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The magnitude of the electric field required to just balance in air a `2 xx 10^(-4) kg` liquid drop carrying a charge of `10 xx 10^(-2) mu C` isA. `10^(4) N//C`B. `2 xx 10^(4) N//C`C. `4 xx 10^(4) N//C`D. `5 xx 10^(4) N//C` |
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Answer» Correct Answer - 2 `qE = mg` `E = (mg)/(q) = ( 2 xx 10^(-4) xx 10)/(10 xx 10^(-2) xx 10^(-6)) = 2 xx 10^(4) N//C` |
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