(a) (i) Draw equipotential surfaces for an electric dipole. (ii) Two point charges q₁ and q₂ are located at ve

Question

Q. (a)

(i) Draw equipotential surfaces for an electric dipole.

(ii) Two point charges

q_1

and

q_2

are located at

\vec{r}_1

and

\vec{r}_2

respectively in an external electric field

\vec{E}

. Obtain an expression for the potential energy of the system.

(iii) The dipole moment of a molecule is

10^{-30}

C·m. It is placed in an electric field

\vec{E}

10^{5}

V/m such that its axis is along the electric field. The direction of

\vec{E}

is suddenly changed by

60^{\circ}

at an instant. Find the change in the potential energy of the dipole at that instant.

(OR)

(b)

(i) A thin spherical shell of radius

R

has a uniform surface charge density

\sigma

. Using Gauss's law, deduce an expression for the electric field

(i) outside and

(ii) inside the shell.

(ii) Two long straight thin wires AB and CD have linear charge densities

10\,\mu C/m

and

-20\,\mu C/m

respectively. They are kept parallel to each other at a distance of 1 m. Find the magnitude and direction of the net electric field at a point midway between them.

CBSECBSE Class XII Board 2024Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Electrostatic Potential Energy of a System of Charges

The potential energy of a system of N point charges is the total work needed to assemble them by bringing each one, in turn, from infinity to its final position, working against the fields of every charge already placed. The general result is a sum over every distinct PAIR of charges: $U=\dfrac{1}{4\pi\epsilon_0}\sum_{\text{pairs }j,k}\dfrac{q_jq_k}{r_{jk}}$ -- for just two charges this reduces to the simple two-charge formula $U=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r_{12}}$ . …

Part (b)Concept understanding — Gauss Law

Gauss's Law is a shortcut. Instead of adding up the Coulomb contribution of every charge — a nightmare of vectors and integrals — it lets you read the field straight off the symmetry of a problem. The whole idea rests on one quantity: electric flux.

Flux — field-lines counted through a surface. For a uniform field E crossing a flat area A, the flux is Φ = E·A = EA cosθ, where θ is the angle between the field and the normal to the surface. Picture the field as a bundle of lines; flux is how many pierce the surface. When E is edge-on (θ = 90°) nothing passes through and Φ = 0; when E is ⊥ to the surface (θ = 0) the count is maximal. For a closed surface, Φ is the net number of lines leaving it — lines that enter and exit cancel.

The law itself. Gauss's law states that the net flux out of any closed surface depends only on the charge trapped inside:

Φ = q_enclosed / ε₀.

Two consequences do most of the work. First, only enclosed charge counts — a charge outside sends as many lines in as out, so its net contribution is exactly zero. Second, the surface's shape is irrelevant; move the charge around inside or deform the surface, and Φ never changes.

Why symmetry makes it powerful. By itself Φ = q/ε₀ has E buried in an integral. It becomes a tool only when you pick a Gaussian surface matched to the symmetry — one where E is constant and everywhere either ⊥ to the surface (so Φ = EA) or ∥ to it (contributing nothing). Then E slides out and you solve in one line. This works for exactly three geometries:

1 — Infinite line charge (linear density λ). Use a coaxial cylinder: E = λ / 2πε₀r, falling off as ∝ 1/r.

2 — Infinite sheet (surface density σ). Use a pillbox pierced through the sheet: E = σ / 2ε₀ — uniform and completely independent of distance. The field near a large charged plane simply doesn't weaken as you step back. …

Why this formula?

Gauss's Law: Why It Holds

Gauss's Law is one of the four Maxwell's equations and a cornerstone of electromagnetism. Let's build the understanding from the ground up — not just the formula, but the why.

1. The Core Idea: Flux as "Flow" of Field

Imagine an electric field $\vec{E}$ passing through a small patch of area $dA$ . The electric flux through that patch is:

d\Phi_E = \vec{E} \cdot d\vec{A} = E \, dA \cos\theta

where $\theta$ is the angle between $\vec{E}$ and the outward normal to the surface.

