Question

A charged oil drop od mass $2.5\times {10}^{-7}kg$ is in space between the two plates, each of area $2\times {10}^{-2}{m}^2$ of a parallel plate capacitor. When the upper plate has a charge of $5\times {10}^{-7}C$ and the lower plate has an equal negative charge then the oil remains stationery. The charge of the oil drop is (take, $g=10{m/s}^2$)

• A. $9\times {10}^{-1}C$
• B. $9\times {10}^{-6}C$
• C. $8.85\times {10}^{-13}C$
• D. $1.8\times {10}^{-14}C$

Answer 1

The correct option is A $9\times {10}^{-1}C$

Mass $=2.5\times {10}^{-7}kg$

$qE=mg$

or, $q\times \frac{Q}{A{ϵ}_0}=mg$

$q=\frac{mgA{ϵ}_0}{Q}$

$=\frac{2.5\times {10}^{-2}\times 2\times {10}^{-2}\times 8.854\times {10}^{-12}\times 10}{5\times C7\times 1}=9\times {10}^{-14}C$