A free pith ball A of 8g carries a positive charge 5×10^-8C.What must be the nature and magnitude of charge so that be given to a second pith ball B fixed 5 cm below thw former ball so that the upper ball is stationary?
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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.
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.
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Answered by
57
Answer:
here force=mass×acceleration
the pith ball B must be positive charge that is same nature as that of A so that the upward force of repulsion balances the weight of pith ball A
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