Why this definition?

If $\vec{E}$ is perpendicular to the surface ( $\theta = 0$ ), maximum field "flows through".
If $\vec{E}$ is parallel ( $\theta = 90^\circ$ ), no flux — the field just slides along the surface.

Total flux through a closed surface $S$ is:

\Phi_E = \oint_S \vec{E} \cdot d\vec{A}

2. The Key Insight: Flux Depends Only on Enclosed Charge

Consider a single point charge $+q$ at the centre of a spherical surface of radius $r$ .

By Coulomb's law, at every point on the sphere: $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$ , radially outward.
The area vector $d\vec{A}$ is also radially outward.
So $\vec{E} \cdot d\vec{A} = E \, dA$ everywhere.

The total flux:

\Phi_E = \oint E \, dA = E \oint dA = \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \right) (4\pi r^2) = \frac{q}{\varepsilon_0}

Notice: The $r^2$ cancels! The flux is independent of the radius.

3. Why Shape Doesn't Matter

Now, what if the surface is not spherical but any closed shape enclosing the charge?

Draw a small cone from the charge to the surface.
The flux through a patch $dA$ at distance $r$ is $d\Phi = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \cos\theta \, dA$ .
But $\frac{\cos\theta \, dA}{r^2}$ is exactly the solid angle $d\Omega$ subtended by $dA$ at the charge.

So $d\Phi = \frac{q}{4\pi\varepsilon_0} d\Omega$ .

Integrating over the entire closed surface: $\oint d\Omega = 4\pi$ (total solid angle around a point).

Hence:

\Phi_E = \frac{q}{4\pi\varepsilon_0} \cdot 4\pi = \frac{q}{\varepsilon_0}

Result: For any closed surface enclosing $q$ , the flux is $\frac{q}{\varepsilon_0}$ .

4. Multiple Charges: Superposition

If there are many charges $q_1, q_2, \dots, q_n$ inside the surface, the total electric field is the vector sum of individual fields:

\vec{E} = \vec{E}_1 + \vec{E}_2 + \dots + \vec{E}_n

Flux is linear: …

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(a) Dipole equipotentials are $\perp$ to field lines (bisecting plane is $V=0$ ); system energy $U=q_1V(\vec r_1)+q_2V(\vec r_2)+\frac{q_1q_2}{4\pi\varepsilon_0 r_{12}}$ ; turning the field by $60^\circ$ changes the dipole energy by $\Delta U=+5\times10^{-26}$ J. (b) For a shell $E=\frac{Q}{4\pi\varepsilon_0 r^2}$ outside and $0$ inside; the net field midway between the two wires is $1.08\times10^{6}$ N/C towards the negative wire.

Part (a)

(i) Equipotential surfaces of an electric dipole

An equipotential surface is one on which the potential is constant, so no work is done moving a charge over it — hence it is always perpendicular to the electric field lines. For a dipole the field lines run from $+q$ to $-q$ ; the equipotentials are a family of curved surfaces orthogonal to them. The perpendicular bisector plane of the dipole axis is the $V=0$ surface (every point on it is equidistant from the two equal and opposite charges). Surfaces are close together (strong field) near the charges and far apart (weak field) at large distances.

(ii) Potential energy of a two-charge system in an external field

Bring $q_1$ from infinity to $\vec r_1$ : work $=q_1V(\vec r_1)$ , where $V$ is the external-field potential. Bring $q_2$ to $\vec r_2$ : it now moves through the external field and the field of $q_1$ :

W_2=q_2V(\vec r_2)+\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}}.

Total potential energy of the system:

\boxed{U = q_1 V(\vec r_1) + q_2 V(\vec r_2) + \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}}

(iii) Change in dipole energy

The energy of a dipole of moment $p$ in a uniform field $E$ is $U=-\vec p\cdot\vec E=-pE\cos\theta$ .

