Electrostatics> A free pith ball of mass 8g carries a pos...

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?

s maheshwara srinivas , 11 Years ago

Grade 11

3 Answers

Apoorva Arora

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.
$F=0.08N=\frac{9\times 10^{9}\times 5\times 10^{-8}\times q}{(0.05)^{2}}$
hence,
$q=4.4\times 10^{-7}C$ .

Last Activity: 11 Years ago

Yash Chourasiya

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

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

Last Activity: 5 Years ago