Question

Three point charges $2q$, $-2q$ and $q$ are kept at the vertices of an equilateral triangle of side $l$. The potential energy of the system is

• A. zero
• B. $\frac{-2q^{2}}{\pi\epsilon_{0}l}$
• C. $\frac{q^{2}}{2\pi\epsilon_{0}l}$
• D. $\frac{-q^{2}}{\pi\epsilon_{0}l}$

Hint:

For multi-charge systems, always compute pairwise energies only—never treat it like a single charge system.

Answer 1

The Correct Option is D

Concept: The electrostatic potential energy of a system of point charges is given by the sum of potential energies of all distinct pairs: \[ U = \frac{1}{4\pi\epsilon_0} \sum \frac{q_i q_j}{r_{ij}} \] Since the charges are placed at vertices of an equilateral triangle, every separation is equal to $l$.

Step 1: Identify all charge pairs The three charges are: \[ 2q,\quad -2q,\quad q \] Pairs: - $(2q, -2q)$ - $(2q, q)$ - $(-2q, q)$

Step 2: Compute each interaction energy 1. Between $2q$ and $-2q$: \[ U_1 = \frac{1}{4\pi\epsilon_0}\cdot \frac{(2q)(-2q)}{l} = \frac{-4q^2}{4\pi\epsilon_0 l} \] 2. Between $2q$ and $q$: \[ U_2 = \frac{1}{4\pi\epsilon_0}\cdot \frac{2q\cdot q}{l} = \frac{2q^2}{4\pi\epsilon_0 l} \] 3. Between $-2q$ and $q$: \[ U_3 = \frac{1}{4\pi\epsilon_0}\cdot \frac{-2q\cdot q}{l} = \frac{-2q^2}{4\pi\epsilon_0 l} \]

Step 3: Add all contributions \[ U = \frac{1}{4\pi\epsilon_0 l}(-4q^2 + 2q^2 - 2q^2) \] \[ U = \frac{-4q^2}{4\pi\epsilon_0 l} \] \[ U = \frac{-q^2}{\pi\epsilon_0 l} \] Final Answer: \[ \boxed{\frac{-q^2}{\pi\epsilon_0 l}} \]