JEE Mains Physics · Electrostatics

Electric Potential, Conductors and Work

Potential is work per unit charge and a scalar, so potentials add with signs and no directions; the field is minus its slope, and the work to move a charge is q times the potential difference.

All Electrostatics notes JEE Mains Physics strategy

Why this matters

Thirty-six PYQs, twenty-nine of them multiple choice, and eleven from 2026, more than any other page in the chapter. Fourteen add potentials: point charges, rings and half rings, concentric shells and potential graphs. Ten are about conductors: charged drops that merge, and spheres joined by a wire. Twelve find the field from a potential, or the work done in moving a charge and the energy of a group of charges.

Concept 1 of 3: Potential of charges, rings and shells

Potential is a number at each point, not an arrow, so adding potentials needs only signs. Every piece of a ring is the same distance from its centre, so the ring's potential there is simply kQ/R, even though the field there is zero. A charged shell has no field inside, so the potential inside does not change: it stays at the surface value.

Definition

$V = \dfrac{kq}{r}$ , sign included. Several charges: $V = \sum \dfrac{kq_i}{r_i}$ .
Ring or arc of total charge Q at its centre: $V = \dfrac{kQ}{R}$ . A half ring of density λ: $V = k\lambda\pi$ , the same for any radius.
Ring on its axis at distance z: $V = \dfrac{kQ}{\sqrt{R^{2} + z^{2}}}$ .
Shell of charge Q: $V = \dfrac{kQ}{R}$ everywhere inside, $\dfrac{kQ}{r}$ outside. The graph is flat, then falls as 1/r.
Concentric shells, at radius r: a shell inside r adds $\dfrac{kq_i}{r}$ ; a shell outside r adds $\dfrac{kq_i}{R_i}$ , using its own radius.
With only positive charges, V is positive at every point, though E can be zero somewhere. Only potential differences can be measured.

Point charges and a shell

V = \sum_i \frac{kq_i}{r_i}, \qquad V_{\text{shell}}(r) = \begin{cases} kQ/R, & r \le R \\ kQ/r, & r \ge R \end{cases}

Worked example

Two concentric thin shells have radii 10 cm and 20 cm and carry

+4

nC and

-2

nC. Find the potential at 5 cm, 15 cm and 30 cm from the centre.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q28Moderate

Example 1 · Electrostatics · Electric Potential, Conductors and Work

Three concentric spherical metallic shells X, Y and

Z

of radius

a,b

and

c

respectively

\lbrack a < b < c\rbrack

have surface charge densities

\sigma, - \sigma

and

\sigma

, respectively. The shells

X

and

Z

are at same potential. If the radii of

X\& Y

are

2\text{ }cm

and

3\text{ }cm

, respectively. The radius of shell

Z

is____ cm.

Potential has no direction

Add the potentials of the charges as signed numbers. Resolving them into components, as for fields, is an error.

Inside a charged shell V is not zero

The field inside is zero, so the potential is constant there, equal to its value at the surface.

Outer shells use their own radius

At a point inside a shell, that shell contributes kq/R, not kq/r. Using r for every shell overcounts the outer ones.

Drill 13 more on potential of charges, rings and shells

Concept 2 of 3: Drops that merge and spheres joined by a wire

A conductor's whole body is at one potential. When charged drops merge, the charge adds but the radius grows only as the cube root of the number of drops, so the potential rises. When two spheres are joined by a long wire, charge flows until their potentials are equal: the larger sphere takes more charge, but the smaller one ends with the higher surface density and the stronger field.

Definition

n identical drops (radius r, potential V) merge: volume is conserved, so $R = n^{1/3}r$ . Charge × n, potential × $n^{2/3}$ , surface density × $n^{1/3}$ , capacitance × $n^{1/3}$ , stored energy × $n^{5/3}$ .
Spheres joined by a long wire: equal potentials, so $q \propto R$ , while σ and the surface field go as $1/R$ .
Common potential after joining: $V = \dfrac{q_1 + q_2}{4\pi\varepsilon_0(R_1 + R_2)}$ . Total charge is conserved.
A conductor is an equipotential. The field just outside is normal to its surface, of size σ/ε₀, and charge crowds at sharp points.

