in electron volts and in joules, and its speed. ( $m_{e} = 9.1 \times 10^{- 31} k g$ , $e = 1.6 \times 10^{- 19} C$ )

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

2.8.2 Potential energy of a system of two charges in an external field

Now let two charges, $q_{1}$ at $𝐫_{1}$ and $q_{2}$ at $𝐫_{2}$ , sit in the same external field, and build the arrangement one charge at a time. Placing $q_{1}$ costs $q_{1} V (𝐫_{1})$ , as for any single charge. By the time $q_{2}$ arrives, $q_{1}$ is already there, so $q_{2}$ has to be pushed against the field of $q_{1}$ as well as the external one. Fields superpose, and so the two contributions to the work simply add:

W_{2} = q_{2} V (𝐫_{2}) + \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}

(2.28)

The total is the potential energy of the pair in the field:

Two charges in an external field

U = q_{1} V (𝐫_{1}) + q_{2} V (𝐫_{2}) + \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}

(2.29)

Read the three terms as two kinds of energy: each charge's energy in the external field, plus the mutual energy of the pair from Section 2.7.

Example 2.5NCERT Example 2.5 · p. 59

Two beads on a rod in a uniform field

Two small beads are fixed to the ends of an insulating rod $10 c m$ long. Bead 1 carries $q_{1} = + 2.0 μ C$ and bead 2 carries $q_{2} = - 3.0 μ C$ . (a) What is the mutual potential energy of the beads? (b) The rod is laid along a uniform external field of $4.0 \times 10^{5} N / C$ pointing along $+ x$ , with bead 1 at the origin and bead 2 at $x = + 10 c m$ . Taking the external potential to be zero at the origin, so that $V (x) = - E_{0} x$ , find the total electrostatic energy of the pair. (c) The rod is then turned end for end about its midpoint, so that the beads swap places. How much work does this take?

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

2.8.3 Potential energy of a dipole in an external field

A dipole in a uniform field feels no net force but a torque $𝝉 = 𝐩 \times 𝐄$ (Chapter 1), which tries to swing $𝐩$ round into line with $𝐄$ . Turning it the other way takes work, and that work is stored as potential energy that depends on the angle $θ$ between $𝐩$ and $𝐄$ (Fig. 2.14).

2/4Turned through $60^{\circ}$ : the couple tries to swing it back; $U = - p E \cos 60^{\circ} = - \frac{1}{2} p E$ .

Fig. 2.14

U (θ) = - p E \cos θ

: lowest when

𝐩

is along

𝐄

, highest when it points the opposite way.NCERT Fig. 2.16 · p. 60

On the left, horizontal field lines point to the right and a dipole sits in the middle with a force arrow on each charge. On the right, a graph shows U(theta) rising smoothly from minus pE at 0 degrees, through zero at 90 degrees, to plus pE at 180 degrees, with a dot marking the current state. In the first step the dipole points along the field, the forces act along one line and the dot sits at the minimum, labelled stable. In the second step the dipole is turned to 60 degrees and the dot has climbed to minus one half pE. In the third step the dipole is at right angles to the field and the dot is at zero. In the fourth step the dipole points against the field and the dot is at the maximum, labelled unstable.

DerivationEnergy of a dipole at angle

θ

The electric torque on the dipole is

Torque on a dipole

𝝉 = 𝐩 \times 𝐄

(2.30)

with size $p E \sin θ$ ; it always acts to shrink $θ$ . Suppose you grip the dipole and open the angle from $θ_{0}$ to $θ_{1}$ at a crawl, never letting it spin. At each angle your twist cancels the electric one, and over a small turn $d θ$ you do work $p E \sin θ d θ$ . Summing the small turns,

W = \int_{θ_{0}}^{θ_{1}} p E \sin θ d θ = p E (\cos θ_{0} - \cos θ_{1})

(2.31)

That work becomes potential energy, so $U (θ_{1}) - U (θ_{0}) = p E (\cos θ_{0} - \cos θ_{1})$ . As always, the formula fixes differences only. If we agree that $U = 0$ when the dipole stands across the field ( $θ_{0} = 90^{\circ}$ ), the starting angle drops out and the result is compact:

Energy of a dipole in a uniform field

U (θ) = p E (\cos \frac{π}{2} - \cos θ) = - p E \cos θ = - 𝐩 \cdot 𝐄

(2.32)

The same result from the charges. Treat the dipole as two separate charges and apply Eq. (2.29). The mutual term, $- q^{2} / (4 π ε_{0} \cdot 2 a)$ , belongs to the rigid dipole and never changes as it turns:

U^{'} (θ) = q [V (𝐫_{1}) - V (𝐫_{2})] - \frac{q^{2}}{4 π ε_{0} \cdot 2 a}

(2.33)

Here $𝐫_{1}$ locates $+ q$ and $𝐫_{2}$ locates $- q$ . A uniform field lowers the potential by $E$ volts for each metre travelled along it, and the $+ q$ end is $2 a \cos θ$ further 'downstream' than the $- q$ end, so $V (𝐫_{1}) - V (𝐫_{2}) = - E (2 a \cos θ)$ and

U^{'} (θ) = - p E \cos θ - \frac{q^{2}}{4 π ε_{0} \cdot 2 a}

(2.34)

Apart from the fixed mutual term, which we are free to leave out, this is Eq. (2.32) again. It also shows why $90^{\circ}$ is the natural zero. A dipole lying across the field has both ends on one equipotential, so the field gives one end exactly as much energy as it takes from the other.

The energy is lowest, $- p E$ , when $𝐩$ points along $𝐄$ : a stable equilibrium. It is highest, $+ p E$ , when $𝐩$ points against $𝐄$ : an unstable equilibrium. Turning a dipole from aligned to reversed costs $2 p E$ .

Example 2.6NCERT Example 2.6 · p. 61

A dipole that swings and settles

A light insulating rod $2.0 c m$ long carries charges of $+ 5.0 n C$ and $- 5.0 n C$ at its two ends and can turn freely about its centre in a uniform field of $3.0 \times 10^{4} N / C$ . It is held at $120^{\circ}$ to the field, pointing partly against it, and released. (a) Find its dipole moment and its potential energy at $120^{\circ}$ . (b) How much kinetic energy does it have as it swings through the direction of the field? (c) Air resistance eventually brings it to rest along the field. How much heat has been produced? (d) How much work would it then take to turn it round until it points exactly against the field?

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

Quick check

1 of 4

A proton starts from rest and is accelerated through

500 V

. What is its kinetic energy, in joules? (

e = 1.6 \times 10^{- 19} C

)

Show answer

Answer 8e-17 J

K = e Δ V = (1.6 \times 10^{- 19}) (500) J = 8.0 \times 10^{- 17} J

, which is simply

500 e V

2.9 Electrostatics of Conductors

Touch a charged metal sphere to an uncharged one and the charge spreads over both almost at once. The reason is that in a metal a few electrons from every atom are not tied to any one atom: they roam through the whole crystal, while the positive ions stay locked in place. What do these roaming electrons do once the metal has been left alone and all motion has stopped? The answer gives metals some remarkable electrical properties.

