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The plates of a parallel plate capacitor have an area of $90 c m^{2}$ each and are separated by 2 mm. The capacitor is charged by connecting it to a 400 V supply. Then the energy density of the energy stored $(i n J m^{- 3})$ in the capacitor is $(t a k e ε_{0} = 8.8 \times 10^{- 12} F m^{- 1})$

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Detailed Solution

$A = 90 c m^{2}, d = 2 m m$

$E = \frac{V}{d}, V = 400 V$

$u = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} ε_{0} {(\frac{V}{d})}^{2}$

$= \frac{1}{2} (8.85 \times 10^{- 12}) {(\frac{400}{2 \times 10^{- 3}})}^{2} = 0.177 J m^{- 3}$

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