[figure] 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= 14 0, the difference between the potential energies of configuration (2) and (1) is given by :
- Kq02a (4-2 2 )
- Kq02a (4 2-2 )
- Kq02a (3- 2 )
- Kq02a (3 2-2 )
Answer
(4) Kq02a (3 2-2 )
Solution
U= Kq02rij over the 6 pairs.
[figure]
(1): 4 sides a, 2 diagonals 2a: U1= Kq02a (4+ 2 )
(2): 4 sides a 2, 2 diagonals a: U2= Kq02a (4 2+2 )
U2-U1= Kq02a (3 2-2 )
Answer: (4)