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  4. The magnitude of electric field E required to balance an oil drop of mass m, carrying charge q is (g = acceleration due to gravity)

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Q. The magnitude of electric field $E$ required to balance an oil drop of mass $m$, carrying charge $q$ is ($g$ = acceleration due to gravity)

J & K CETJ & K CET 2009Electric Charges and Fields

Solution:

To balance the oil drop the electric force should be equal to the weight of the oil drop. ie,
$ qE=mg $ or $ E=\frac{mg}{q} $