Initially the axis is along $\vec E$ : $\theta_i=0$ , $U_i=-pE$ .
The field turns by $60^\circ$ : $\theta_f=60^\circ$ , $U_f=-pE\cos 60^\circ=-\tfrac12 pE$ . …

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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 49 on this concept.

CBSE 2026Set 55/2/11 markMCQ
$2q$
›Reveal solutionSolution
The potential energy of a system of point charges is the sum of the potential energies of every distinct pair. For the given charges $2q$ , $-2q$ , and $q$ at the vertices of an equilateral triangle of side $l$ , the total potential energy comes out to $-\dfrac{q^2}{\pi\varepsilon_0 l}$ , which corresponds to option (D).

The concept here is electric potential energy of a system of point charges. This is not about the potential at a point, but about the work done to assemble the charges from infinity to their positions. For any pair of charges $q_i$ and $q_j$ separated by distance $r_{ij}$ , the potential energy of that pair is
$U_{ij} = \frac{1}{4\pi\varepsilon_0} \frac{q_i q_j}{r_{ij}}.$
The total potential energy of the system is simply the sum over all distinct pairs. Since the triangle is equilateral, every side is $l$ , so the distances are all equal — that simplifies the arithmetic.

A common mistake is to forget the sign of the charges or to double-count pairs. Let’s be careful.

Identify the three distinct pairs.

The charges are at vertices A, B, C. Let’s label them:

$q_A = 2q$

$q_B = -2q$

$q_C = q$

The three pairs are: (A,B), (B,C), and (C,A). Each pair is separated by distance $l$ .

Write the potential energy for each pair.

Using $U = \frac{1}{4\pi\varepsilon_0} \frac{q_i q_j}{l}$ :

Pair (A,B): $U_{AB} = \frac{1}{4\pi\varepsilon_0} \frac{(2q)(-2q)}{l} = \frac{1}{4\pi\varepsilon_0} \frac{-4q^2}{l}$

Pair (B,C): $U_{BC} = \frac{1}{4\pi\varepsilon_0} \frac{(-2q)(q)}{l} = \frac{1}{4\pi\varepsilon_0} \frac{-2q^2}{l}$

Pair (C,A): $U_{CA} = \frac{1}{4\pi\varepsilon_0} \frac{(q)(2q)}{l} = \frac{1}{4\pi\varepsilon_0} \frac{2q^2}{l}$

Sum them up.

$U_{\text{total}} = U_{AB} + U_{BC} + U_{CA} = \frac{1}{4\pi\varepsilon_0 l} \left( -4q^2 - 2q^2 + 2q^2 \right)$ …
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CBSE 2026Set A1 markMCQ
$Q. S.I. unit of electric flux is (A) Vm (B) Vm^2 (C) Jm (D) NC^-1$
›Reveal solutionSolution

Φ = E·A → (V/m)(m²) = V·m.

Electric flux is $\Phi = \vec{E}\cdot\vec{A}$ .

Unit of electric field $E$ = N/C = V/m (volt per metre).

Unit of area $A$ = m².

…

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CBSE 2026Set A1 markMCQ
$Q. The surface charge densities on the surface of two conducting spheres of radii r1 and r2 are equal. The ratio of electric field intensities on the surfaces is (A) r1/r2 (B) r1^2/r2^2 (C) r2^2/r1^2 (D) 1 : 1$
›Reveal solutionSolution

Just outside a charged conductor E = σ/ε₀; with equal σ the fields are equal (1:1).

The electric field just outside the surface of a charged conductor is

$E = \frac{\sigma}{\varepsilon_0},$

which depends only on the local surface charge density σ, not on the radius.

…

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CBSE 2026Set A1 markMCQ
$Q. When an alpha-particle is brought towards another alpha-particle, the potential energy of the system (A) increases (B) decreases (C) remains unchanged (D) none of these$
›Reveal solutionSolution

Two positive alpha particles repel; pushing them together raises the potential energy.

