Question

Figure shows a parallel plate capacitor having square plates of edge length $a$ and separation $d$. The gap between the plates is filled with a dielectric of dielectric constant $K$ which varies parallel to an edge as, $K=K_o+\alpha x$
Where, $K$ and $\alpha$ are constants and $x$ is the distance from the left end. Calculate the capacitance of system.

• A. $\frac{{ϵ}_{o}a{K}_o\alpha }{2d}$
• B. $\frac{{ϵ}_o{a}^2{K}_o}{d}$
• C. $\frac{{ϵ}_o{a}^2(K_o+a\alpha )}{2}$
• D. $\frac{{ϵ}_o{a}^2}{d}(K_o+\frac{a\alpha }{2})$

Answer 1

The correct option is D $\frac{{ϵ}_o{a}^2}{d}(K_o+\frac{a\alpha }{2})$


Consider a small strip of width $dx$ at a separation $x$ from the left end.

This small strip forms a small capacitor of plate area

$dA=adx$

The capacitance of small capacitor,

$dC=\frac{(K_o+\alpha x){ϵ}_o\left(adx\right)}{d}$

$[\because C=\frac{KA{ϵ}_o}{d}]$

Arrangement of such small strips starting from $x=0$ to $x=a$ will give rise to the entire capacitor.

And, since all small capacitors will be connected in parallel $($ because their end terminals are connected at same points i.e., $\text{A \& B})$

Therefore,

$C_{net}=\int dC$

$C_{net}=\underset{0}{\overset{a}{\int }}\frac{(K_o+\alpha x){ϵ}_{o}adx}{d}$

$\Rightarrow C_{net}=\frac{K_o{ϵ}_{o}a}{d}{\left[x\right]}_o^a+\frac{\alpha a{ϵ}_o}{d}{\left[\frac{x^2}{2}\right]}_o^a$

$\Rightarrow C_{net}=\frac{K_o{ϵ}_{o}a}{d}[a-0]+\frac{\alpha a{ϵ}_o}{2d}[a^2-0]$

$\therefore C_{net}=\frac{{ϵ}_o{a}^2}{d}[K_o+\frac{a\alpha }{2}]$

Therefore, right option is $\left(b\right)$

Why this question? Tip: It intends to challenge the understanding of combination of capacitors. If we look carefully we can infer that all small capacitors are connected in parallel, hence integration will be used to obtain summation of all capacitance. $or,C_{net}=\int dC$