Question
When 1.0 × 1012 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 × 1012
Net charge Q = 1 × 1012 × 1.6 × 10-19 = 1.6 × 10-7C
$\therefore$ The net potential difference = 10L.
$\therefore$ Capacitance
$\text{C}=\frac{\text{q}}{\text{v}}$
$=\frac{1.6\times10^{-7}}{10}=1.6\times10^{-8}\text{F}.$
Number of electron = 1 × 1012
Net charge Q = 1 × 1012 × 1.6 × 10-19 = 1.6 × 10-7C
$\therefore$ 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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