Gujarati
Hindi
2. Electric Potential and Capacitance
easy

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

A

$\frac{q}{{Vd}}$

B

$q\frac{d}{V}$

C

$q\,Vd$

D

$q\frac{V}{d}$

Solution

$W = qE = q\frac{V}{d}$

Standard 12
Physics

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