Question

a free pith ball of 8 g carries a positive charge of 5×10^-8c ,what must be the nature and magnitude of charge that should be given to a second pith ball b fixed 5 CM below the former ball so that the upper ball is stationary​

Answer 1

Explanation:

To keep the first pith ball stationary, the net force acting on it should cancel out to be zero.

The force of gravity acting downward =Mg=

1000

8×10

=0.08N

So, this should be the force acting on it along the upward direction due to other charged pith ball.

Hence, the pith ball placed vertically downward should exert a repulsive force on the first pith ball.

F=0.08N=

(0.05)

2

9×10

9

×5×10

−8

×q

Hence, q=4.4×10

−7

C