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When a small uncharged conducting ball of radius a = 1 cm and mass m = 50 g is dropped from a height h above the center of another large conducting sphere of radius b( = 1 m) having charge Q( =(100 mu C). It rises to a height h_1 (= 2 m) after the collision. Find the value of h. Assume that during the impact there is no dissipation of energy. . |
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Answer» When SMALL ball COMES in contact with bigger, their potentials will become same. So `(k q)/a = (k (Q - q))/b` Neglecting q in comparision to Q, we get `q = (Q a)/b = 1 mu C` Apply conservation of energy after collision `(1)/(2) m (sqrt(2 g h))^2 + ( k qQ)/b = mg h_1 + ( kq Q)/(b + h_1)` or `h = (4)/(5) m`. |
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