Question
When $1.0 \times 10^{12}$ electrons are transferred from one conductor to another, a potential difference of $10V$ appears between the conductors. Calculate the capacitance of the two $-$ conductor system.

Answer

Given that Number of electron $= 1 \times 10^{12}$
​​​​​​​Net charge $Q = 1 \times 10^{12} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-7}C$
$\therefore$ The net potential difference $= 10L$.
$\therefore$ Capacitance $\text{C}=\frac{\text{q}}{\text{v}}$
$=\frac{1.6\times10^{-7}}{10}=1.6\times10^{-8}\text{F}.$

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