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JEE Physics 2025 Question with Solution

Two square configurations of four identical charges q0. In configuration 1 charges are at corners A, B, C, D of a square with midpoints G, E, H, F and center O. In configuration 2 charges are at midpoints G, E, H, F of the same square.

In the first configuration (1) as shown in the figure, four identical charges q0q_0 are kept at the corners AA, BB, CC and DD of square of side length aa. In the second configuration (2), the same charges are shifted to mid points GG, EE, HH, and FF of the square. If K=14πϵ0K = \frac{1}{4\pi \epsilon_0}, the difference between the potential energies of configuration (2) and (1) is given by:

  • A

    Kq02a(42−2)\frac{Kq_0^2}{a} (4\sqrt{2} - 2)

  • B

    Kq02a(4−2)\frac{Kq_0^2}{a} (4 - \sqrt{2})

  • C

    Kq02a(32−2)\frac{Kq_0^2}{a} (3\sqrt{2} - 2)

  • D

    Kq02a(3−2)\frac{Kq_0^2}{a} (3 - \sqrt{2})

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Four identical charges q0q_0 are placed on a square of side aa. We need the difference in electrostatic potential energy, ΔU=U2−U1\Delta U = U_2 - U_1.

Find: The expression for the difference between the potential energies of configuration (2) and configuration (1).

For a system of point charges,

U = \sum_{i

For configuration (1), charges are at the four corners of the square.

  • There are 4 adjacent pairs, each separated by distance aa.
  • There are 2 diagonal pairs, each separated by distance 2a\sqrt{2}a.

Therefore,

U1=4×Kq02a+2×Kq022aU_1 = 4 \times \frac{K q_0^2}{a} + 2 \times \frac{K q_0^2}{\sqrt{2}a}

For configuration (2), charges are at the midpoints of the sides.

  • There are 4 adjacent pairs, each separated by distance a2\frac{a}{\sqrt{2}}.
  • There are 2 opposite pairs, each separated by distance aa.

Therefore,

U2=4×Kq02a/2+2×Kq02aU_2 = 4 \times \frac{K q_0^2}{a/\sqrt{2}} + 2 \times \frac{K q_0^2}{a}

Now take the difference:

ΔU=U2−U1\Delta U = U_2 - U_1

Substituting,

ΔU=(4×Kq02a/2+2×Kq02a)−(4×Kq02a+2×Kq022a)\Delta U = \left(4 \times \frac{K q_0^2}{a/\sqrt{2}} + 2 \times \frac{K q_0^2}{a}\right) - \left(4 \times \frac{K q_0^2}{a} + 2 \times \frac{K q_0^2}{\sqrt{2}a}\right)

Using 1a/2=2a\frac{1}{a/\sqrt{2}} = \frac{\sqrt{2}}{a},

ΔU=(42Kq02a+2Kq02a)−(4Kq02a+2Kq022a)\Delta U = \left(\frac{4\sqrt{2}K q_0^2}{a} + \frac{2K q_0^2}{a}\right) - \left(\frac{4K q_0^2}{a} + \frac{2K q_0^2}{\sqrt{2}a}\right)

This simplifies to

ΔU=Kq02a(42−2)\Delta U = \frac{K q_0^2}{a}(4\sqrt{2} - 2)

Therefore, the difference in potential energies is Kq02a(42−2)\frac{Kq_0^2}{a}(4\sqrt{2} - 2), so the correct option is A.

Pair Counting in Both Configurations

Given: The electrostatic potential energy of each pair is Kq02r\frac{K q_0^2}{r}.

Find: Compare all six pairs in the two configurations.

A set of 4 charges gives

(42)=6\binom{4}{2} = 6

pairs in total.

In configuration (1):

  • 4 pairs contribute Kq02a\frac{Kq_0^2}{a} each.
  • 2 pairs contribute Kq022a\frac{Kq_0^2}{\sqrt{2}a} each.

So,

U1=4Kq02a+2Kq022aU_1 = 4\frac{Kq_0^2}{a} + 2\frac{Kq_0^2}{\sqrt{2}a}

In configuration (2):

  • 4 pairs contribute Kq02a/2\frac{Kq_0^2}{a/\sqrt{2}} each.
  • 2 pairs contribute Kq02a\frac{Kq_0^2}{a} each.

So,

U2=4Kq02a/2+2Kq02aU_2 = 4\frac{Kq_0^2}{a/\sqrt{2}} + 2\frac{Kq_0^2}{a}

Since

1a/2=2a\frac{1}{a/\sqrt{2}} = \frac{\sqrt{2}}{a}

we get

U2=42Kq02a+2Kq02aU_2 = \frac{4\sqrt{2}Kq_0^2}{a} + \frac{2Kq_0^2}{a}

Now,

ΔU=U2−U1\Delta U = U_2 - U_1

Therefore,

ΔU=(42Kq02a+2Kq02a)−(4Kq02a+2Kq022a)\Delta U = \left(\frac{4\sqrt{2}Kq_0^2}{a} + \frac{2Kq_0^2}{a}\right) - \left(\frac{4Kq_0^2}{a} + \frac{2Kq_0^2}{\sqrt{2}a}\right)

Using

22=2\frac{2}{\sqrt{2}} = \sqrt{2}

this becomes

ΔU=Kq02a(42−2)\Delta U = \frac{Kq_0^2}{a}(4\sqrt{2} - 2)

Hence, the correct option is A.

Common mistakes

  • Counting only the 4 side pairs and forgetting the 2 diagonal pairs is incorrect because a system of four charges has 6 total pairs. Always count all distinct charge pairs before summing potential energy.

  • Using the midpoint separation as a2\frac{a}{2} instead of a2\frac{a}{\sqrt{2}} is wrong. Adjacent midpoints form a smaller square, and the side of that square is obtained from geometry.

  • Taking the difference as U1−U2U_1 - U_2 instead of U2−U1U_2 - U_1 changes the sign of the answer. Read carefully which configuration is subtracted from which.

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