In Millikan's experiment,an oil drop having c | vedclass

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(D) For the oil drop to be stationary,the electric force acting on it must balance its weight.The electric force $F_e$ on a charge $q$ in an electric

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$q\frac{V}{d}$

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(D) For the oil drop to be stationary,the electric force acting on it must balance its weight.The electric force $F_e$ on a charge $q$ in an electric field $E$ is given by $F_e = qE$.The electric field $E$ between two plates separated by a distance $d$ with a potential difference $V$ is given by $E = \frac{V}{d}$.Equating the electric force to the weight of the drop $(mg)$:$mg = qE$$mg = q\left(\frac{V}{d}\right)$Therefore,the weight of the drop is $q\frac{V}{d}$.

List-$I$	List-$II$
$A$. Figure of merit $(k)$	$P$. in series
$B$. Current sensitivity $(S_I)$	$Q$. to get power loss minimized
$C$. Deflection is made half by using resistance	$R$. in shunt/parallel
$D$. High resistance box connected	$S$. current per unit deflection
	$T$. deflection per unit current

In Millikan's experiment,an oil drop having charge $q$ becomes stationary upon applying a potential difference $V$ between two plates separated by a distance $d$. The weight of the drop is:

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