A parallel plate capacitor having a plate sep | vedclass

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Question

(b) 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}

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(b) The energy density of parallel plate capacitor is given by $U = \frac{1}{2}{\varepsilon _0}{E^2} = \frac{1}{2}{\varepsilon _0}{\left( {\frac{V}{d}} ight)^2}$
$ = \frac{1}{2} 	imes 8.85 	imes {10^{ - 12}}\,{C^2}/N{m^2} 	imes {\left( {\frac{{300\,volt}}{{2 	imes {{10}^{ - 3}}\,m}}} ight)^2}$$ = 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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