Question
A parallel-plate capacitor having plate area 25cm2 and separation 1.00mm is connected to a battery of 6.0V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?

Answer


$\text{A}=25\text{cm}^2=2.5\times10^{-3}\text{cm}^2$

$\text{d}=1\text{mm}=0.01\text{m}$

$\text{V}=\text{6V},\ \text{Q}=1$

$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{8.854\times10^{-12}\times2.5\times10^{-3}}{0.01}$

$\text{Q}=\text{CV}=\frac{8.854\times10^{-12}\times2.5\times10^{-3}}{0.01}\times6$

$=1.32810\times10^{-10}\text{C}$

$\text{W}=\text{Q}\times\text{V}=1.32810\times10^{-10}\text{C}\times6$

$=8\times10^{-10}\text{J}.$

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