Question

49W

Answer 1

Answer:

16

Explanation:

The magnitude of the electrostatic force between two charged particles is given by

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the particles

In this problem, the initial force between the particles is F.

Later, the distance between the two particles is increased by four, so

r' = 4r

So, the new force between the particles will be

$F'=\frac{kq_1 q_2}{(4r)^2}=\frac{kq_1 q_2}{16r^2}=\frac{1}{16}F$

So, the new force decreases by a factor of 16.