Question

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 1

Correct Answer - `(4)/(5) m`
Potential of bigger sphere is `V = (kQ)/b`
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`.