Fig. 2.15 follows a metal sphere dropped into a uniform field. The field drags free electrons towards one side, leaving the other side short of them. The charges piling up on the two sides make a field of their own, which points the opposite way inside the metal. The drift goes on only as long as the two fields fail to cancel. In a good conductor the whole adjustment is over in far less than a microsecond, and six results follow from the settled state.

3/3The two fields cancel inside the metal. Outside, the field lines now meet the surface at right angles.

Fig. 2.15Free electrons shift until the field of the induced charges cancels the applied field inside; outside, the lines meet the surface at right angles.NCERT Fig. 2.20 · p. 65

A metal sphere sits in a uniform field whose lines run from left to right. In the first step the straight field lines pass right through the sphere and free electrons inside have arrows pointing left, against the field. In the second step minus signs have gathered on the left side of the sphere and plus signs on the right side, and dashed orange arrows inside show the field of these induced charges pointing left, opposing the applied field. In the third step the field lines outside bend so that they end on the minus charges and start again from the plus charges, meeting the surface at right angles, and the inside is labelled E equals 0.

PropertyResult 1: No field inside a conductor

In electrostatic equilibrium the electric field is zero at every point within the material of a conductor, whether or not the conductor carries a net charge and whatever field surrounds it.

PropertyResult 2: Field lines meet the surface at right angles

Wherever field lines leave or enter a conductor in equilibrium, they meet its surface at right angles.

Split the field just outside the surface into a part along the surface and a part straight out of it. The free electrons at the surface can slide sideways as easily as they move inside, so any sideways part would push them along until it vanished. The perpendicular part only presses the surface charges into the metal or tugs them outwards, and since they cannot escape from the metal it moves nothing along the surface. So in equilibrium only the perpendicular part survives.

PropertyResult 3: Excess charge sits on the surface

Inside the material of a conductor in equilibrium there is no net charge at all; any extra charge given to it ends up on its surface.

Suppose some small region deep inside the metal did hold a net charge. Surround it with a tiny closed surface lying wholly within the metal. By Result 1 the field is zero at every point of that surface, so the flux through it is zero, and Gauss's law then says it encloses no net charge, contradicting the supposition. Only at the surface of the metal, where no Gaussian surface can be drawn entirely inside the material, can excess charge settle.

PropertyResult 4: The whole conductor is at one potential

A conductor in equilibrium has a single potential: every point inside it and on its surface is at the same value of $V$ .

Carry a test charge from any point of the metal to any other, travelling through the interior or along the surface. Inside, Result 1 says there is no force on it; on the surface, Result 2 says the force is perpendicular to the path. Either way no work is done, so the two points share one potential. On a map of $V$ , a conductor is a flat plateau, however irregular its shape, and $V$ changes only once you step off it into the field outside. Two separate conductors may sit on plateaus at different heights.

DerivationResult 5: The field just outside a charged conductor

How strong is the field just outside? It depends only on how much charge sits on the nearby patch of surface, measured by the surface charge density $σ$ :

Field just outside a conductor

𝐄 = \frac{σ}{ε_{0}} \hat{𝐧}

(2.35)

Here $\hat{𝐧}$ is the outward unit normal. The proof uses a Gaussian surface shaped like a flat coin standing across the surface, called a 'pillbox' (Fig. 2.16). One face, of area $δ S$ , lies just outside the metal and the other just inside, and the rim is so short that it hardly pokes out.

Fig. 2.16Only the outer face of the pillbox carries flux.NCERT Fig. 2.17 · p. 63

A zoomed-in patch of a conductor surface is drawn as a horizontal line with plus signs along it, with the metal shaded below. Field arrows point straight up out of the surface. A dashed cylinder crosses the surface: its top face of area delta S is just outside, where a field arrow passes through it, and its bottom face of area delta S is inside the metal, where there is no field.

Count the flux face by face. The face inside the metal gets none, because $𝐄 = 0$ there. The short rim gets none either: its part inside the metal sees no field, and its part outside sees a field running parallel to it (Result 2). That leaves the outer face, where the field is normal and effectively uniform over the tiny area, so the flux is $E δ S$ for a positively charged patch. The pillbox encloses the charge $σ δ S$ , and Gauss's law gives

E δ S = \frac{σ δ S}{ε_{0}} \Rightarrow E = \frac{σ}{ε_{0}}

(2.36)

Adding the direction from Result 2 gives Eq. (2.35), and it works for either sign: where $σ < 0$ the vector points into the metal. Notice the factor. An isolated charged sheet produces only $σ / 2 ε_{0}$ on each side. On a conductor, the rest of the charge arranges itself to cancel the patch's field inside the metal, and in doing so it doubles the field just outside.

PropertyResult 6: Electrostatic shielding

An empty cavity inside a conductor is always free of electric field. No choice of cavity shape, charge on the conductor or surrounding field can change this, and the cavity wall itself carries no charge.

Result 4 makes the cavity wall a single equipotential. Now suppose some field did exist in the empty cavity. A field line must begin and end on charges, and the cavity holds none, so the line would have to leave the wall and return to it. But the potential falls steadily along a field line, so its two ends would be at different potentials on a wall that has only one. That is impossible, so the cavity has no field, and a pillbox on the wall then shows that the wall has no charge. (For a spherical shell, Chapter 1 reached the same result by symmetry; this argument needs no symmetry at all.)

1/3Field lines end on induced charges on the outer surface; the cavity inside stays field-free.

Fig. 2.17A closed conductor shields its cavity from outside fields, but a charge placed inside the cavity is still felt outside.NCERT Fig. 2.18 · p. 64

A square metal box with a square cavity sits in the middle. In the first step horizontal field lines from the left end on minus charges induced on the left outer wall and start again from plus charges on the right outer wall; the empty cavity is labelled E equals 0. In the second step extra plus charges are added along the top and bottom outer faces as well, and the cavity is still labelled E equals 0. In the third step the outside field is gone and a positive charge sits in the cavity: field lines run from it to minus charges on the cavity wall, plus charges appear on the outer surface, and field lines spread outward from the box in all directions.

This is electrostatic shielding: a closed metal enclosure keeps outside electric fields away from whatever it contains. Sensitive instruments sit in metal cases, the signal wire of a TV or network cable is wrapped in a metal braid, and the metal body of a car or an aircraft protects the people inside when lightning strikes. The protection works one way only, though, as the last step of Fig. 2.17 shows: a charge placed inside the cavity makes its presence felt outside.

Try thisNCERT p. 64

Silence a phone with a steel tiffin box

You’ll need

a mobile phone
a second phone to call it
a steel tiffin box or biscuit tin with a tight metal lid
a cardboard box of similar size

Predict first

Will the phone inside the closed metal tin ring when you call it?

