Question

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. $\left(K+2\right)\frac{C}{4}$
• B. $\left(K+3\right)\frac{C}{4}$
• C. $\left(K+1\right)\frac{C}{4}$
• D. None of these

Answer 1

Answer: (B)

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
$C_{\text {net }} =C_{1}+C_{2} $
$=\frac{3 \varepsilon_{0} A}{4 d}+\frac{\varepsilon_{0} A K}{4 d} $
$=\frac{\varepsilon_{0} A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]=\frac{C}{4}(K+3)$