(a) Obtain an expression for the electrostatic potential energy of an equilateral triangle of side $a$ with three charges $q$, $2q$ and $-3q$ placed at its vertices. OR (b) Two small conducting balls A and B of radius $r_1$ and $r_2$ have charges $q_1$ and $q_2$ respectively. They are connected by a wire. Obtain the expression for the charges on A and B, in equilibrium.

(a) Obtain an expression for the electrostatic potential energy of an equilateral triangle of side a with thre

Question

a

(OR)

(b) Two small conducting balls A and B of radius

r_1

and

r_2

have charges

q_1

and

q_2

respectively. They are connected by a wire. Obtain the expression for the charges on A and B, in equilibrium.

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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 — Charge Sharing Between Conductors

Charge Sharing Between Conductors

Imagine you have two buckets of water at different heights. If you connect them with a pipe at the bottom, water flows from the higher bucket to the lower one until both reach the same water level. That's exactly what happens with charge and conductors — except the "height" is electric potential, and the "water" is charge.

When two conductors are connected by a thin wire, charge flows from the one at higher potential to the one at lower potential. The flow stops the instant both conductors reach the same potential. At that moment, the system is in electrostatic equilibrium.

Note

The connecting wire is assumed to have negligible capacitance, so it doesn't store any charge itself — it's just a path for charge to move.

The Precise Physics

Let conductor 1 have capacitance $C_1$ and initial charge $Q_1$ , and conductor 2 have capacitance $C_2$ and initial charge $Q_2$ . Before connection, their potentials are:

V_1 = \frac{Q_1}{C_1}, \quad V_2 = \frac{Q_2}{C_2}

If $V_1 \neq V_2$ , charge flows. After connection, the two conductors become a single conductor (electrically), so they must share a common potential $V_f$ . The total charge is conserved:

Q_1 + Q_2 = Q_1' + Q_2'

where $Q_1'$ and $Q_2'$ are the final charges. Since both are now at the same potential $V_f$ :

V_f = \frac{Q_1'}{C_1} = \frac{Q_2'}{C_2}

From these two equations, you can solve for the final charges:

Q_1' = \frac{C_1}{C_1 + C_2} (Q_1 + Q_2), \quad Q_2' = \frac{C_2}{C_1 + C_2} (Q_1 + Q_2)

And the common potential is:

V_f = \frac{Q_1 + Q_2}{C_1 + C_2}

V_f = \frac{Q_{\text{total}}}{C_{\text{total}}}

This is the fundamental result: the final potential is simply the total charge divided by the total capacitance — exactly as if the two conductors had been combined into one.

What Changes and What Doesn't

Conserved: Total charge. Charge is neither created nor destroyed, only redistributed.

Not conserved: Total energy. Some energy is always lost as heat in the connecting wire (or as electromagnetic radiation). You can calculate the energy loss:

\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2

This is always positive unless $V_1 = V_2$ initially. So charge sharing is an irreversible process — you cannot get back the original separated charges without doing work.

Watch out

A common mistake: assuming total energy is conserved. It is not. Only charge is conserved. The lost energy goes into heating the wire or radiating.

A Concrete Example

Take a $2\ \mu\text{F}$ capacitor charged to $100\ \text{V}$ and an uncharged $3\ \mu\text{F}$ capacitor. Connect them.

Initial charges: $Q_1 = 200\ \mu\text{C}$ , $Q_2 = 0$ .

Total capacitance: $C_1 + C_2 = 5\ \mu\text{F}$ .

Final potential: $V_f = \frac{200}{5} = 40\ \text{V}$ .

Final charges: $Q_1' = 2 \times 40 = 80\ \mu\text{C}$ , $Q_2' = 3 \times 40 = 120\ \mu\text{C}$ . …

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(a) The triangle's electrostatic PE is $U=-\dfrac{7q^2}{4\pi\varepsilon_0 a}$ . (b) Two conductors joined by a wire share their total charge in the ratio of their radii: $Q_A=\dfrac{(q_1+q_2)r_1}{r_1+r_2}$ , $Q_B=\dfrac{(q_1+q_2)r_2}{r_1+r_2}$ .

Part (a)

Electrostatic potential energy belongs to the whole configuration and equals the sum of the interaction energies of every distinct pair (no factor of $\tfrac12$ when pairs are listed once):

U=\frac{1}{4\pi\varepsilon_0}\sum_{\text{pairs}}\frac{q_iq_j}{r_{ij}}.