Only if the tin is connected to earthYes; a thin tin makes no differenceUsually not, or only weakly

Check that the phone shows good signal. Put it in the cardboard box, close the lid and call it from the second phone.
Now put it in the metal tin, press the lid firmly shut and call again.
Open the lid a little and call once more.

Through cardboard the call connects normally. In a well-closed metal tin the call usually fails or the signal bars drop sharply. The phone's signal is a rapidly changing electric (and magnetic) field; the free electrons in the metal rearrange to cancel it inside, just as they cancel a static field. A gap in the lid lets the field leak in. A microwave oven uses the same idea the other way round: its metal walls and the mesh in its door keep the microwaves inside.

Example 2.7NCERT Example 2.7 · p. 64

Static electricity in daily life

Explain each of the following. (a) A plastic ruler rubbed with a dry cloth makes a thin stream of water from a tap bend towards it, whichever sign of charge the ruler has picked up. (b) Technicians who assemble phone circuit boards wear a wrist strap joined by a wire to earth, and work on a conducting mat that is earthed too. (c) A person sitting inside a car is relatively safe if lightning strikes the car. (d) On a dry winter day you walk across a nylon carpet and reach for a metal door handle: a spark jumps to your finger. Touch the handle again at once and nothing happens.

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

(a) The charged ruler's field acts on the nearby water. Tap water conducts a little and the tap is connected to the plumbing, so charge of the opposite sign flows into the part of the stream nearest the ruler; the polar water molecules also turn partly towards the field. The opposite charge ends up closer to the ruler than any charge of the same sign, and the field is stronger closer in, so the attraction wins and the stream bends. Reverse the ruler's charge and every induced charge reverses too, so the stream still bends towards it.
(b) Walking and moving about can charge a body by friction to thousands of volts. If a charged finger touched a chip, the charge would rush through it in a tiny spark, and the insulating layers inside modern chips, only nanometres thick, break down easily. The strap and the mat hold the technician, the bench and the board at one common potential (earth), so there is no potential difference to drive a spark. (The strap contains a large resistor, so it cannot carry a dangerous current.)
(d) Rubbing against the carpet leaves your body charged and at a potential far from that of the handle. When your finger comes close, the field in the small air gap exceeds the dielectric strength of air and charge jumps across until you and the handle reach the same potential (Result 4 for the joined conductors). At the second touch there is no potential difference left, so no charge flows and there is no spark. Charge flows only between points at different potentials.

2.10 Dielectrics and Polarisation

A dielectric is an insulator: glass, mica, rubber, plastic, dry paper, pure water. Its charges are bound to their atoms and molecules and cannot wander about. Even so, an electric field does not pass through a dielectric unchanged. Within each molecule the positive and negative charges can shift a little, or the whole molecule can turn, and the combined effect weakens the field inside.

Compare

A conductor and a dielectric in the same field

Conductor

Free electrons drift through the whole piece
Induced charges appear on its surfaces
Their field exactly cancels the applied field inside
Inside: $E = 0$

Dielectric

Bound charges shift only within each molecule
Bound (polarisation) charges appear on its surfaces
Their field only partly cancels the applied field
Inside: $E$ reduced, but not zero

How a molecule responds depends on its shape. In water ( $H_{2} O$ ), ammonia ( $N H_{3}$ ) or hydrogen chloride ( $H C l$ ), one atom pulls the shared electrons harder than its partners, and the shape is lopsided. One end of the molecule is therefore permanently a little negative and the other a little positive. These are polar molecules, each with a built-in dipole moment. In a symmetric molecule such as $N_{2}$ , $C O_{2}$ or $C H_{4}$ the pulls cancel and the centres of positive and negative charge coincide. Such a molecule has no dipole moment of its own: it is non-polar.

2/4Each electron cloud shifts against the field (an induced dipole); the polar molecules turn partly towards the field.

Fig. 2.18Non-polar molecules become dipoles and polar ones partly line up; in a slab this leaves bound charge on the two faces, which weakens the field inside.NCERT Fig. 2.22 · p. 66

On the left are a row of non-polar atoms, each a small positive nucleus inside a round electron cloud, and a row of polar molecules drawn as ovals with an arrow for their dipole moment. In the first step there is no field: each nucleus sits at the centre of its cloud and the polar molecules point in random directions. In the second step a field points to the right: every electron cloud has shifted to the left of its nucleus and the polar molecules have turned to point mostly to the right. In the third step a slab on the right is filled with aligned dipoles, each plus end next to the minus end of its neighbour. In the fourth step only the two faces of the slab are highlighted, minus charges on the left face and plus charges on the right face, and arrows inside show the applied field, the opposing field of the face charges and the smaller net field.

Start with the non-polar case, the top row of Fig. 2.18. An applied field tugs the nucleus of each atom one way and its electron cloud the other. The cloud does not fly off, because the attraction of its own nucleus holds it, so it settles slightly off-centre. Each atom now carries a small induced dipole along the field, and in most materials a field twice as strong produces a shift twice as large. In this chapter we deal only with such linear isotropic dielectrics, which respond in proportion to the field and in the same way whatever its direction.

Polar molecules carry dipoles already, but in a sample with no field thermal jostling scrambles their directions, and on average they cancel. Picture a tray of compass needles on a table that is being shaken while a magnet is held nearby. The magnet tries to line the needles up, the shaking keeps undoing its work, and the result is only a partial, average alignment. A field acts on polar molecules in the same way, since a dipole along $𝐄$ has lower energy, $- 𝐩 \cdot 𝐄$ (Section 2.8.3). The colder the sample and the stronger the field, the better the alignment.

Induced or aligned, the molecular dipoles now lean the same way, so every small volume of the material has a net dipole moment. The material has become polarised, and we measure how strongly by its polarisation:

Definition

Polarisation (

𝐏

)

The net dipole moment per unit volume of a dielectric. Its SI unit is $C m^{- 2}$ .

ExampleFor a linear isotropic dielectric, $𝐏$ points along the field and is proportional to it.

Linear isotropic dielectric

𝐏 = ε_{0} χ_{e} 𝐄

(2.37)

where

$χ_{e}$: the electric susceptibility of the material, a dimensionless constant
$𝐄$: the field inside the dielectric

A large $χ_{e}$ means a material that polarises readily. For our purposes it is simply a measured property of each material, like its density.

What does all this do to the field? Put a slab of dielectric in a uniform field $𝐄_{0}$ that meets two opposite faces squarely (Fig. 2.18, last two steps), and follow a single row of molecules from one face to the other. It reads $- + - + - + \dots$ : wherever two neighbours meet, a $+$ end sits against a $-$ end and, seen from outside, the pair looks neutral. The only unmatched charges are at the two ends of the row, a $-$ on the face where the field enters and a $+$ on the face where it leaves.

Every row ends the same way, so the slab behaves like two thin sheets of charge, $- σ_{p}$ on one face and $+ σ_{p}$ on the other, with a neutral interior. Between those sheets their own field runs from $+$ to $-$ , backwards compared with $𝐄_{0}$ , so the field inside the material is weaker than the field around it. This surface charge is bound charge: it belongs to the molecules and, unlike the free charge on a metal, cannot be led away by a wire.

