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An oil drop carrying charge `Q` is held in equilibrium by a potential difference of `600V` between the horizontal plates.In order to hold another drop of radius in equilibrium a potential drop of `1600V` had to be maintained .The charge on the second drop isA. `(Q)/(2)`B. `2Q`C. `(3Q)/(2)`D. `3Q` |
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Answer» Correct Answer - 4 `(V_(1))/(V_(2))=((R_(1))/(R_(2)))^(3),(Q_(2))/(Q_(1))` |
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