Question

Two charges, Q₁ and Q₂, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force between them if the original situation is changed as described.

(a) The magnitude of charge Q₁ is reduced to Q₁/5.  N
(b) The distance between the charges is reduced to 2·cm.  N
(c) The magnitude of charge Q₂ is increased to 5Q₂ and the distance is increased to 30·cm.  N

Answer 1

a) The force between charges is proportional to magnitude of charge

Therefore, if magnitude of charge Q1 is reduced to Q1/5, the foece will also be reduced to (1/5)*25=5 N

b)initial distance = 6 cm

new distance = 2 cm

force between two charges is inversely proportional to square of distance between charges

Therefore ,new charge = 25*6²/2² = 225 N

c) force is directly proportional to magnitude of charge and inversely proportional to square of distance between them

Therefore, new force = 25*5*6²/30² = 5 N