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A parallel plate capacitor with air as the dielectric has capacitance $C$. A slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be
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Verified Answer
The correct answer is:
$(K+3) \frac{C}{4}$
The condenser with air as the dielectric has capacitance
$$
C_{1}=\frac{\varepsilon_{0}}{d}\left(\frac{3 A}{4}\right)=\frac{3 \varepsilon_{0} A}{3 d}
$$
Similarly, the condenser with $K$ as the dielectric constant has capacitance
$$
C_{2}=\frac{\varepsilon_{0} K}{d}\left(\frac{A}{4}\right)=\frac{\varepsilon_{0} A K}{4 d}
$$
Since, $C_{1}$ and $C_{2}$ are in parallel
$$
\begin{aligned}
C_{n e t} &=C_{1}+C_{2} \\
&=\frac{3 \varepsilon_{0} A}{4 d}+\frac{\varepsilon_{0} A K}{4 d} \\
&=\frac{\varepsilon_{0} A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]=\frac{C}{4}(K+3)
\end{aligned}
$$
$$
C_{1}=\frac{\varepsilon_{0}}{d}\left(\frac{3 A}{4}\right)=\frac{3 \varepsilon_{0} A}{3 d}
$$
Similarly, the condenser with $K$ as the dielectric constant has capacitance
$$
C_{2}=\frac{\varepsilon_{0} K}{d}\left(\frac{A}{4}\right)=\frac{\varepsilon_{0} A K}{4 d}
$$
Since, $C_{1}$ and $C_{2}$ are in parallel
$$
\begin{aligned}
C_{n e t} &=C_{1}+C_{2} \\
&=\frac{3 \varepsilon_{0} A}{4 d}+\frac{\varepsilon_{0} A K}{4 d} \\
&=\frac{\varepsilon_{0} A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]=\frac{C}{4}(K+3)
\end{aligned}
$$
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