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(B) The energy density $u$ of a parallel plate capacitor is given by the formula: $u = \frac{1}{2} \varepsilon_0 E^2$.Since the electric field $E = \f

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(B) The energy density $u$ of a parallel plate capacitor is given by the formula: $u = \frac{1}{2} \varepsilon_0 E^2$.Since the electric field $E = \frac{V}{d}$,we can write the energy density as $u = \frac{1}{2} \varepsilon_0 \left( \frac{V}{d} ight)^2$.Given values: $\varepsilon_0 = 8.85 	imes 10^{-12} \, F/m$,$V = 300 \, V$,and $d = 2 	imes 10^{-3} \, m$.Substituting these values: $u = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes \left( \frac{300}{2 	imes 10^{-3}} ight)^2$.$u = 0.5 	imes 8.85 	imes 10^{-12} 	imes (1.5 	imes 10^5)^2$.$u = 4.425 	imes 10^{-12} 	imes 2.25 	imes 10^{10}$.$u \approx 9.956 	imes 10^{-2} \approx 0.1 \, J/m^3$.

$A$ parallel plate capacitor having a plate separation of $2\, mm$ is charged by connecting it to a $300\, V$ supply. The energy density is.....$J/m^3$

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