Question

A charged oil drop of mass $2.5 \times 10^{-7}$ kg is in space between the two piates, each of area $2 \times 10^{-2} \, m^2$ of a parallel piate capacitor. When the upper plate has a charge of $5 \times 10^{-7} \, C$ and the lower plate has an equal negative charge, the oil remains stationary. 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

Answer: (C)

We know that
$ qE = mg$
$ \frac{qQ}{\varepsilon_0 A} = mg \, \, or \, \, q = \frac{\varepsilon_0 A \, mg}{Q}$
$ = \frac{8.85 \times 10^{-12} \times 2 \times 10^{-2} \times 2.5 \times 10^{-7} \times 10}{5 \times 10^{-7}}$
$ = 8.85 \times 10^{-13} \, C$