A free pith ball of mass 8g carries a positive charge of 5*10-8 c . what must be the magnitude of charge that shuold be given to a second pith ball fixed 5 cm vertically below the former pith ball so that the upper pith ball is stationary?

Apoorva Arora IIT Roorkee
10 years ago
To keep the first pith ball stationary, the net force acting on it should cancel aout to be zero.
The force of gravity acting downwards= Mg=8*10/1000=0.08 N
So, this should be the force acting on it along the upward direction due to the other charged pith ball.
Hence, the pith ball placed vertically downward should exert a repulsive force on the first pithbull.

hence,
.
Yash Chourasiya
4 years ago
Dear Student

Mass of the pith ball, m = 8 g = .008 kg

Charge on it is, q = 5 × 10-8C

Let Q be the charge on the other pith ball placed at 5 cm or 0.05 m below.

The weight of the upper pith ball is balanced by the electrostatic repulsion between the balls.

mg = kqQ/r2

=> (0.008)(9.8) = (9 × 109)( 5 × 10-8)Q/(0.052)

=> Q = 4.4 × 10-7C