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The plates of a parallel plate capacitor have an area of $90 \mathrm{~cm}^2$ each and area separated by $2.5 \mathrm{~mm}$. The capacitor is charged by connecting it to a $400 \mathrm{~V}$ supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$, Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$, Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.
Solution:
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Verified Answer
(a) Given, $\mathrm{A}=90 \mathrm{~cm}^2=90 \times 10^{-4} \mathrm{~m}^2$, $\mathrm{d}=2.5 \mathrm{~mm}=2.5 \times 10^{-3} \mathrm{~m}, \mathrm{~V}=400 \mathrm{~V}, \mathrm{C}=$ ?
By formula, capacitance
$$
\begin{aligned}
\mathrm{C} &=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{1 \times 90 \times 10^{-4}}{4 \pi \times 9 \times 10^9 \times 2.5 \times 10^{-3}} \\
&=\frac{10^{-8}}{4 \times 3.14 \times 25}=\frac{1}{3.14} \times 10^{-10} \mathrm{~F} \text { and } \\
\mathrm{E} &=\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times \frac{1}{3.14} \times 10^{-10} \times 400 \times 400 \\
&=\frac{8}{3.14} \times 10^{-6}=2.55 \times 10^{-6} \mathrm{~J}
\end{aligned}
$$
(b) Volume of medium between parallel plates,
$$
\begin{aligned}
\mathrm{V} &=\mathrm{A} \times \mathrm{d}=90 \times 10^{-4} \times 2.5 \times 10^{-3} \\
&=225 \times 10^{-7} \mathrm{~m}^3
\end{aligned}
$$
Hence, energy per unit volume,
$$
\mathrm{u}=\frac{2.55 \times 10^{-6}}{225 \times 10^{-7}}=0.113 \mathrm{Jm}^{-3} \text {. }
$$
By formula, capacitance
$$
\begin{aligned}
\mathrm{C} &=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{1 \times 90 \times 10^{-4}}{4 \pi \times 9 \times 10^9 \times 2.5 \times 10^{-3}} \\
&=\frac{10^{-8}}{4 \times 3.14 \times 25}=\frac{1}{3.14} \times 10^{-10} \mathrm{~F} \text { and } \\
\mathrm{E} &=\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times \frac{1}{3.14} \times 10^{-10} \times 400 \times 400 \\
&=\frac{8}{3.14} \times 10^{-6}=2.55 \times 10^{-6} \mathrm{~J}
\end{aligned}
$$
(b) Volume of medium between parallel plates,
$$
\begin{aligned}
\mathrm{V} &=\mathrm{A} \times \mathrm{d}=90 \times 10^{-4} \times 2.5 \times 10^{-3} \\
&=225 \times 10^{-7} \mathrm{~m}^3
\end{aligned}
$$
Hence, energy per unit volume,
$$
\mathrm{u}=\frac{2.55 \times 10^{-6}}{225 \times 10^{-7}}=0.113 \mathrm{Jm}^{-3} \text {. }
$$
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