Label $q$ at A, $2q$ at B, $-3q$ at C; every pair is separated by $a$ . With $k=\dfrac{1}{4\pi\varepsilon_0}$ :

$U_{AB}=\frac{k(q)(2q)}{a}=\frac{2kq^2}{a},\quad U_{BC}=\frac{k(2q)(-3q)}{a}=-\frac{6kq^2}{a},\quad U_{CA}=\frac{k(-3q)(q)}{a}=-\frac{3kq^2}{a}.$ …

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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 15 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
$Q. When an alpha-particle is brought towards another alpha-particle, the potential energy of the system (A) increases (B) decreases (C) remains unchanged (D) none of these$
›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
$Q. [Case study] Consider the charges q1 and q2 initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q1 is brought from infinity to the point r1⃗. There is no external field against which work needs to be done, so work done in bringing q1 from infinity to r1⃗ is zero. From the definition of potential, work done in bringing charge q2 from infinity to the point r2⃗ is q2 times the potential at r2⃗ due to q1. (Figure: two point charges q1 and q2 joined by a line of length r12.)$
(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
$q_1$
(a) zero
(b) less than before
(c) more than before
(d) same as before
›Reveal solutionSolution

After touching, the charge redistributes equally between the identical balls; the AM-GM inequality $\left(\frac{q_1+q_2}{2}\right)^2 \ge q_1q_2$ shows the resulting force is generally larger than before.

Working

Original force: $F_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ .

Since the balls are identical conductors, on touching, the total charge redistributes equally: each now carries $q' = \dfrac{q_1+q_2}{2}$ .

After separating back to $r$ : $F_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q'^2}{r^2}$ .

Comparing:

$q'^2 - q_1q_2 = \dfrac{(q_1+q_2)^2}{4}-q_1q_2 = \dfrac{(q_1-q_2)^2}{4} \ge 0$

…

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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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CBSE 2025Set D1 markMCQ
$Q. Which of the following statements is true for two point charges of opposite sign? (A) The potential energy is always negative (B) The potential energy is always positive (C) The potential energy can be either positive or negative (D) The potential energy is zero$
›Reveal solutionSolution

Potential energy of two charges is U = kq₁q₂/r; for opposite signs the product q₁q₂ is negative, so U is always negative.

The mutual potential energy of two point charges is

$U = \frac{k q_1 q_2}{r}$

For two charges of opposite sign, one is positive and one is negative, so the product $q_1 q_2 < 0$ . Since k and r are positive, U is negative for every finite separation.

…

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CBSE 2024Set IMPROVEMENT1 mark
$Q. What is the change in electric potential energy when one proton is brought near to another proton?$
›Reveal solutionSolution

Bringing two like (positive) charges closer increases their electrostatic potential energy.

The electric potential energy of a pair of point charges is $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}$ . Both protons carry positive charge, so as one proton is brought closer to the other, the separation $r$ decreases, and since $U\propto 1/r$ for like charges, $U$ increases. Physically, work has to be done against the mutually repuls …

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CBSE 2024Set A1 markMCQ
$Q. The electrostatic energy of the system made by two electric dipoles kept at a distance 'r' is proportional to (A) r^2 (B) r^3 (C) r^4 (D) none of these$
›Reveal solutionSolution

Dipole–dipole interaction energy ∝ 1/r³, not r², r³ or r⁴ → 'none of these'.

A single dipole's field falls off as 1/r³. The potential energy of a second dipole placed in that field is therefore proportional to $\dfrac{1}{r^3}$ :

$U \propto \frac{p_1 p_2}{r^3}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$q_1$
(a) more than before
(b) less than before
(c) zero
(d) same as before
›Reveal solutionSolution

After touching, the charge redistributes equally between the identical balls; since the AM-GM inequality gives $\left(\dfrac{q_1+q_2}{2}\right)^2 \ge q_1q_2$ , the resulting force is generally larger than before (equal only when $q_1=q_2$ ).

Working

Original force between the two balls, separated by $r$ :

$F_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$

Since the balls are identical conducting spheres, when brought into contact the total charge $q_1+q_2$ redistributes equally between them (identical spheres share charge equally regardless of the individual initial values). Each now carries

$q' = \dfrac{q_1+q_2}{2}$

After separating back to distance $r$ , the new force is

$F_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q'^2}{r^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(q_1+q_2)^2}{4r^2}$

Comparing $q'^2$ to $q_1q_2$ :

$q'^2 - q_1q_2 = \dfrac{(q_1+q_2)^2}{4} - q_1q_2 = \dfrac{(q_1-q_2)^2}{4} \ge 0$ …

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CBSE 2023Set F1 markMCQ
$Q. If the distance between the two charges is increased, then the electrostatic potential energy of the charges (A) decreases (B) increases (C) may increase or decrease (D) remains the same$
›Reveal solutionSolution
The sign of the effect depends on whether the charges are like or unlike, so U may increase or decrease.

The electrostatic potential energy of two charges is

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

Like charges ( $q_1q_2>0$ ): $U$ is positive and $\propto 1/r$ , so increasing $r$ decreases $U$ . …
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Q.(a) Obtain an expression for the electrostatic potential energy of an equilateral triangle of side aaa with three charges qqq, 2q2q2q and −3q-3q−3q placed at its vertices.

Part (a)

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Q.(a) Obtain an expression for the electrostatic potential energy of an equilateral triangle of side $a$ with three charges $q$ , $2q$ and $-3q$ placed at its vertices.

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