Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes :
- F
- 16F9
- 9F16
- 4F3
Answer
(3) 9F16
Solution
Initially, F = k(Q)(Q)r2 = kQ2r2 (attractive)
Charge transferred from A to B = 25100Q = Q4
qA = Q - Q4 = 3Q4 and qB = -Q + Q4 = - 3Q4
F' = k ( 3Q4 ) ( 3Q4 )r2
= 916 kQ2r2 = 9F16
Answer: (3)