A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R $\gg$ r), such that their surface charge densities are equal. Derive the expression for the potential at the common centre. OR Three concentric metallic shells A, B and C of radii a, b and c (a < b < c) have surface charge densities $+\sigma$, $-\sigma$ and $+\sigma$ respectively as shown. (a) Obtain the expressions for the potential of three shells A, B and C. (b) If shells A and C are at the same potential, obtain the relation between a, b and c.

A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R gg r), such t

Question

\gg

(OR)

+\sigma

(a) Obtain the expressions for the potential of three shells A, B and C.

(b) If shells A and C are at the same potential, obtain the relation between a, b and c.

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Electric Potential

Electric Potential

The Idea

When you lift a book onto a shelf you do work against gravity, and the book gains gravitational potential energy that depends on where it sits. Electric charges behave the same way in an electric field. Electric potential is the electrical analogue of "height" in a gravitational field: it tells you the potential energy a unit charge would have at a given point.

Formally, the electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, slowly (without giving it kinetic energy), against the electric field.

V = \frac{W}{q_0}

Here $W$ is the work done to move a small test charge $q_0$ from infinity to the point. Because both $W$ and $q_0$ are scalars, electric potential is a scalar quantity — it has magnitude but no direction. Its SI unit is the volt:

1\ \text{V} = 1\ \text{J/C}

Potential Due to a Point Charge

For a single point charge $q$ , the potential at a distance $r$ from it is

V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}

Notice it falls off as $1/r$ , whereas the electric field of a point charge falls off as $1/r^2$ . The potential is taken as zero at infinity, our chosen reference. A positive charge makes the potential around it positive; a negative charge makes it negative.

Potential Difference

Usually we care about the potential difference between two points $A$ and $B$ :

V_A - V_B = \frac{W_{B\to A}}{q_0}

the work per unit charge needed to move a charge from $B$ to $A$ . This is the quantity a voltmeter reads and the "voltage" that drives current in a circuit.

Relation Between Field and Potential

Field and potential are two views of the same thing. The electric field is the negative rate of change of potential with distance:

E = -\frac{dV}{dr}

The minus sign says the field points from high potential toward low potential — a positive charge, left free, rolls "downhill" in potential. Where the potential changes steeply, the field is strong.

Superposition

Because potential is a scalar, the potential due to several charges is just the algebraic sum of their individual potentials — no vectors, no angles:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots\right)

This makes potential far easier to compute than the field, which needs vector addition. Once you have $V$ everywhere, you can get the field by differentiating.

Equipotential Surfaces

An equipotential surface is a surface on which the potential has the same value everywhere.

No work is done in moving a charge along an equipotential surface (since $\Delta V = 0$ ).
The electric field is always perpendicular to an equipotential surface. …

Part (b)Concept understanding — Electric Potential

Electric Potential

The Idea

Formally, the electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, slowly (without giving it kinetic energy), against the electric field.

V = \frac{W}{q_0}

1\ \text{V} = 1\ \text{J/C}

Potential Due to a Point Charge

For a single point charge $q$ , the potential at a distance $r$ from it is

V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}

Potential Difference

Usually we care about the potential difference between two points $A$ and $B$ :

V_A - V_B = \frac{W_{B\to A}}{q_0}

the work per unit charge needed to move a charge from $B$ to $A$ . This is the quantity a voltmeter reads and the "voltage" that drives current in a circuit.

Relation Between Field and Potential

Field and potential are two views of the same thing. The electric field is the negative rate of change of potential with distance:

E = -\frac{dV}{dr}

Superposition

Because potential is a scalar, the potential due to several charges is just the algebraic sum of their individual potentials — no vectors, no angles:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots\right)

This makes potential far easier to compute than the field, which needs vector addition. Once you have $V$ everywhere, you can get the field by differentiating.

Equipotential Surfaces

An equipotential surface is a surface on which the potential has the same value everywhere.

No work is done in moving a charge along an equipotential surface (since $\Delta V = 0$ ).
The electric field is always perpendicular to an equipotential surface. …

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Centre potential of two equal- $\sigma$ spheres $=\dfrac{\sigma}{\varepsilon_0}(r+R)=\dfrac{Q}{4\pi\varepsilon_0}\dfrac{r+R}{r^2+R^2}$ . For three shells, $V_A=\dfrac{\sigma}{\varepsilon_0}(a-b+c)$ etc., and $V_A=V_C$ gives $c=a+b$ .

