Question

If the distance between 2 charged particles is double, the force between them changes by a factor of?

Answer 1

See explanation

Step-by-step explanation:

Given that;

q1 and q2 are the magnitudes of the two charges while r1 is the distance between the charges.

F1 =kq1q2/r1^2

Then,

F2 = kq1q2/(2r1)^2

F2 = kq1q2/4r1^2

So,

F2 = 1/4 F1

The force between the charges changes by a factor of 1/4.