Question

The distance between two charged objects is doubled. What happens to the electrostatic force between the two?a)It will double.b)It will be unchanged.c)It will be one-half of the original force.d)It will be cut to a fourth of the original force.

Answer 1

Answer:

d) It will be cut to a fourth of the original force.

Explanation:

The magnitude of the electrostatic force between the charged objects is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the separation between the two objects

In this problem, the initial distance is doubled, so

r' = 2r

Therefore, the new electrostatic force will be

$F=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}(k\frac{q_1 q_2}{r^2})=\frac{1}{4}F$

So, the force will be cut to 1/4 of the original value.