On two like charge a repulsive force of 1.6 N is acting, when the distance between them is 0.04 m. If distance is changed to 0.02 m, then force between them will be

Options

A. 2.3 N
B. 4 N
C. 6.4 N
D. 4.6 N

Answer

C

Explanation

We know that, `F prop (1)/( r^2) rArr (F_2)/( F_1) = (( r_1)/( r_2) )^2`
`(r_1)/( r_2) = (0.04)/( 0.02) = 2`
`therefore (F_2)/( F_1) = (2)^2 = 4`
`therefore F_2 = 4 F_1 = 1.6 xx 4 = 64N`