A ball of mass 9x10^-5kg carries a charge is 5x10^-6C . What must be the magnitude and sign of the charge on another ball held directly 2cm above the first ball so that the first ball remains stationary .

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Solution

To keep the first ball stationary, the net force acting on it should cancel aout to be zero.
The force of gravity acting downwards= Mg=9×10^-510/1000=9×10^-7N
So, this should be the force acting on it along the upward direction due to the other charged ball.
Hence, the ball placed vertically upward should exert a attractive force on the first ball
so
9×10^-7=9×10^95×10^-6q/(.02)^2
solving this q=.02^210^-1/(10^9*5)
. 008×10^-10C

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A ball of mass 9x10-5kg carries a charge is 5x10-6 C . What must be the magnitude and sign of the charge on another ball held directly 2cm above the first ball so that the first ball remains stationary .

A ball of mass 9x10^-5kg carries a charge is 5x10^-6C . What must be the magnitude and sign of the charge on another ball held directly 2cm above the first ball so that the first ball remains stationary .