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a free pith ball of 8g 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 fixed 5cm vertically elow the former pith ball so that the upper pith ball is stationary?

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

Charge on it is, q = 5 × 10-8 C

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 × 10-7 C

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here weight [F'] of upper ball is acting downwards.To make it stationary the electric force provided by lower ball should be repulsive in nature.also magnituDE of this force should be equal to F'.                                                                                    

Thus,                    F' = k qQ/ R2

                        => mg =kqQ/r

                                Q=mgr/kq = 8 x 10-3 x 9.8 x 5 x10-2 /  5 x 10-8 x 9x109  

                                      =8.7 x 10-8 C

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