JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Electrostatics

In the first configuration (1) as shown in the figure, four identical charges (

q_{0}

) are kept at the corners

A, B, C

and D of square of side length ' a '. In the second configuration (2), the same charges are shifted to mid points

G, E, H

and F , of the square, If

K = \frac{1}{4 π ε _{0}}

, the difference between the potential energies of configuration (2) and (1) is given by :

Answer: (D)

Solution

Configuration 1: four equal charges

q_{0}

sit at the corners of a square of side

a

. There are 4 nearest-neighbour pairs at distance

a

and 2 diagonal pairs at distance

a 2

U_{1} = 4 \cdot \frac{K q _{0}^{2}}{a} + 2 \cdot \frac{K q _{0}^{2}}{a 2} = \frac{K q _{0}^{2}}{a} (4 + 2) .

Configuration 2: the charges move to the midpoints of the sides, forming a smaller square (rotated 45°) of side

\frac{a}{2}

. Again there are 4 nearest-neighbour pairs, now at distance

\frac{a}{2}

, and 2 diagonal pairs at distance equal to the diagonal of this smaller square, which is

a

U_{2} = 4 \cdot \frac{K q _{0}^{2}}{a / 2} + 2 \cdot \frac{K q _{0}^{2}}{a} = \frac{K q _{0}^{2}}{a} (42 + 2) .

The difference is

U_{2} - U_{1} = \frac{K q _{0}^{2}}{a} [(42 + 2) - (4 + 2)] = \frac{K q _{0}^{2}}{a} (32 - 2) .

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option