An alpha particle carries charge $+2e$ . The electrostatic potential energy of two point charges is:

$U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Electric flux is a$
(a) scalar quantity
(b) vector quantity
(c) scalar or vector quantity
(d) constant quantity
›Reveal solutionSolution

Electric flux is a scalar quantity, even though it is defined using two vectors.

Electric flux through a surface is defined as

$\Phi = \oint \vec{E}\cdot d\vec{A}$

Although $\vec{E}$ (electric field) and $d\vec{A}$ (area vector, normal to the surface element) are both vectors, their dot product $\vec{E}\cdot d\vec{A} = E\,dA\cos\theta$ is a single number (magnitude only, with a sign depending on $\theta$ ) — it …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A charge Q, is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will$
(a) decrease to half
(b) increase two times
(c) remain unchanged
(d) increase four times
›Reveal solutionSolution

Gauss's law: flux through any closed surface = Q_enclosed / epsilon_0, and this does NOT depend on the surface's size or shape.

Gauss's law states that for any closed (Gaussian) surface,

flux (phi) = Q_enclosed / epsilon_0

Here the same charge Q sits at the centre of the sphere both before and after the radius is doubled - the enclosed charge Q_enclosed is unchanged. Since flux depends ONLY on Q_enclosed and the permittivity of free space epsilon_0 (both unchanged here), the flux does not change even though the surface area (4piR^2) has increased fourfold. Doubling R spreads the same total flux over 4 ti …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. [Case study] Consider the charges q1 and q2 initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q1 is brought from infinity to the point r1⃗. There is no external field against which work needs to be done, so work done in bringing q1 from infinity to r1⃗ is zero. From the definition of potential, work done in bringing charge q2 from infinity to the point r2⃗ is q2 times the potential at r2⃗ due to q1. (Figure: two point charges q1 and q2 joined by a line of length r12.)$
(i) To bring q2 from infinity to r2. The work done in this step is
(a) W2 = 1/(4πε₀) × 2q1q2/r12³
(b) W2 = 1/(4πε₀) × q1q2/r12³
(c) W2 = 1/(4πε₀) × q1q2/r12²
(d) W2 = 1/(4πε₀) × q1q2/r12
›Reveal solutionSolution

The work done to bring $q_2$ from infinity is just $q_2$ times the potential already set up by $q_1$ at that point — no squared or cubed distance term.

Since $q_1$ is already placed at $\vec r_1$ , it sets up an electrostatic potential everywhere in space. At the point $\vec r_2$ (distance $r_{12}$ away from $q_1$ ), this potential is:

$V_1(\vec r_2) = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_{12}}$

The work done in bringing charge $q_2$ from infinity to $\vec r_2$ against this potential (by definition of electric potential, $V = W/q$ ) is:

$W_2 = q_2 \, V_1(\vec r_2) = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. [Case study, continued] Let us calculate the potential energy of a system of three charges q1, q2 and q3 located at r1⃗, r2⃗, r3⃗ respectively as shown in the figure (triangle with sides r12, r13, r23). To bring q1 first from infinity to r1⃗, no work is required (W1 = 0).$
(iii) The work done in bringing q3 from infinity to the point r3 is -
(a) W3 = 1/(4πε₀) × (2q1q3/r13 + 2q2q3/r23)
(b) W3 = 1/(4πε₀) × (q1q2/r12 + q1q3/r13)
(c) W3 = 1/(4πε₀) × (q1q3/r13 + q2q3/r23)
(d) W3 = 1/(4πε₀) × (q1q3/r12 - q2q3/r23)
›Reveal solutionSolution

The work to bring the third charge is $q_3$ times the potential already created at that point by BOTH earlier charges combined.