Drops and joined spheres

V_{\text{big}} = n^{2/3}V, \qquad \frac{q_1}{q_2} = \frac{R_1}{R_2}, \qquad \frac{\sigma_1}{\sigma_2} = \frac{E_1}{E_2} = \frac{R_2}{R_1}

Worked example

125 identical charged drops, each at 2 V, merge into one drop. Find its potential, and the factor by which the surface charge density changes.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q9Moderate

Example 2 · Electrostatics · Electric Potential, Conductors and Work

Two charged conducting spheres

S_{1}

and

S_{2}

of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on

S_{1}

and

S_{2}

spheres are

E_{S1}

and

E_{S2}

respectively. The value of

\frac{E_{S1}}{E_{S2}}

\_\_\_\_

Volume is conserved, not radius

n drops make a drop of radius n^(1/3) r, not n r. Using nr makes the potential stay the same.

Joined spheres share potential, not charge

Charge splits in proportion to radius. Splitting it equally is right only for identical spheres.

Less charge, stronger field

The smaller of two joined spheres holds less charge but has the larger surface density and surface field.

Drill 9 more on drops that merge and spheres joined by a wire

Concept 3 of 3: Field from potential, work and potential energy

The field points the way the potential falls fastest, and its size is how fast it falls: E is minus the slope of V. Going the other way, the potential difference is the field added up along a path. Since the electrostatic force is conservative, the work between two points depends only on their potentials, never on the path, and it is zero along an equipotential.

Definition

$\vec E = -\nabla V$ : $E_x = -\dfrac{\partial V}{\partial x}$ , and so on. In a uniform field, $E = \dfrac{\Delta V}{d}$ along the field.
$V_B - V_A = -\displaystyle\int_A^B \vec E \cdot d\vec r$ .
Work by an external agent, moving q slowly: $W = q(V_B - V_A)$ . Work by the field: $q\displaystyle\int_A^B \vec E \cdot d\vec r = -q(V_B - V_A)$ .
Equipotential surfaces are perpendicular to the field lines; no work is done moving along one.
Potential energy of a group: $U = \sum_{\text{pairs}} \dfrac{kq_iq_j}{r_{ij}}$ , each pair once. In an external field, add $\sum q_iV(\vec r_i)$ .

Gradient, work and energy

\vec E = -\nabla V, \qquad W_{\text{ext}} = q(V_B - V_A), \qquad U = \sum_{i<j} \frac{kq_iq_j}{r_{ij}}

Worked example

The potential in a region is

V = 4x^{2}y - 3z

volts, with x, y, z in metres. Find the field at (1, 2, 0) m.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q24Moderate

Example 3 · Electrostatics · Electric Potential, Conductors and Work

A three coulomb charge moves from the point (

0, - 2, - 5

) to the point (

5,1,2

) in an electric field expressed as

\overrightarrow{E} = 2x\widehat{i} + 3y^{2}\widehat{j} + 4\widehat{k}N/C

. The work done in moving the charge is

\_\_\_\_

Work by the field or by an agent

The two differ by a sign. The agent's work is q(V_B − V_A); the field's work is the negative of that. Read which one is asked.

Keep the minus sign in E = −∇V

The field points towards falling potential. Dropping the sign reverses every component.

Count each pair once

For three charges there are three pairs, not six. Summing over i and j without care doubles the energy.

Drill 11 more on field from potential, work and potential energy

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Potential of charges, rings and shells
Point charges and a shell
$V = \sum_i \frac{kq_i}{r_i}, \qquad V_{\text{shell}}(r) = \begin{cases} kQ/R, & r \le R \\ kQ/r, & r \ge R \end{cases}$
Drops that merge and spheres joined by a wire
Drops and joined spheres
$V_{\text{big}} = n^{2/3}V, \qquad \frac{q_1}{q_2} = \frac{R_1}{R_2}, \qquad \frac{\sigma_1}{\sigma_2} = \frac{E_1}{E_2} = \frac{R_2}{R_1}$
Field from potential, work and potential energy
Gradient, work and energy
$\vec E = -\nabla V, \qquad W_{\text{ext}} = q(V_B - V_A), \qquad U = \sum_{i<j} \frac{kq_iq_j}{r_{ij}}$

Watch out for (9)

Potential has no direction
→ Potential of charges, rings and shells
Inside a charged shell V is not zero
→ Potential of charges, rings and shells
Outer shells use their own radius
→ Potential of charges, rings and shells
Volume is conserved, not radius
→ Drops that merge and spheres joined by a wire
Joined spheres share potential, not charge
→ Drops that merge and spheres joined by a wire
Less charge, stronger field
→ Drops that merge and spheres joined by a wire
Work by the field or by an agent
→ Field from potential, work and potential energy
Keep the minus sign in E = −∇V
→ Field from potential, work and potential energy
Count each pair once
→ Field from potential, work and potential energy

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