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A capacitor of capacitance $9 \mathrm{nF}$ having dielectric slab of $\varepsilon_r=2.4$, dielectric strength $20 \mathrm{MV} \mathrm{m}^{-1}$, and potential difference $=20 \mathrm{~V}$. Calculate area of plates.
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The correct answer is:
$4.2 \times 10^{-4} \mathrm{~m}^2$
Here, $C=9 \mathrm{nF}, \varepsilon_r=2.4, V=20$ volt
Dielectric strength $=20 \mathrm{MV} \mathrm{m}^{-1}$
Let separation between plates $=d$
$\begin{aligned} & E=\frac{V}{d} \Rightarrow 20 \times 10^6=\frac{20}{d} \\ & d=10^{-6} \mathrm{~m}\end{aligned}$
Now, $C=\frac{\varepsilon_0 A \varepsilon_r}{d}$
$\begin{aligned} & 9 \times 10^{-9}=\frac{8.85 \times 10^{-12} \times A \times 2.4}{10^{-6}} \\ & A=\frac{9 \times 10^{-15}}{8.85 \times 2.4 \times 10^{-12}}=4.2 \times 10^{-4} \mathrm{~m}^2\end{aligned}$
Dielectric strength $=20 \mathrm{MV} \mathrm{m}^{-1}$
Let separation between plates $=d$
$\begin{aligned} & E=\frac{V}{d} \Rightarrow 20 \times 10^6=\frac{20}{d} \\ & d=10^{-6} \mathrm{~m}\end{aligned}$
Now, $C=\frac{\varepsilon_0 A \varepsilon_r}{d}$
$\begin{aligned} & 9 \times 10^{-9}=\frac{8.85 \times 10^{-12} \times A \times 2.4}{10^{-6}} \\ & A=\frac{9 \times 10^{-15}}{8.85 \times 2.4 \times 10^{-12}}=4.2 \times 10^{-4} \mathrm{~m}^2\end{aligned}$
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