wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A free pith-ball of 8 g carries a positive charge of 5×108C. What must be the nature and magnitude of charge that should be given to a second pith-ball fixed 5 cm vertically below the former pith-ball so that the upper pith-ball is stationary?

Open in App
Solution

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=8×101000=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=9×109×5×108×q(0.05)2
Hence, q=4.4×107C

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Coulomb's Law - Children's Version
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon