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A parallel plated capacitor has area $2 \mathrm{~m}^{2}$ separated by 3 dielectric slabs. Their relative permittivity is $2,3,6$ and thickness is $0.4 \mathrm{~mm}$, $0.6 \mathrm{~mm}, 1.2 \mathrm{~mm}$, respectively. The capacitance is
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Verified Answer
The correct answer is:
$2.95 \times 10^{-8} \mathrm{~F}$
The given situation is shown below,
$d_{1}=0.4 \mathrm{~mm}=4 \times 10^{-4} \mathrm{~m}$
$d_{2}=0.6 \mathrm{~mm}=6 \times 10^{-4} \mathrm{~m}$
$d_{3}=1.2 \mathrm{~mm}=1.2 \times 10^{-3} \mathrm{~m}$
The given arrangement will be equal to three capacitors connected in series.
i.e. $\quad C_{1}=\frac{\varepsilon_{0} K_{1} A}{d_{1}}, C_{2}=\frac{\varepsilon_{0} K_{2} A}{d_{2}}$
and
$$
C_{3}=\frac{\varepsilon_{0} K_{3} A}{d_{3}}
$$
Equivalent capacitance is given as,
$$
\begin{aligned}
&\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{\varepsilon_{0} A}\left(\frac{d_{1}}{K_{1}}+\frac{d_{2}}{K_{2}}+\frac{d_{3}}{K_{3}}\right) \\
&\frac{1}{C_{\mathrm{eq}}}=\frac{1}{\varepsilon_{0} A}\left[\frac{4 \times 10^{-4}}{2}+\frac{6 \times 10^{-4}}{3}+\frac{1.2 \times 10^{-3}}{6}\right] \\
&\Rightarrow \frac{1}{C_{\mathrm{eq}}}=\frac{1}{\varepsilon_{0} A}\left[6 \times 10^{-4}\right]
\end{aligned}
$$
$$
C_{\mathrm{eq}}=\frac{\varepsilon_{0} A}{6 \times 10^{-4}}=\frac{8.85 \times 10^{-12} \times 2}{6 \times 10^{-4}}=2.95 \times 10^{-8} \mathrm{~F}
$$
$d_{1}=0.4 \mathrm{~mm}=4 \times 10^{-4} \mathrm{~m}$
$d_{2}=0.6 \mathrm{~mm}=6 \times 10^{-4} \mathrm{~m}$
$d_{3}=1.2 \mathrm{~mm}=1.2 \times 10^{-3} \mathrm{~m}$
The given arrangement will be equal to three capacitors connected in series.
i.e. $\quad C_{1}=\frac{\varepsilon_{0} K_{1} A}{d_{1}}, C_{2}=\frac{\varepsilon_{0} K_{2} A}{d_{2}}$
and
$$
C_{3}=\frac{\varepsilon_{0} K_{3} A}{d_{3}}
$$
Equivalent capacitance is given as,
$$
\begin{aligned}
&\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{\varepsilon_{0} A}\left(\frac{d_{1}}{K_{1}}+\frac{d_{2}}{K_{2}}+\frac{d_{3}}{K_{3}}\right) \\
&\frac{1}{C_{\mathrm{eq}}}=\frac{1}{\varepsilon_{0} A}\left[\frac{4 \times 10^{-4}}{2}+\frac{6 \times 10^{-4}}{3}+\frac{1.2 \times 10^{-3}}{6}\right] \\
&\Rightarrow \frac{1}{C_{\mathrm{eq}}}=\frac{1}{\varepsilon_{0} A}\left[6 \times 10^{-4}\right]
\end{aligned}
$$
$$
C_{\mathrm{eq}}=\frac{\varepsilon_{0} A}{6 \times 10^{-4}}=\frac{8.85 \times 10^{-12} \times 2}{6 \times 10^{-4}}=2.95 \times 10^{-8} \mathrm{~F}
$$
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