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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?

 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?

Grade:11

3 Answers

Apoorva Arora IIT Roorkee
askIITians Faculty 181 Points
8 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.
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.
Yash Chourasiya
askIITians Faculty 256 Points
2 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

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya
Rohini Selvaraj
13 Points
one year ago
An object of mass 1kg contains 4×10^20 atoms.if one electron is removed from every atom of the solid,the charge gained by the solid in 1 g is

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