Two charged bodies exert a force of 0.445 N on each other.?

Question

If they are moved so that they are one-fourth as far apart, what force is exerted?

Answer 1

Ok, the force between two charged bodies is :

F = K*q1*q2 / r^2

r = distance

q1 and q2 = charges

K = electric constant

0.445 = K*q1*q2 / r^2

now the distance will be one - fourt = 1/4r

F' = k*q1*q1 / r^2 / 16 = 16*K*q1*q2 / r^2

F' = 16*0.445 = 7.12 Newtons

The force exerted = 7,2 N

The force is higher than the first one, as it was expected.

Hope that might helped you