Two charges 8 C and -6 C are placed with a distance of separation 'd' between them and exert a force of magnitude F on each other. If a charge 8 C is added to each of these and they are brought nearer by a distance `(d)/(3)`, the magnitude of force between them will be -

Options

A. `(1)/(3) F`
B. `(9)/(4) F`
C. `(3)/(2) F`
D. `(2)/(3) F`

Answer

C

Explanation

`F=(K(8)(6))/(d^(2))rArr F' =(K(16)(2))/(((2d)/(3))^(2))=(3F)/(2)`