2.11 Capacitors and Capacitance

Look at the circuit board of a phone or the inside of an old radio and you will find dozens of small components, many of them capacitors. Strip away the packaging and every capacitor is the same thing: a pair of conductors kept apart by an insulator (Fig. 2.19). Connect the two conductors to the terminals of a battery and it moves charge from one to the other until they hold equal and opposite charges, $+ Q$ and $- Q$ , at a potential difference $V$ . When we speak of 'a capacitor charged to $50 μ C$ ' we mean the charge on its positive conductor; the capacitor as a whole still has zero net charge.

Fig. 2.19Any pair of conductors with an insulator between them makes a capacitor. The symbols for a fixed and a variable capacitor are on the right.NCERT Fig. 2.24 · p. 67

On the left, two irregular conductors face each other across a shaded insulating gap. The left one carries plus charges on its facing side and is labelled plus Q; the right one carries minus charges and is labelled minus Q. Field lines run across the gap from the positive to the negative conductor. On the right are two circuit symbols: a fixed capacitor, drawn as two short parallel lines, and a variable capacitor, the same symbol crossed by a diagonal arrow.

Charge a capacitor from a $3 V$ cell, then from $6 V$ and from $9 V$ , and measure the charge each time: the results come out in the ratio $1 : 2 : 3$ . Superposition explains why. Any pattern of charge on the conductors, scaled up by some factor, produces a field scaled by the same factor at every point, and so a potential difference scaled by that factor too. $Q$ and $V$ therefore always change in step, and their ratio depends only on the conductors themselves:

Capacitance

C = \frac{Q}{V}

(2.38)

Definition

Capacitance (

C

)

The ratio of the charge on either conductor of a capacitor to the potential difference between them, $C = Q / V$ . Its SI unit is the farad: $1 F = 1 C / V$ .

ExampleA capacitor that holds $\pm 5 μ C$ at $10 V$ has $C = 0.5 μ F$ .

Like the volume of a bucket, capacitance is fixed when the device is made. Larger conductors placed closer together have a larger $C$ , and Section 2.13 shows that the filling between them matters too; the amount of charge you happen to put on does not. A farad is enormous: practical capacitors are rated in microfarads ( $1 μ F = 10^{- 6} F$ ), nanofarads ( $10^{- 9} F$ ) and picofarads ( $10^{- 12} F$ ). A lone metal sphere also has a capacitance, found by imagining its partner conductor spread out at infinity; the example below works one out.

Think of a capacitor as a tank for charge, with $V$ playing the part of the pressure. A wide tank (large $C$ ) holds a lot at low pressure; squeezing the same charge into a small $C$ needs a high $V$ , and a high $V$ means a strong field in the insulator. Past a certain field the insulator fails. In air, stray ions are flung about fast enough to knock more electrons free, an avalanche of charge carriers forms, and a spark jumps the gap, dumping the stored charge. The largest field an insulator can take is its dielectric strength.

Definition

Dielectric strength

The largest electric field an insulating material can withstand before it breaks down and starts to conduct. For dry air it is about $3 \times 10^{6} V / m$ .

ExampleA static spark that jumps $5 m m$ from a finger to a door handle needs roughly $15 k V$ .

That is why a capacitor's case shows a maximum working voltage as well as its capacitance, for example ' $470 μ F$ , $25 V$ '. If a design needs more stored charge, the safe way to get it is a larger $C$ , since raising $V$ soon runs into breakdown. Capacitors also have a second job beyond storage: in alternating-current circuits (Chapter 7) they block steady currents while letting changing ones through, which is how they filter and shape signals.

Example

How much charge can a sphere hold in air?

A metal sphere of radius $15 c m$ stands on an insulating support in dry air. (a) What is the largest charge it can hold before the air next to it breaks down? (b) What is its potential then? (c) What is its capacitance, treating the second conductor as being at infinity?

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

Quick check

1 of 3

A capacitor stores

\pm 24 μ C

when connected to

12 V

. What is its capacitance, in µF?

µF

Show answer

Answer 2 µF

C = Q / V = 24 μ C / 12 V = 2.0 μ F

2.12 The Parallel Plate Capacitor

The simplest capacitor to analyse is two flat metal plates of area $A$ facing each other across a narrow gap $d$ (Fig. 2.20). Take the gap to be empty (air behaves almost the same) and put $+ Q$ on one plate and $- Q$ on the other, so the charge densities are $\pm σ$ with $σ = Q / A$ . If the plates are wide compared with the gap ( $d^{2} ≪ A$ ), a point between them sees each plate as an endless sheet. Chapter 1 gave the field of such a sheet: $σ / 2 ε_{0}$ on either side, pointing away from positive charge and towards negative.

3/4Outside the plates the two fields cancel; between them they add to $σ / ε_{0}$ .

Fig. 2.20Outside the plates the two sheet fields cancel; between them they add to

σ / ε_{0}

, a uniform field except near the edges.NCERT Fig. 2.25 · p. 68

Two vertical plates face each other: the left one is positive, the right one negative. In the first step the negative plate is drawn faintly and blue arrows show the field of the positive plate alone, pointing away from it on both sides with size sigma over 2 epsilon nought everywhere. In the second step dashed arrows add the field of the negative plate, pointing towards it on both sides. In the third step only the sum is drawn: thick arrows between the plates, of size sigma over epsilon nought, and the regions outside labelled E equals 0. In the fourth step curved lines bulge outward at the top and bottom edges of the plates, and dashed vertical lines between the plates mark flat equipotential planes.

DerivationField and capacitance of parallel plates

Call the region beyond the positive plate I and the region beyond the negative plate II. In each outer region the two sheet fields point in opposite directions and cancel:

E_{I} = \frac{σ}{2 ε_{0}} - \frac{σ}{2 ε_{0}} = 0

(2.39)

E_{II} = \frac{σ}{2 ε_{0}} - \frac{σ}{2 ε_{0}} = 0

(2.40)

Between the plates both fields point from the positive plate towards the negative one, so they add:

E = \frac{σ}{2 ε_{0}} + \frac{σ}{2 ε_{0}} = \frac{σ}{ε_{0}} = \frac{Q}{ε_{0} A}

(2.41)

(Result 5 of Section 2.9 gives the same value directly: just outside the inner face of the positive plate the field is $σ / ε_{0}$ .) Because $E$ is the same all the way across, the potential falls by equal amounts for equal steps, and across the whole gap it falls by

V = E d = \frac{Q d}{ε_{0} A}

(2.42)

and the capacitance is

Parallel plate capacitor

C = \frac{Q}{V} = \frac{ε_{0} A}{d}

(2.43)

Notice that neither $Q$ nor $V$ survives in the answer: only the plate area and the gap. Larger plates hold more charge at the same voltage, and a narrower gap needs less voltage for the same charge.

