Q119 / 120 · VITEEE 2013 · Physics
The capacitance of a parallel plate capacitor with air as dielectric is $C$. If a slab of dielectric constant $K$ and of the same thickness as the separation between the plates is introduced so as to fill $1 / 4$ th of the capacitor (shown in figure), then the new capacitance is
- A$(K+2) \frac{C}{4}$
- Correct answerB$(K+3) \frac{C}{4}$
- C$(K+1) \frac{C}{4}$
- DNone of these
Solution
Capacitance, $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$
As one-fourth of capacitor is filled with dielectric of constant $\mathrm{K}$, then,
$\mathrm{C}_{1}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A} / 4}{\mathrm{~d}}$
and $C_{2}=\frac{\varepsilon 3 A / 4}{d}$
Both $C_{1}$ and $C_{2}$ are in parallel.
$\begin{array}{l}
\therefore \mathrm{C}_{\mathrm{P}}=\mathrm{C}_{1}+\mathrm{C}_{2}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{4 \mathrm{~d}}+\frac{3 \varepsilon_{0} \mathrm{~A}}{4 \mathrm{~d}} \\
=(\mathrm{K}+3) \frac{\varepsilon_{0} \mathrm{~A}}{4 \mathrm{~d}}=(\mathrm{K}+3) \frac{\mathrm{C}}{4}
\end{array}$
Hence, the correct answer is option: 'B'.