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.
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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 $\text{C}=\frac{\text{q}}{\text{v}}$
$=\frac{1.6\times10^{-7}}{10}=1.6\times10^{-8}\text{F}.$
$\therefore$ Capacitance $\text{C}=\frac{\text{q}}{\text{v}}$
$=\frac{1.6\times10^{-7}}{10}=1.6\times10^{-8}\text{F}.$
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