Question

A dielectric slab is partially introduced between two square plates of area $A$ of a parallel plate capacitor as shown in the figure. Dielectric constant of slab is ${ϵ}_r$. The total capacitance of the system is :

• A. $\frac{{\in }_0\sqrt{Ax}}{d}$
• B. $\frac{{\in }_0}{d}({\in }_r\sqrt{A}x-A-\sqrt{A}x)$
• C. $\frac{{\in }_0}{d}(A-\sqrt{A}x+{\in }_r\sqrt{A}x)$
• D. $\frac{{\in }_0{\in }_r}{d}(A-\sqrt{A}x+A+{\in }_r\sqrt{A}x)$

Answer 1

Answer: (C)

$\frac{{\in }_0}{d}(A-\sqrt{A}x+{\in }_r\sqrt{A}x)$

As the plates are square so length $\left(l\right)=$ widith $\left(w\right)$.
here, $A=lw=l^2\to l=\sqrt{A}$
There are two capacitors - one is air filled and other is dielectric filled and they are in parallel.
For air filled capacitor: area $A_1=(l-x)l=(\sqrt{A}-x)\sqrt{A}=(A-\sqrt{A}x)$ and
Capacitance, $C_1=\frac{A_1{ϵ}_0}{d}=\frac{(A-\sqrt{A}x){ϵ}_0}{d}$
For dielectric field capacitor: area $A_2=lx=\sqrt{A}x$ and
Capacitance , $C_2=\frac{A_2{ϵ}_r{ϵ}_0}{d}=\frac{\sqrt{A}x{ϵ}_r{ϵ}_0}{d}$
The net capacitance is $C=C_1+C_2=\frac{(A-\sqrt{A}x){ϵ}_0}{d}+\frac{\sqrt{A}x{ϵ}_r{ϵ}_0}{d}=\frac{{ϵ}_0}{d}(A-\sqrt{A}x+{ϵ}_r\sqrt{A}x)$