Potential is a scalar, so we simply add each shell's contribution. Inside a shell of radius $s$ and charge $q$ , the potential is constant at $\dfrac{q}{4\pi\varepsilon_0 s}$ ; outside it is $\dfrac{q}{4\pi\varepsilon_0 (\text{distance})}$ .

Part (a) — potential at the common centre

Equal surface densities give $q_1=\sigma\cdot4\pi r^2$ and $q_2=\sigma\cdot4\pi R^2$ . At the common centre each shell contributes its surface value:

V_{centre}=\frac{1}{4\pi\varepsilon_0}\Big(\frac{q_1}{r}+\frac{q_2}{R}\Big)=\frac{1}{4\pi\varepsilon_0}\big(4\pi\sigma r+4\pi\sigma R\big)=\frac{\sigma}{\varepsilon_0}(r+R).

Since the total charge is $Q=q_1+q_2=4\pi\sigma(r^2+R^2)$ , we can write $\sigma=\dfrac{Q}{4\pi(r^2+R^2)}$ and

$V_{centre}=\frac{Q}{4\pi\varepsilon_0}\cdot\frac{r+R}{r^2+R^2}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 44 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$r_1$
›Reveal solutionSolution
When two widely separated conducting spheres are connected by a wire, they reach the same electric potential. Since surface field $E = \frac{kQ}{r^2}$ and potential $V = \frac{kQ}{r}$ , combining these gives $E \propto 1/r$ . Therefore the ratio of surface fields is $E_A/E_B = r_2/r_1$ , which corresponds to option (B).

The key insight here is about what happens when conductors are connected by a wire. Charge flows until both spheres are at the same electric potential — that's the fundamental condition for electrostatic equilibrium in a conductor. Once you grasp that, the rest is just algebra.

Let's think about why potential equality is the right starting point. A conducting wire means the two spheres form a single conductor. In electrostatics, the entire surface of a conductor is an equipotential. So spheres A and B must have the same potential $V$ .

Now, for an isolated conducting sphere of radius $r$ carrying charge $Q$ , the potential at its surface (taking infinity as zero) is:
$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$
And the electric field just outside its surface is:
$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$
Notice the relationship: $E = V/r$ . That's a neat shortcut we'll use.

Tip

For any isolated conducting sphere, $E = V/r$ directly. This saves you from carrying the $Q$ through the algebra — just remember it comes from $V = kQ/r$ and $E = kQ/r^2$ .

Let's work through it step by step.

Set potentials equal. Since the wire connects them, $V_A = V_B$ . Using $V = kQ/r$ (where $k = 1/4\pi\epsilon_0$ ):

$k\frac{Q_A}{r_1} = k\frac{Q_B}{r_2}$
Cancel $k$ and rearrange:
$\frac{Q_A}{Q_B} = \frac{r_1}{r_2}$

Write the surface field ratio. For each sphere, $E = kQ/r^2$ . So: $\frac{E_A}{E_B} = \frac{k Q_A / r_1^2}{k Q_B / r_2^2} = \frac{Q_A}{Q_B} \cdot \frac{r_2^2}{r_1^2}$ …
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CBSE 2026Set 55/3/11 markMCQ
$m$
›Reveal solutionSolution
Work done by a constant electric field equals force times displacement; since the particle starts from rest, all that work converts to kinetic energy, giving $K = q E_0 x$ .

The heart of this problem is the work-energy theorem: the work done by all forces on a particle equals its change in kinetic energy. When a charged particle moves through an electric field, the field exerts a force that does work, and if the particle starts from rest, every joule of work becomes kinetic energy.

A uniform electric field $\vec{E} = E_0\,\hat{i}$ exerts a force $\vec{F} = q\vec{E}$ on a charge $q$ . This force is constant in magnitude and direction, so the work done is simply force times displacement along the direction of the force.

Step-by-step reasoning

Identify the force on the particle. The electric force on a charge $q$ in field $\vec{E}$ is

$\vec{F} = q\vec{E} = q E_0\,\hat{i}$
The magnitude is $F = q E_0$ , directed along the positive $x$ -axis.