Real plates are not endless. Near their rims some field lines bow outwards into the space beyond (the last step of Fig. 2.20), an effect called fringing, and the charge is slightly denser near the edges than in the middle. How much does this matter? The disturbed region is only about one gap-width wide, so for plates many times wider than the gap almost all of the space between them has the uniform field of Eq. (2.41).

Some numbers show how big a farad is. Two metal sheets the size of an A4 page ( $0.21 m \times 0.297 m$ ), held $0.10 m m$ apart in air, give

C = \frac{(8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}) (0.0624 m^{2})}{1.0 \times 10^{- 4} m} \approx 5.5 \times 10^{- 9} F = 5.5 n F

Turn the question round: how thin would the gap have to be for plates as large as a cricket field, about $150 m \times 150 m$ , to give $1 F$ ? The answer, $d = ε_{0} A / C \approx 2 \times 10^{- 7} m$ , is a fifth of a micrometre, far thinner than any sheet you could hold apart. (To check the units, note that $1 F = 1 C / V = 1 C^{2} / J = 1 C^{2} N^{- 1} m^{- 1}$ , so $ε_{0} A / C$ comes out in metres.) Supercapacitors reach such values only by using insulating layers a few molecules thick on electrodes with an enormous internal surface (see the box in Section 2.15).

Example

A home-made parallel plate capacitor

Two square aluminium plates, $20 c m$ on a side, are held $1.5 m m$ apart in air and connected to a $9.0 V$ battery. Find (a) the capacitance, (b) the charge on each plate, (c) the field between the plates and (d) the surface charge density on the plates.

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

Let's explore

A kitchen-foil capacitor

You’ll need

two sheets of kitchen aluminium foil, about 15 cm square
a clear plastic file sleeve
a few heavy books
a digital multimeter with a capacitance range

Predict first

If you double the thickness of plastic between the foils, the measured capacitance will…

doublestay the sameroughly halvefall to zero

Lay one foil sheet flat, cover it with the plastic sleeve, and place the second foil on top so that the foils overlap but do not touch anywhere.
Press the stack flat under the books. Touch the meter leads to the two foils and note the capacitance.
Slide the top foil so that only half its area overlaps the bottom one, press it flat again and measure.
Restore the full overlap, but put two layers of plastic between the foils, and measure once more.

Table 2.3Your measurements

Arrangement	C measured (nF)	Ratio to the first
Full overlap, one layer
Half overlap, one layer
Full overlap, two layers

Your entries are saved automatically.

Quick check

1 of 4

A parallel plate capacitor has

C = 40 p F

. Its plate separation is doubled and its plate area tripled. What is the new capacitance, in pF?

Show answer

Answer 60 pF

C \propto A / d

40 p F \times 3 / 2 = 60 p F

2.13 Effect of Dielectric on Capacitance

Fill the gap of a capacitor with a dielectric such as mica or glass and its capacitance rises, often several times over. The reason is the polarisation of Section 2.10 (Fig. 2.21).

3/3Most field lines end on the bound charges, so fewer cross the dielectric: $E = E_{0} / K$ . For the same $Q$ , $V$ falls by $K$ and $C$ rises by $K$ .

Fig. 2.21Bound charges on the faces of the dielectric cancel part of the field of the free charges, so for the same

Q

the voltage falls and the capacitance rises.NCERT p. 70

Two vertical plates face each other; plus charges line the inner face of the left plate and minus charges the inner face of the right plate. In the first step eight field lines cross the empty gap from plate to plate. In the second step a slab slides up into the gap; minus signs line its left face next to the positive plate, plus signs line its right face, small dipoles fill its interior, and the eight lines of the applied field still show faintly through it. In the third step all eight lines cross the thin air spaces next to the plates, but only two continue through the slab; the others end on the minus face and start again from the plus face, and a label reads E equals E0 over K, V equals V0 over K and C equals K C0.

DerivationCapacitance with a dielectric

Keep the free charge on the plates fixed at $\pm Q$ throughout, so $σ = Q / A$ never changes. With nothing in the gap, Section 2.12 gives

E_{0} = \frac{σ}{ε_{0}}, V_{0} = E_{0} d, C_{0} = \frac{Q}{V_{0}} = \frac{ε_{0} A}{d}

(2.46)

Now slide in a slab that fills the gap. Next to the positive plate lies the slab's $- σ_{p}$ face, and next to the negative plate its $+ σ_{p}$ face (Fig. 2.21). Seen from inside the slab, each plate and the face beside it act together like a single sheet of density $σ - σ_{p}$ , so in the material

E = \frac{σ - σ_{p}}{ε_{0}}

(2.47)

V = E d = \frac{(σ - σ_{p}) d}{ε_{0}}

(2.48)

Double the free charge and, in a linear dielectric, the polarisation doubles as well, so $σ_{p}$ doubles too. The uncancelled part $σ - σ_{p}$ is therefore always the same fraction of $σ$ for a given material. Calling that fraction $1 / K$ ,

σ - σ_{p} = \frac{σ}{K}

(2.49)

Bound charge can weaken the effect of the free charge but never overturn it, so $0 < σ_{p} < σ$ and $K > 1$ . The potential difference and the capacitance become

V = \frac{σ d}{ε_{0} K} = \frac{Q d}{ε_{0} K A}

(2.50)

C = \frac{Q}{V} = \frac{K ε_{0} A}{d}

(2.51)

It is convenient to fold $K$ into a constant of the material. The permittivity of a medium is $ε = ε_{0} K$ , and the pure number $K$ , the ratio of the two permittivities, is its dielectric constant or relative permittivity:

ε = ε_{0} K

(2.52)

K = \frac{ε}{ε_{0}}

(2.53)

Empty space is the reference, with $K = 1$ exactly and $ε = ε_{0}$ . Putting Eqs. (2.46) and (2.51) side by side gives the result in its simplest form:

Effect of a dielectric

C = K C_{0}

(2.54)

Definition

Dielectric constant (

K

)

The factor by which the capacitance of a capacitor increases when the space between its conductors is completely filled with the material: $K = C / C_{0} = ε / ε_{0}$ . It is a pure number greater than 1.

ExampleMica has $K \approx 6$ , pure water about 80.

Eq. (2.54) is more general than our flat-plate derivation suggests: completely filling the space around any pair of conductors with a material multiplies their capacitance by the same $K$ . That makes a capacitor a handy measuring instrument, and the values in Table 2.4 come from comparisons of $C$ with and without the material.

Table 2.4Dielectric constants and approximate dielectric strengths at room temperature

Material	Dielectric constant K	Dielectric strength (MV/m)
Vacuum	1 (exactly)	—
Dry air	1.0006	3
Teflon (PTFE)	2.1	60
Polystyrene	2.6	24
Paper	3.5	16
Pyrex glass	5.6	14
Mica	5.4–6	—
Water (20 °C)	80	—
Strontium titanate	310	8

InteractiveCapacitor lab

A side view of a parallel plate capacitor. Sliders set the plate area, the gap between the plates, the battery voltage and the thickness of a dielectric slab, and a menu chooses the slab material. A switch connects or disconnects the battery; once disconnected, the charge on the plates stays fixed. The drawing shows plus and minus charges on the plates, bound charges on the faces of the slab and field lines whose spacing shows the field strength. Readouts give the capacitance, charge, potential difference, the field in the air and in the slab, and the stored energy, with a warning if the air would break down.

