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.....$J/m^3$
Medium
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(b) The energy density of parallel plate capacitor is given by $U = \frac{1}{2}{\varepsilon _0}{E^2} = \frac{1}{2}{\varepsilon _0}{\left( {\frac{V}{d}} \right)^2}$
$ = \frac{1}{2} \times 8.85 \times {10^{ - 12}}\,{C^2}/N{m^2} \times {\left( {\frac{{300\,volt}}{{2 \times {{10}^{ - 3}}\,m}}} \right)^2}$$ = 0.1\,J/{m^3}$
$ = \frac{1}{2} \times 8.85 \times {10^{ - 12}}\,{C^2}/N{m^2} \times {\left( {\frac{{300\,volt}}{{2 \times {{10}^{ - 3}}\,m}}} \right)^2}$$ = 0.1\,J/{m^3}$
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