Question

A parallel plate capacitor has capacitance C when no dielectric between the plates. Now a slab of dielectric constant K, having 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

• A. $(K+3)\frac{C}{4}$
• B. $(K+2)\frac{C}{4}$
• C. $(K+1)\frac{C}{4}$
• D. $\frac{KC}{4}$

Answer 1

The correct option is A $(K+3)\frac{C}{4}$

Given system can be considered as a combination of two capacitors in parallel. One capacitor is filled with dielectric while other is not
According to the given condition,
$C_1=\frac{\epsilon }{0}\left(\frac{3A}{4}\right)=\frac{3A{\epsilon }_0}{4d}$

$C_2=\left(\frac{A}{4}\right)\left(\frac{{\epsilon }_0}{4}\right)=\frac{A{\epsilon }_{0}k}{4d}$

Now,
$C_{eq}=C_1+C_2=\frac{3A{\epsilon }_{0}K}{4d}+\frac{A{\epsilon }_{0}K}{4d}$

$=\frac{A{\epsilon }_0}{d}[\frac{3}{4}+\frac{K}{4}]=\frac{C}{4}(K+3)$

Hence, correct option is (C).