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
$C_1=\frac{{\epsilon }_0}{d}\left(\frac{3A}{4}\right)=\frac{3{\epsilon }_{0}A}{3d}$
Similarly, the condenser with $K$ as the dielectric constant has capacitance
$C_2=\frac{{\epsilon }_{0}K}{d}\left(\frac{A}{4}\right)=\frac{{\epsilon }_{0}AK}{4d}$
Since, $C_1$ and $C_2$ are in parallel
$C_{\text{net }}=C_1+C_2$
$=\frac{3{\epsilon }_{0}A}{4d}+\frac{{\epsilon }_{0}AK}{4d}$
$=\frac{{\epsilon }_{0}A}{d}\left[\frac{3}{4}+\frac{K}{4}\right]=\frac{C}{4}(K+3)$