Question

A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the 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. $\frac{(K+3)C}{4}$
• B. $\frac{(K+2)C}{4}$
• C. $\frac{(K+1)C}{4}$
• D. $\frac{KC}{4}$

Answer 1

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

Capacitance $=\frac{ϵA}{d}$

Initially for air: $C=\frac{{ϵ}_{o}A}{d}$

Finally for air: $C=\frac{{ϵ}_o\frac{3A}{4}}{d}=\frac{3C}{4}$

For dielectric: $C_k=\frac{K{ϵ}_o\frac{A}{4}}{d}=\frac{KA}{4}$

We can consider them as two capacitors in series which gives,

$C_{net}=\frac{3C}{4}+\frac{KC}{4}=\frac{(K+3)C}{4}$