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 1/4th of the capacitor (shown in figure), then the new capacitance is

• A. $(K+2)\frac{C}{4}$
• B. $(K+3)\frac{C}{4}$
• C. $(K+1)\frac{C}{4}$
• D. None of these

Answer 1

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

Given,

Capacitance of a parallel plate capacitor with air as dielectric is $=C$

Dielectric constant of the slab$=K$

Sepration between the plates is $={\frac{1}{4}}^{th}$

Capacitance, $C=\frac{{ϵ}_{0}A}{d}$
As one-fourth of capacitor is filled with dielectric of constant K, then,
$C_1=\frac{K{ϵ}_{0}A/4}{d}$
and $C_2=\frac{ϵ3A/4}{d}$
Both $C_1$ and $C_2$ are in parallel.
$\therefore C_p=C_1+C_2=\frac{K{ϵ}_{0}A}{4d}+\frac{3{ϵ}_{0}A}{4d}$
$=(K+3)\frac{{ϵ}_{0}A}{4d}=(K+3)\frac{C}{4}$

So, The new capacitance $=(K+3)\frac{C}{4}$.