A parallel plate capacitor with air between the plates has a capacitance of \(9 \mathrm{pF}\). The…
A parallel plate capacitor with air between the plates has a capacitance of \(9 \mathrm{pF}\). The separation between its plates is ' \(d\) '. The space between the plates is now filled with two dielectrics. One dielectric has dielectric constant \(K_{1}=3\) and thickness \(\frac{d}{3}\) while the other one has dielectric constant \(K_{2}=6\) and thickness \(\frac{2 d}{3} .\) Capacitance of the capacitor is now
\(40.5 \mathrm{pF}\)
\(20.25 \mathrm{pF}\)
\(1.8 \mathrm{pF}\)
\(45 \mathrm{pF}\)
Solution
Given $C_{0}=\frac{\varepsilon_{0} A}{d}=9 \mathrm{pF}$
$\begin{aligned}
C_{1}&=\frac{k_{1} \varepsilon_{0} A}{d / 3}=\frac{3 \varepsilon_{0} A}{d / 3}=\frac{9 \varepsilon_{0} A}{d}=9 C_{0} \\
C_{2}&=\frac{k_{2} \varepsilon_{0} A}{2 d / 3}=\frac{6 \varepsilon_{0} A}{2 d / 3}=\frac{9 \varepsilon_{0} A}{d}=9 C_{0}
\end{aligned}$
Capacitors $C_{1}$ and $C_{2}$ are in series. The equivalent capacitance is
$\begin{aligned}
\mathrm{C} &=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{9 C_{0} \times 9 C_{0}}{18 C_{0}}=\frac{9}{2} C_{0} \\
&=\frac{9}{2} \times 9 \mathrm{pF}=40.5 \mathrm{pF}
\end{aligned}$
.
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