Question
A parallel-plate capacitor with the plate area $100cm^2$ and the separation between the plates 1.0cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.
✓
Answer
$A = 100cm^2 = 10^{-2}m^2$
$d = 1cm = 10^{-2}m$
$V = 24V_0$
$\therefore$ The capacitance $\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{8.85\times10^{-12}\times10^{-2}}{10^{-2}}=8.85\times10^{-12}$
$\therefore$ The energy stored $\text{C}_1=\Big(\frac{1}{2}\Big)\text{CV}^2$
$=\Big(\frac{1}{2}\Big)\times10^{-12}\times(24)^2$
$=2548.8\times10^{-12}$
$\therefore$ The forced attraction between the plates $=\frac{\text{C}_1}{\text{d}}=\frac{2548.8\times10^{-12}}{10^{-2}}=2.54\times10^{-7}\text{N.}$
$d = 1cm = 10^{-2}m$
$V = 24V_0$
$\therefore$ The capacitance $\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{8.85\times10^{-12}\times10^{-2}}{10^{-2}}=8.85\times10^{-12}$
$\therefore$ The energy stored $\text{C}_1=\Big(\frac{1}{2}\Big)\text{CV}^2$
$=\Big(\frac{1}{2}\Big)\times10^{-12}\times(24)^2$
$=2548.8\times10^{-12}$
$\therefore$ The forced attraction between the plates $=\frac{\text{C}_1}{\text{d}}=\frac{2548.8\times10^{-12}}{10^{-2}}=2.54\times10^{-7}\text{N.}$
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