Fig. 2.22Change the plate area, the gap and the dielectric, with the battery connected or disconnected, and watch

C

Q

V

E

and the stored energy.NCERT p. 70

Example 2.8NCERT Example 2.8 · p. 71

How thick must the sheet be?

An air-filled parallel plate capacitor has plates of area $0.020 m^{2}$ held $5.0 m m$ apart by fixed spacers, giving about $35 p F$ . A circuit needs $1.5$ times this value, about $53 p F$ , but the plates cannot be moved, so the designer slips a sheet of mica ( $K = 6.0$ ), with the same area as the plates, into the gap. (a) Find the capacitance with a sheet of thickness $t$ in a gap $d$ . (b) How thick must the mica sheet be? (c) Does it matter where in the gap the sheet sits?

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

2.14 Combination of Capacitors

Capacitors are sold in standard values, so circuit designers often combine several to get the capacitance they need. Any combination can be replaced by one equivalent capacitor that draws the same charge $Q$ from the supply at the same potential difference $V$ , so that $C = Q / V$ . The two basic ways of connecting capacitors lead to two simple rules (Fig. 2.23).

1/4In series the inner plates form an isolated island with zero net charge, so every capacitor carries the same $\pm Q$ .

Fig. 2.23In series every capacitor carries the same charge and the voltages add; in parallel every capacitor has the same voltage and the charges add.NCERT Fig. 2.26 · p. 71

At the top, two capacitors C1 and C2 are joined end to end between two terminals: a series connection. At the bottom, the same two capacitors sit on parallel branches between the two terminals. In the first step the series row is highlighted: each capacitor shows plus Q and minus Q, and a dashed box around the two inner plates and the wire between them is labelled net 0. In the second step brackets under the series row mark V1 and V2 adding up to V. In the third step the parallel circuit is highlighted, with a bracket showing the same V across both branches. In the fourth step the parallel capacitors are labelled Q1 equals C1 V and Q2 equals C2 V, with the total Q equal to Q1 plus Q2.

2.14.1 Capacitors in series

DerivationEquivalent capacitance in series

Connect $C_{1}$ and $C_{2}$ one after the other across a supply, which puts $+ Q$ on the outer plate of $C_{1}$ and $- Q$ on the outer plate of $C_{2}$ . Look at $C_{1}$ first, and draw a Gaussian box whose two ends lie in the metal of its plates. The field is zero in the metal and runs along the box's sides in the gap, so no flux crosses the box. The facing surfaces must therefore carry equal and opposite charge: $- Q$ on the inner plate of $C_{1}$ . That inner plate is wired to the inner plate of $C_{2}$ . Together they form an island of metal that started neutral and touches nothing else, so the $- Q$ on one side leaves $+ Q$ on the other. So every capacitor in a series chain carries the same charge.

The potential differences across the capacitors add up to the supply voltage:

V = V_{1} + V_{2} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}}

(2.55)

\frac{V}{Q} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

(2.56)

An equivalent capacitor would hold the same $Q$ at the same $V$ :

C = \frac{Q}{V}

(2.57)

Comparing with Eq. (2.56),

\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

(2.58)

Nothing changes with more capacitors: each carries $Q$ and the voltages add,

V = V_{1} + V_{2} + \dots + V_{n} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \dots + \frac{Q}{C_{n}}

(2.59)

Capacitors in series

\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + \dots + \frac{1}{C_{n}}

(2.60)

The equivalent capacitance of a series chain is smaller than the smallest capacitor in it: in effect, the gaps add up. The voltage divides in inverse proportion to capacitance, $V_{i} = Q / C_{i}$ , so the smallest capacitor takes the largest share of the voltage, which matters when capacitors have voltage ratings.

2.14.2 Capacitors in parallel

DerivationEquivalent capacitance in parallel

Now connect $C_{1}$ and $C_{2}$ side by side, so that each spans the same pair of junctions. Wired to the same two points, both are at the same potential difference $V$ , and each takes the charge its own capacitance allows:

Q_{1} = C_{1} V, Q_{2} = C_{2} V

(2.61)

The supply has delivered the total charge

Q = Q_{1} + Q_{2}

(2.62)

and an equivalent capacitor must hold this charge at the same $V$ :

Q = C V = C_{1} V + C_{2} V

(2.63)

C = C_{1} + C_{2}

(2.64)

For $n$ capacitors in parallel the charges add in the same way:

Q = Q_{1} + Q_{2} + \dots + Q_{n}

(2.65)

C V = C_{1} V + C_{2} V + \dots + C_{n} V

(2.66)

Capacitors in parallel

C = C_{1} + C_{2} + \dots + C_{n}

(2.67)

In parallel the plate areas effectively add, so the equivalent capacitance is larger than the largest capacitor. Notice that both rules are the reverse of those for resistors in Chapter 3.

Compare

Series and parallel at a glance

Series

Same charge $Q$ on each
Voltages add: $V = V_{1} + V_{2} + \dots$
$1 / C = 1 / C_{1} + 1 / C_{2} + \dots$
$C$ is smaller than the smallest $C_{i}$

Parallel

Same voltage $V$ across each
Charges add: $Q = Q_{1} + Q_{2} + \dots$
$C = C_{1} + C_{2} + \dots$
$C$ is larger than the largest $C_{i}$

Example 2.9NCERT Example 2.9 · p. 73

Reducing a network

In the network of Fig. 2.24, $C_{1} = 2.0 μ F$ and $C_{2} = 4.0 μ F$ are in parallel between A and B, and this pair is in series with $C_{3} = 3.0 μ F$ between B and D. An $18 V$ supply is connected across A and D. Work out (a) the single capacitance that could replace the whole network, (b) the charge stored by each capacitor and (c) the voltage across each one.

Fig. 2.24The network for Example 2.9.NCERT p. 73

Between junctions A and B two capacitors sit on parallel branches: C1 of 2 microfarads above and C2 of 4 microfarads below. From B a single capacitor C3 of 3 microfarads leads to D. A cell of 18 volts, with its positive terminal on the A side, connects D back to A along the bottom of the figure.

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

Quick check

1 of 4

Capacitors of

3.0 μ F

and

6.0 μ F

are connected in series. What is the equivalent capacitance, in µF?

µF

Show answer

Answer 2 µF

1 / C = 1 / 3.0 + 1 / 6.0 = 1 / 2.0

, so

C = 2.0 μ F

, smaller than either.

2.15 Energy Stored in a Capacitor

Charging a capacitor means moving charge from one conductor to the other, against a potential difference that grows as the charge builds up. The work done is stored as electrostatic potential energy, and a capacitor can return it very quickly: the flash of a camera, the jolt from a defibrillator and the burst of power in some electric buses all come from stored capacitor energy.