Calculate the work done by this force. The particle moves a distance $x$ along the $x$ -axis, in the same direction as the force. Work done by a constant force is

$W = F \cdot d = q E_0 \cdot x = q E_0 x$

Apply the work-energy theorem. The particle starts from rest, so initial kinetic energy $K_i = 0$ . The work-energy theorem states

$W = \Delta K = K_f - K_i$
Therefore,
$K_f = W = q E_0 x$
The kinetic energy after travelling distance $x$ is simply the work done by the electric field. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. SI unit of electric potential is:$
(a) Ohm
(b) Volt
(c) Coulomb
(d) Ampere
›Reveal solutionSolution

Electric potential is defined as work done per unit charge, so its SI unit is the Volt.

Electric potential at a point is $V = \dfrac{W}{q}$ , i.e. the work done in bringing a unit positive charge from infinity to that point. Since work is measured in joules (J) and charge in coulombs (C), the unit o …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. [Case study, continued] The charges q1 and q2 produce a potential, which at any point P will be$
(a) V1,2 = 1/(4πε₀) × (q1/r1P² + q2/r2P²)
(b) V1,2 = 1/(4πε₀) × (q1/r1P + q2/r2P)
(c) V1,2 = 1/(4πε₀) × (q1/r1P - q2/r2P)
(d) V1,2 = 1/(4πε₀) × (2q1/r1P + 3q2/r2P)
›Reveal solutionSolution

The potential due to a group of point charges at any point is just the plain algebraic (scalar) sum of the potentials each charge produces there — the superposition principle for potential.

Electric potential obeys the superposition principle: the total potential at any point due to several charges equals the SCALAR sum (not vector sum, since potential is a scalar) of the potentials due to each individual charge, each given by $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ (distance to the first power).

For two charges $q_1$ (at distance $r_{1P}$ from P) and $q_2$ (at distance $r_{2P}$ from P):

$V_{1,2} = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_{1P}} + \frac{q_2}{r_{2P}}\right)$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The standard potential of earth is :$
(a) Zero
(b) Infinite
(c) One
(d) None of the above
›Reveal solutionSolution

The Earth is the chosen reference for potential, so its standard potential is taken as zero.

Electric potential is always measured relative to some reference. Because the Earth is a very large conductor whose potential is practically unaffected by adding or removing charge, it is universally chosen as the reference (zero) level of potential.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Match the Column A with Column B and write the correct pair. Column A: Potential gradient. Column B:$
(i) μ₀nI,
(ii) Volt × second,
(iii) μ₀nI/2,
(iv) Volt × meter,
(v) Volt × meter⁻¹,
(vi) Volt.
›Reveal solutionSolution

Potential gradient dV/dx has unit V/m = Volt × metre⁻¹, option (v).

The potential gradient is the rate of change of electric potential with distance, dV/dx. Its SI unit is volt per metre (V/m), i.e. Volt × metre⁻¹. In magnitude it equals th …

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CBSE 2025Set 55/5/11 markMCQ
$\vec{E}=\dfrac{\alpha}{r^{2}}\,\hat{r}$
›Reveal solutionSolution
For a radial electric field $\vec{E} = \frac{\alpha}{r^2}\,\hat{r}$ , integrate $-\vec{E} \cdot d\vec{l}$ along a radial path from infinity to find the potential; the magnitude is $\frac{\alpha}{r}$ .

The connection between electric field and potential is one of the most fundamental relationships in electrostatics. The electric field points in the direction of steepest decrease of potential, and its magnitude tells us how rapidly the potential drops. Mathematically, $\vec{E} = -\nabla V$ , or in one dimension, $E = -\frac{dV}{dr}$ for a radial field.

To find the potential at a point, we integrate the electric field along a path. The potential difference between two points is:
$V(r) - V(r_0) = -\int_{r_0}^{r} \vec{E} \cdot d\vec{l}$
We conventionally choose $r_0 = \infty$ as our reference point where $V(\infty) = 0$ , so:
$V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{l}$
Now let's work through this problem step by step.

Set up the line integral for a radial field.

Since both $\vec{E}$ and the path element $d\vec{l}$ point radially (we choose a radial path for simplicity), we have:

$\vec{E} \cdot d\vec{l} = E_r \, dr = \frac{\alpha}{r^2} \, dr$

Evaluate the integral from infinity to $r$ .