By the time $q_3$ is brought in, both $q_1$ (at $\vec r_1$ ) and $q_2$ (at $\vec r_2$ ) are already in place, having together set up a potential at the location $\vec r_3$ (where $q_3$ will go):

$V_{1,2}(\vec r_3) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_{13}} + \frac{q_2}{r_{23}}\right)$

The work done in bringing $q_3$ from infinity to this point, against this combined potential, is:

$W_3 = q_3\, V_{1,2}(\vec r_3) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right)$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. [Case study, continued] The total work done in assembling the system of three charges q1, q2 and q3 at the given locations is given by$
(a) U = 1/(4πε₀) × (q1q2/r12 + q1q3/r13 + q2q3/r23)
(b) U = 1/(4πε₀) × (q1q2/r12 - q1q3/r13 + q2q3/r23)
(c) U = 1/(4πε₀) × (q1q2/r12 + q1q3/r13 - q2q3/r23)
(d) U = 1/(4πε₀) × (q1q2/r13 + q1q3/r22 + q2q3/r12)
›Reveal solutionSolution

The total potential energy of the three-charge system is the sum of all THREE distinct pairwise interaction terms, each with a plain '+' sign.

Assembling the three charges one at a time (as in parts i–iii): $q_1$ costs no work ( $W_1=0$ ), $q_2$ costs $W_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ , and $q_3$ costs $W_3 = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1q_3}{r_{13}} + \dfrac{q_2q_3}{r_{23}}\right)$ .

The total work done to assemble the configuration — which equals the total electrostatic potential energy of the system, since this work is stored as PE — is:

$U = W_1 + W_2 + W_3 = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right)$

…

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CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) increases as $r$ increases for $r<R$ and for $r>R$
(b) is zero as $r$ increases for $r<R$ and decreases as $r$ increases for $r>R$
(c) is zero as $r$ increases for $r<R$ and increases as $r$ increases for $r>R$
(d) decreases as $r$ increases for $r<R$ and for $r>R$
›Reveal solutionSolution

A charged conducting (hollow metal) sphere carries all its charge on the outer surface. Gauss's law gives $E=0$ inside and $E\propto 1/r^2$ (decreasing) outside.

Setting up Gauss's law

For a hollow, uniformly charged conducting sphere of radius $R$ and total charge $Q$ , all the charge resides on the outer surface (a fundamental property of conductors in electrostatic equilibrium — free charges repel each other and move to the surface where the electric field inside the conducting material is zero).

Take a concentric spherical Gaussian surface of radius $r$ .

Case 1: $r<R$ (inside the shell)

The Gaussian surface of radius $r$ encloses no charge, because all the charge $Q$ lies on the surface at radius $R>r$ .

$\oint \vec E\cdot d\vec A = \frac{Q_{\text{enc}}}{\varepsilon_0} = 0 \implies E=0$

This is true for every $r<R$ — the field is zero throughout the interior, it does not "increase" or "decrease," it is simply zero.

Case 2: $r>R$ (outside the shell)

Now the Gaussian surface encloses the entire charge $Q$ . By spherical symmetry, $\vec E$ is radial and has the same magnitude everywhere on the Gaussian sphere, so

…

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

Electric flux φ_E = E·A; unit = (V/m)(m²) = V·m, option (iv).

Electric flux through a surface is φ_E = E·A (for a uniform field perpendicular to area A). The SI unit of electric field E is volt per metre (V/m) = N/C, and area is in m². Therefore the unit of electric flux is

…

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CBSE 2025Set D1 markMCQ
$Q. Two equal positive point charges of 1 μC charge are kept at a distance of 1 metre in air. The electric potential energy of the system will be (A) 1 joule (B) 1 eV (C) 9 \times 10^-3 joule (D) zero$
›Reveal solutionSolution

The mutual potential energy of two point charges is U = kq₁q₂/r; plugging in gives 9×10⁻³ J.

The electric potential energy of a system of two point charges is

$U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r} = \frac{k q_1 q_2}{r}$

Here both charges are equal and positive: $q_1 = q_2 = 1\,\mu C = 1\times10^{-6}\,C$ , the separation is $r = 1\,m$ , and $k = 9\times10^{9}\,N\,m^2/C^2$ .

…

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Q.(a)

Part (a)

(i) Equipotential surfaces of an electric dipole

(ii) Potential energy of a two-charge system in an external field

(iii) Change in dipole energy

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