Question

A parallel plate capacitor with air as a dielectric has capacitance $C$. 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. $(K+1)\frac{C}{4}$
• C. $(K+2)\frac{C}{4}$
• D. $\frac{KC}{4}$

Answer 1

Answer: (A)

The two condensers with $K$ and with air are in parallel. With air, $C_1=\frac{{ϵ}_0}{d}\left(\frac{3A}{4}\right)=\frac{3{ϵ}_{0}A}{4d}$
With medium, $C_2=\frac{{ϵ}_{0}K}{d}\left(\frac{A}{4}\right)=\frac{{ϵ}_{0}AK}{4d}$
$\therefore C^{\prime }=C_1+C_2$
or $C^{\prime }=\frac{3{ϵ}_{0}A}{4d}+\frac{{ϵ}_{0}AK}{4d}=\frac{{ϵ}_{0}A}{d}[\frac{3}{4}+\frac{K}{4}]$
or $C^{\prime }=\frac{C}{4}(K+3)[\because C=\frac{A{ϵ}_0}{d}]$.