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How many electrons should be removed from a coin of mass `1.6g`, so that it may float in an electric field of intensity of `10^(9)NC^(-1)` directed upward. (take `g= 10m//s^(2)`)A. `10^(6)`B. `10^(7)`C. `10^(9)`D. `10^(8)` |
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Answer» Correct Answer - d Let `n` be the number of electrons removed from the coin. Then, charge on coin, `q= +n e` Now, `qE= mg` Or `(n e)E=mg` or `n=(mg)/(eE)= (1.6xx10^(-3)xx10)/(1.6xx10^(-19)xx10^(9))= 10^(8)` |
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