Question

A parallel plate capacitor has plates of area $ A $separated by distance $ ‘d’$ between them. It is filled with a dielectric which has a dielectric constant varies as $ k \left(x\right) = k(1 + \alpha x)$, where $ ‘x’$ is the distance measured from one of the plates. If $ (\alpha d <<1)$, the total capacitance of the system is best given by the expression:

/I /)

• A. $\begin{array}{l}\frac{A{ϵ}_{0}k}{d}\left[1+(\frac{\alpha d}{2}{)}^2\right]\end{array}$
• B. $\begin{array}{l}\frac{A{ϵ}_{0}k}{d}\left[1+\left(\frac{\alpha d}{2}\right)\right]\end{array}$
• C. $\begin{array}{l}\frac{A{ϵ}_{0}k}{d}\left[1+\left(\frac{{\alpha }^{2}d}{2}\right)\right]\end{array}$
• D. $\begin{array}{l}\frac{A{ϵ}_{0}k}{d}\left[1+\alpha d)\right]\end{array}$

Answer 1

The correct option is B

$\begin{array}{l}\frac{A{ϵ}_{0}k}{d}\left[1+\left(\frac{\alpha d}{2}\right)\right]\end{array}$

Given, $k\left(x\right)=k(1+\alpha x)$

[where $‘x’$ is the distance measured from one of the plates]

$dC=\frac{A\epsilon 0k}{dx}$

Since all capacitance is in series, we can apply

$\begin{array}{rcl}\frac{1}{C_{eq}}& =& \int \frac{1}{dc}\\ & =& {\int }_0^d\frac{dx}{k\left(1+\alpha x\right){\epsilon }_{0}A}\\ & =& {\left[\frac{\ln \left(1+\alpha x\right)}{k{\epsilon }_{0}A\alpha }\right]}_0^d\end{array}$

On putting the limits from $0$ to $d$

$\frac{1}{C_{eq}}=\frac{\ln (1+\alpha d)}{k{\epsilon }_{0}A\alpha }$

Using expression, $\ln (1+x)=x–x^2/2+.......$

And putting $x=\alpha d$ where, $x$ approaches to $0$

$$\begin{gathered} \begin{array}{l}\frac{1}{C}=\frac{d}{k{ϵ}_{0}A\alpha }\left[\alpha d-{{\alpha }^{2}d}^2/2)\right]\end{array} \\ \begin{array}{l}\Rightarrow C=\frac{A{ϵ}_{0}k}{d\left(1-\frac{\alpha d}{2}\right)}\end{array} \\ \begin{array}{l}\Rightarrow C=\frac{{ϵ}_{0}kA}{d}\left(1+\frac{\alpha d}{2}\right)\end{array} \end{gathered}$$

Hence, option $\left(B\right)$ is correct.