3/3Adding up the strips fills the triangle under the line: $W = \frac{1}{2} Q V$ .

Fig. 2.25The work to charge a capacitor is the area under its

V

–

Q

line:

W = \frac{1}{2} Q V

.NCERT Fig. 2.30 · p. 74

A graph has charge along the horizontal axis and voltage up the vertical axis. In the first step a straight line rises from the origin to the final point where the charge is Q and the voltage is V, labelled V prime equals Q prime over C. In the second step a thin orange strip of width delta Q prime stands under the line at an intermediate charge, labelled delta W equals V prime delta Q prime. In the third step the whole triangle under the line is shaded and labelled W equals one half Q V.

DerivationEnergy stored in a capacitor

Charging is a matter of lifting charge, a little at a time, from one conductor to the other. Midway through, the conductors hold $\pm Q^{'}$ and stand at a potential difference $V^{'} = Q^{'} / C$ , so lifting the next small parcel $δ Q^{'}$ from the negative conductor to the positive one costs

δ W = V^{'} δ Q^{'} = \frac{Q^{'}}{C} δ Q^{'}

(2.68)

which is the area of a thin strip under the line $V^{'} = Q^{'} / C$ in Fig. 2.25. Adding up the strips from $Q^{'} = 0$ to $Q^{'} = Q$ gives the area of the triangle, or equivalently the integral

W = \int_{0}^{Q} \frac{Q^{'}}{C} d Q^{'} = \frac{Q^{2}}{2 C}

Using $Q = C V$ , the stored energy can be written in three equivalent ways:

Energy stored in a capacitor

U = \frac{Q^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V

(2.69)

Why the half? The charge was not all carried across at the full voltage $V$ : early on the voltage was small, and on average each bit crossed at $V / 2$ .

Connect a charged capacitor across a flash tube, a small motor or a plain resistor and the stored energy comes out as light, motion or heat. How much comes out depends only on the charge the capacitor held, not on its history. A battery that charged it in a millisecond and a hand-cranked generator that took a minute leave the same $U$ behind, if they leave the same $Q$ .

Where, exactly, is the energy? The charges sit on the plates, but the field fills the gap. It turns out to be very fruitful to say that the energy is stored in the field, spread through the space it occupies. The parallel plate capacitor, with its uniform field, lets us put a number on this.

DerivationEnergy density of the electric field

Write the stored energy in terms of the surface charge density $σ = Q / A$ and the capacitance $ε_{0} A / d$ :

U = \frac{Q^{2}}{2 C} = \frac{(σ A)^{2}}{2} \times \frac{d}{ε_{0} A}

(2.70)

The field between the plates is

E = \frac{σ}{ε_{0}}

(2.71)

so $σ = ε_{0} E$ , and

U = \frac{1}{2} ε_{0} E^{2} \times (A d)

(2.72)

Here $A d$ is the volume of the gap, the only place where the field exists. The energy per unit volume of field is therefore

Energy density of an electric field

u = \frac{1}{2} ε_{0} E^{2}

(2.73)

The capacitor was only a convenient laboratory for Eq. (2.73). The same energy density belongs to every electric field, whether it surrounds a thundercloud, a charged comb or an atom: wherever there is a field there is energy, $\frac{1}{2} ε_{0} E^{2}$ in every cubic metre. (Inside a dielectric the same argument gives $\frac{1}{2} ε E^{2}$ .) This energy density is a first hint that fields are physical things in their own right, a theme that returns with electromagnetic waves in Chapter 8.

Definition

Energy density (

u

)

The energy stored per unit volume in an electric field, $u = \frac{1}{2} ε_{0} E^{2}$ in vacuum. Its SI unit is $J / m^{3}$ .

ExampleIn air close to breakdown, about $40 J$ in every cubic metre.

Example

Energy in air about to spark

(a) What is the energy density of the field in air just below breakdown, where $E = 3.0 \times 10^{6} V / m$ ? (b) A thundercloud and the ground below it behave roughly like a parallel plate capacitor with plates of area $1.0 \times 10^{7} m^{2}$ , $1.0 k m$ apart. Estimate the most energy this 'capacitor' could store before the air breaks down.

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

Example 2.10NCERT Example 2.10 · p. 75

Where does the energy go?

A $4.0 μ F$ capacitor is charged to $50 V$ and disconnected from the supply. Its terminals are then joined to those of an uncharged $1.0 μ F$ capacitor (Fig. 2.26). Find (a) the energy stored at first, (b) the common potential difference after the charge has shared out, and (c) the final stored energy. Account for the difference.

Fig. 2.26Sharing charge conserves

Q

but not the stored energy.NCERT p. 75

Two panels. Before: a 4 microfarad capacitor charged to 50 volts holds plus and minus 200 microcoulombs, next to an empty 1 microfarad capacitor; an energy bar reads 5.0 millijoules. After: the two capacitors are joined in parallel at a common 40 volts, holding 160 and 40 microcoulombs; the energy bar is shorter, 4.0 millijoules, and a dashed box marks the 1.0 millijoule that was lost.

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

Quick check

1 of 4

10 μ F

capacitor is charged to

20 V

. How much energy does it store, in mJ?

Show answer

Answer 2 mJ

U = \frac{1}{2} C V^{2} = \frac{1}{2} (1.0 \times 10^{- 5}) (20)^{2} J = 2.0 \times 10^{- 3} J = 2.0 m J

Table 2.5Quantities met in this chapterNCERT p. 78

Quantity	Symbol	SI unit	Dimensions	Typical values
Electrostatic potential	$V$	volt, $V = J / C$	$[M L^{2} T^{- 3} A^{- 1}]$	$1.5 V$ (torch cell) to about $10^{8} V$ (lightning)
Capacitance	$C$	farad, $F = C / V$	$[M^{- 1} L^{- 2} T^{4} A^{2}]$	picofarads in radio circuits to thousands of farads in supercapacitors
Polarisation	$𝐏$	$C m^{- 2}$	$[L^{- 2} T A]$	about $4 \times 10^{- 6} C m^{- 2}$ in glass in a field of $10^{5} V / m$
Dielectric constant	$K$	none	dimensionless	1 for vacuum, about 80 for water, thousands in some ceramics
Energy density	$u$	$J m^{- 3}$	$[M L^{- 1} T^{- 2}]$	about $40 J m^{- 3}$ in air just below breakdown

