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(B) The energy density $(u)$ of a parallel plate capacitor is given by the formula:$u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 \left( \fr

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$0.1$

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(B) The energy density $(u)$ of a parallel plate capacitor is given by the formula:$u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 \left( \frac{V}{d} ight)^2$Given:$V = 300 \ V$$d = 2 \ mm = 2 	imes 10^{-3} \ m$$\epsilon_0 = 8.85 	imes 10^{-12} \ F/m$Substituting the values:$u = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes \left( \frac{300}{2 	imes 10^{-3}} ight)^2$$u = 0.5 	imes 8.85 	imes 10^{-12} 	imes (1.5 	imes 10^5)^2$$u = 0.5 	imes 8.85 	imes 10^{-12} 	imes 2.25 	imes 10^{10}$$u \approx 0.1 \ J/m^3$

$A$ parallel plate capacitor has a plate separation of $2 \ mm$ and is charged by connecting it to a $300 \ V$ battery. What is the energy density in $J/m^3$?

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