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A large hollow metal sphere of radius R has a small opening at the top. Small drops of mercury each of radius r and charged to a potential V fall into the sphere. The potential of the sphere becomes V' after N drops fall into it. Then, (a) V' < V for all N (b) V' = V for N = 1 (c) V' = V for N = R/r (d) V' = V for N = (R/r)1/3 |
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Answer» Correct Answer is: (a) V' < V for all N Capacitance of each mercury drop is 4πɛ0r. Charge on each mercury drop is 4πɛ0r VN. Capacitance of the hollow sphere is 4πɛ0R. When it acquires Q charge, its potential is V' = Q/4πɛ0R. For V' = V, 4πɛ0RV = 4πɛ0rVN or N = R/r. |
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