$V(r) = -\int_{\infty}^{r} \frac{\alpha}{r^2} \, dr$
Reversing the limits to make the calculation cleaner:
$V(r) = \int_{r}^{\infty} \frac{\alpha}{r^2} \, dr$

Perform the integration.

$V(r) = \alpha \int_{r}^{\infty} \frac{1}{r^2} \, dr = \alpha \left[-\frac{1}{r}\right]_{r}^{\infty}$ $V(r) = \alpha \left(0 - \left(-\frac{1}{r}\right)\right) = \frac{\alpha}{r}$

Interpret the result. …
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CBSE 2025Set IMPROVEMENT1 markMCQ
$Q. Electron volt is the unit of which of the following?$
(a) Electric Potential
(b) Energy
(c) Electric Field
(d) Electric Current
›Reveal solutionSolution

Electron volt (eV) is a unit of energy, not of potential, field or current.

…

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CBSE 2025Set D1 markMCQ
$Q. The speed of an electron accelerated from rest under a potential difference V is (A) proportional to V (B) proportional to \sqrtV (C) proportional to 1/V (D) proportional to V^2$
›Reveal solutionSolution

The kinetic energy gained equals eV; since ½mv² = eV, the speed v ∝ √V.

An electron accelerated from rest through a potential difference V gains kinetic energy equal to the work done by the field:

$eV = \frac{1}{2}mv^2$

Solving for the speed:

$v = \sqrt{\frac{2eV}{m}}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Two charges 1C and -1C are placed 1 m apart. The potential at the centre of the line joining the two charges will be$
(a) 2 V
(b) -2 V
(c) zero
(d) 0.5 V
›Reveal solutionSolution

At the midpoint of the line joining equal-and-opposite charges, the two charges are equidistant, so their potential contributions are equal in magnitude but opposite in sign and cancel exactly.

Electric potential due to a point charge q at distance r is V = kq/r (k = 1/4-pi-epsilon0).

The two charges are q1 = +1 C and q2 = -1 C, separated by 1 m, so each is r = 0.5 m from the midpoint.

V = k(+1)/0.5 + k(-1)/0.5 = 2k - 2k = 0

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. When a photon is accelerated (from rest) through a potential difference of one volt, the kinetic energy gained by it is equal to$
(a) 1837eV
(b) 1 eV
(c) $\frac{1}{1837}$ eV
(d) none of the above.
›Reveal solutionSolution

Energy gained by a charge accelerated through a potential difference depends only on charge × voltage, not on mass — so it is 1 eV regardless of which particle is being accelerated.

Work-energy theorem for a charge q accelerated from rest through potential difference V:

$KE = qV$

For an elementary charge $q = e = 1.6\times10^{-19}\ \text{C}$ and $V = 1\ \text{V}$ :

$KE = (1.6\times10^{-19})(1) = 1.6\times10^{-19}\ \text{J} = 1\ \text{eV}$ (by definition of the electron-volt)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Electric potential 'V' at a distance 'r' from a point charge is directly proportional to ______.$
(a) r
(b) r²
(c) 1/r
(d) 1/r²
›Reveal solutionSolution

The electric potential of a point charge falls off as the inverse of the distance from it.

The electric potential at a distance $r$ from a point charge $q$ is

$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$

…

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Q.A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R $\gg$ r), such that their surface charge densities are equal. Derive the expression for the potential at the common centre.

Part (a)Concept understanding — Electric Potential

Electric Potential

The Idea

Potential Due to a Point Charge

Potential Difference

Relation Between Field and Potential

Superposition

Equipotential Surfaces

Part (b)Concept understanding — Electric Potential

Electric Potential

The Idea

Potential Due to a Point Charge

Potential Difference

Relation Between Field and Potential

Superposition

Equipotential Surfaces

Potential of concentric charged shells

Part (a) — two concentric spheres, equal surface density

Part (a) — potential at the common centre

Step-by-step reasoning

Q.A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R ≫\gg≫ r), such that their surface charge densities are equal. Derive the expression for the potential at the common centre.

Part (a) — potential at the common centre

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Q.A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R $\gg$ r), such that their surface charge densities are equal. Derive the expression for the potential at the common centre.

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