In the first configuration (1) as shown in the figure, four identical charges (q₀) 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=1/(4 π ε₀), the difference between the potential energies of configuration (2) and (1) is given by :

Solution

Physics sir · 3 steps

Four charges make six pairs. Find every pair distance in both squares, then subtract.

1. Count the pairs. Four charges form C(4, 2) = 6 pairs: 4 pairs along the sides of a square and 2 pairs along its diagonals. Each pair adds Kq₀²/r to the energy.
2. Inner square size. The midpoints G, E, H, F form a smaller square. Neighbouring midpoints, such as G and E, are √((a/2)²+(a/2)²) = a/√2 apart. Opposite midpoints, such as G and H, are a apart.

   Configuration	4 side pairs	2 diagonal pairs	Total energy
   (1) corners	r = a, gives 4Kq₀²/a	r = √2a, gives √2Kq₀²/a	Kq₀²/a(4+√2)
   (2) midpoints	r = a/√2, gives 4√2Kq₀²/a	r = a, gives 2Kq₀²/a	Kq₀²/a(4√2+2)

   U₂ - U₁ = Kq₀²/a[(4√2+2)-(4+√2)] = Kq₀²/a(3√2-2)

3. So. The charges sit closer together in configuration (2), so its energy is higher by Kq₀²/a(3√2-2). That is option (d).

The idea to keep. For a system of point charges, list every pair once. For n charges there are n(n−1)/2 pairs, and in a square the diagonal pairs count just as much as the side pairs.

Answer from the NTA final answer key. Solution by PrepWiser.