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A parallel plate capacitor having a plate separation of $2 \mathrm{~mm}$ is charged by connecting it to a $300 \mathrm{~V}$ supply. The energy density is
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Verified Answer
The correct answer is:
$0.1 \mathrm{~J} / \mathrm{m}$
The energy density of parallel plate capacitor
is given by $U=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2$
$=\frac{1}{2} \times 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2 \times\left(\frac{300 \mathrm{volt}}{2 \times 10^{-3} \mathrm{~m}}\right)^2=0.1 \mathrm{~J} / \mathrm{m}^3$
is given by $U=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2$
$=\frac{1}{2} \times 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2 \times\left(\frac{300 \mathrm{volt}}{2 \times 10^{-3} \mathrm{~m}}\right)^2=0.1 \mathrm{~J} / \mathrm{m}^3$
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