A pith ball of mass 1 g and charge 980 esu is allowed to fall from a height of 1 m on the surface of a sphere of radius 10 cm.what is the value of charge of the sphere
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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.
F=0.08N=9×10^9×5×10^-8×q/(0.05)^2
hence,
q=4.4×10^-7C
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=9×10^9×5×10^-8×q/(0.05)^2
hence,
q=4.4×10^-7C
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