Question

The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value if a dielectric slab of thickness $t=\frac{d}{2}$ is inserted between the plates (where $d$ is the distance of separation between the plates). What is the dielectric constant of the slab?

• A. $K=2$
• B. $K=\frac{!}{2}$
• C. $K=1$
• D. $K=\sqrt{2}$

Answer 1

The correct option is A $K=2$

Original value of parallel plate capacitance $C_o=\frac{A{ϵ}_o}{d}$

Let the dielectric constant of material be $K$.

Capacitance of capacitor with material $C_1=\frac{KA{ϵ}_o}{d/2}$

Capacitance of capacitor without material $C_2=\frac{A{ϵ}_o}{d/2}$

Equivalent capacitance of series combination $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{2K}{K+1}\frac{A{ϵ}_o}{d}=\frac{2K}{K+1}{C}_o$

Given, $C_{eq}=\frac{4}{3}{C}_o$

$\therefore$ We get $\frac{2K}{K+1}=\frac{4}{3}$

$⟹K=2$