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

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

Step-by-step explanation:

The magnitude of the electrostatic force between the charged objects is

`F=k(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(q_1 q_2)/((r')^2)=k(q_1 q_2)/((2r)^2)=(1)/(4)(k(q_1 q_2)/(r^2))=(1)/(4)F`

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