Question

A small oil drop of mass ${10}^{-6}$kg is hanging in at rest between two plates separated by $1$mm having a potential difference of $500$V. The charge on the drop is __________$(g=10m{s}^{-2})$.

• A. $2\times {10}^{-9}$C
• B. $2\times {10}^{-11}$C
• C. $2\times {10}^{-6}$C
• D. $2\times {10}^{-8}$C

Answer 1

The correct option is B $2\times {10}^{-11}$C

Given : $d=1$ mm $=0.001$ m

Gravitational force acting on the drop in downward direction is balanced by the electrostatic force acting upwards.

Electric field between the plates $E=\frac{V}{d}=\frac{500}{0.001}=5\times {10}^5$ $V/m$

$\therefore$ $mg=qE$

$⟹$ $q=\frac{mg}{E}=\frac{{10}^{-6}\times 10}{5\times {10}^5}=2\times {10}^{-11}C$