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Physics Question on electrostatic potential and capacitance

A charged oil drop of mass 2.5×10−72.5 \times 10^{-7} kg is in space between the two piates, each of area 2×10−2 m22 \times 10^{-2} \, m^2 of a parallel piate capacitor. When the upper plate has a charge of 5×10−7 C5 \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/s210 \, m/s^2 )

A

9×10−1 C9 \times 10^{-1} \, C

B

9×10−6 C9 \times 10^{-6} \, C

C

8.85×10−13 C8.85 \times 10^{-13} \, C

D

1.8×10−14 C1.8 \times 10^{-14} \, C

Answer

8.85×10−13 C8.85 \times 10^{-13} \, C

Explanation

Solution

We know that \hspace30mm qE = mg \hspace30mm \frac{qQ}{\varepsilon_0 A} = mg \, \, or \, \, q = \frac{\varepsilon_0 A \, mg}{Q} \hspace30mm = \frac{8.85 \times 10^{-12} \times 2 \times 10^{-2} \times 2.5 \times 10^{-7} \times 10}{5 \times 10^{-7}} \hspace30mm = 8.85 \times 10^{-13} \, C