Flash cards · 13

Formula flashcards

1 / 13

Recap

Electrostatic Potential and Capacitance: key points and summary

The electrostatic force is conservative. Moving a charge $q$ slowly from R to P takes work $q (V_{P} - V_{R})$ , which is stored as potential energy (Section 2.1, Section 2.2).
Potential is work per unit positive charge brought in from infinity; only differences matter. A point charge gives $V = Q / 4 π ε_{0} r$ (Section 2.3).
A dipole's potential, $𝐩 \cdot \hat{𝐫} / 4 π ε_{0} r^{2}$ , depends on direction, falls as $1 / r^{2}$ and is zero on the equatorial plane (Section 2.4).
Potentials of several charges add as signed numbers. A charged shell has $V = q / 4 π ε_{0} r$ outside and the constant $q / 4 π ε_{0} R$ inside (Section 2.5).
Equipotential surfaces join points of equal $V$ . The field crosses them at right angles, points towards lower potential and has size $∣ δ V ∣ / δ l$ (Section 2.6, Section 2.6.1).
The potential energy of a set of charges is the work to assemble it: one term $q_{i} q_{j} / 4 π ε_{0} r_{i j}$ for each pair (Section 2.7).
In an external potential a charge has energy $q V (𝐫)$ , and a dipole in a uniform field has $U = - 𝐩 \cdot 𝐄$ ; $1 e V = 1.6 \times 10^{- 19} J$ (Section 2.8).
In a conductor at equilibrium: no field inside, excess charge only on the surface, one potential throughout, and $𝐄 = (σ / ε_{0}) \hat{𝐧}$ just outside. An empty cavity is shielded from outside fields (Section 2.9).
A dielectric in a field polarises, by induced or partly aligned molecular dipoles; the bound charge on its faces weakens the field inside (Section 2.10).
A capacitor is a pair of conductors with an insulator between them. Its capacitance $C = Q / V$ , in farads, depends only on the geometry and the dielectric (Section 2.11).
A parallel plate capacitor has a uniform field $σ / ε_{0}$ and $C = ε_{0} A / d$ ; filling it with a dielectric multiplies $C$ by $K$ (Section 2.12, Section 2.13).
In series $1 / C = \sum 1 / C_{i}$ (same charge); in parallel $C = \sum C_{i}$ (same voltage) (Section 2.14).
A capacitor stores $U = Q^{2} / 2 C = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$ , which can be pictured as field energy with density $\frac{1}{2} ε_{0} E^{2}$ (Section 2.15).

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercises 2.1	Practice Q1
NCERT Exercises 2.2	Practice Q2
NCERT Exercises 2.3	Practice Q3
NCERT Exercises 2.4	Practice Q4
NCERT Exercises 2.5	Practice Q5
NCERT Exercises 2.6	Practice Q6
NCERT Exercises 2.7	Practice Q7
NCERT Exercises 2.8	Practice Q8
NCERT Exercises 2.9	Practice Q9
NCERT Exercises 2.10	Practice Q10
NCERT Exercises 2.11	Practice Q11

Quick check: MCQs on Electrostatic Potential and Capacitance

2.1 Introduction3 questions
2.2 Electrostatic Potential3 questions
2.3 Potential due to a Point Charge4 questions
2.4 Potential due to an Electric Dipole3 questions
2.5 Potential due to a System of Charges3 questions
2.6.1 Relation between field and potential4 questions
2.7 Potential Energy of a System of Charges4 questions
2.8.3 Potential energy of a dipole in an external field4 questions
2.9 Electrostatics of Conductors4 questions
2.10 Dielectrics and Polarisation4 questions
2.11 Capacitors and Capacitance3 questions
2.12 The Parallel Plate Capacitor4 questions
2.13 Effect of Dielectric on Capacitance4 questions
2.14.2 Capacitors in parallel4 questions
2.15 Energy Stored in a Capacitor4 questions

Glossary · 14

Key terms (English–Hindi)

capacitanceधारिताdhaarita: The ratio $C = Q / V$ of a capacitor's charge to its potential difference; measured in farads.First seen in the chapter →
capacitorसंधारित्रsandhaaritra: Two conductors separated by an insulator, used to store equal and opposite charges and electrical energy.First seen in the chapter →
conservative forceसंरक्षी बलsanrakshi bal: A force whose work between two points does not depend on the path taken, so that a potential energy can be defined for it.First seen in the chapter →

dielectricपरावैद्युतparaavaidyut: An insulating material whose bound charges shift or turn in a field, partly cancelling the field inside it.First seen in the chapter →
dielectric constantपरावैद्युतांकparaavaidyutaank: The factor $K = ε / ε_{0}$ by which filling a capacitor with a material raises its capacitance.First seen in the chapter →
dielectric strengthपरावैद्युत सामर्थ्यparaavaidyut saamarthya: The largest field an insulator can bear before it breaks down; about $3 \times 10^{6} V / m$ for dry air.First seen in the chapter →

electron voltइलेक्ट्रॉन वोल्टelectron volt: The energy change of a charge $e$ moving through $1 V$ : $1 e V = 1.6 \times 10^{- 19} J$ .First seen in the chapter →
electrostatic potentialस्थिरवैद्युत विभवsthirvaidyut vibhav: Work per unit positive test charge to bring the charge from infinity to a point; measured in volts ( $1 V = 1 J / C$ ).First seen in the chapter →
electrostatic potential energyस्थिरवैद्युत स्थितिज ऊर्जाsthirvaidyut sthitij oorja: The work needed to bring a charge, or to assemble a set of charges, from infinity without acceleration.First seen in the chapter →
electrostatic shieldingस्थिरवैद्युत परिरक्षणsthirvaidyut parirakshan: The absence of electric field inside an empty cavity in a conductor, whatever the charges and fields outside.First seen in the chapter →
energy densityऊर्जा घनत्वoorja ghanatva: Energy stored per unit volume of an electric field, $u = \frac{1}{2} ε_{0} E^{2}$ .First seen in the chapter →
equipotential surfaceसमविभव पृष्ठsamvibhav prishth: A surface on which the potential has the same value everywhere; the field crosses it at right angles.First seen in the chapter →

polar moleculeध्रुवीय अणुdhruviya anu: A molecule whose centres of positive and negative charge are permanently separated, giving it a dipole moment of its own.First seen in the chapter →
polarisationध्रुवणdhruvan: The dipole moment per unit volume that a dielectric develops in an electric field.First seen in the chapter →

Electrostatic Potential and Capacitance: Class 12 Physics · Chapter 2

What you will learn in Electrostatic Potential and Capacitance

2.1 Introduction

2.2 Electrostatic Potential

2.3 Potential due to a Point Charge

2.4 Potential due to an Electric Dipole

2.5 Potential due to a System of Charges

2.6 Equipotential Surfaces

2.6.1 Relation between field and potential

2.7 Potential Energy of a System of Charges

2.8 Potential Energy in an External Field

2.8.1 Potential energy of a single charge

2.8.2 Potential energy of a system of two charges in an external field

2.8.3 Potential energy of a dipole in an external field

2.9 Electrostatics of Conductors

2.10 Dielectrics and Polarisation

2.11 Capacitors and Capacitance

2.12 The Parallel Plate Capacitor

2.13 Effect of Dielectric on Capacitance

2.14 Combination of Capacitors

2.14.1 Capacitors in series

2.14.2 Capacitors in parallel

2.15 Energy Stored in a Capacitor

Electrostatic Potential and Capacitance: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Electrostatic Potential and Capacitance

Key terms (English–Hindi)

You reached the end