Question

Two point charges each experience a 1-N electrostatic force when they are 2 cm apart. If they are moved to a new separation of 8 cm, what is the magnitude of the electric force on each of them

Answer 1

Answer 2

SOLUTION :

F is inversely proportional to  squared distance, d^2

=> F = k / d^2 

=> k = F d^2

When F = 1 N and d = 2 cm = 0.02 m

=> k = 1 * (0.02)^2 = 0.0004 N-m^2

So, when distance is 8 cm = 0.08 m :

F2 = k  / (0.5)^2 = 0.0004/ (0.08)^2 = 1 / 16 = 0.0625 N (ANSWER).