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 $(g = 10\, ms^{-2})$

• A. $2\times10^{-9}C$
• B. $2\times10^{-11}C$
• C. $2\times10^{- 6}C$
• D. $2\times10^{- 8}C$

Answer 1

Answer: (B)

Given, the drop rests between the two plates $\therefore $
$\therefore q E=m g$
or$ q \frac{V}{r}=m g \,\,\,\,\,\,\, \left(\because E=\frac{V}{r}\right) $
$\Rightarrow q=\frac{m g r}{V} $
here, $m=10^{-6} \,kg , g=10 \,m / s ^{2}, r=1\, mm =10^{-3}\, m$
and $V=500\, V$
Substituting all the values, we get
$q=\frac{10^{-6} \times 10 \times 10^{-3}}{500}, q=2 \times 10^{-11}\, C$