Question

A capacitor is formed by two square metal plates of edge $1\text{m}$, separated by a distance $1\text{mm}$. Dielectrics of dielectric constant $K_1$ and $K_2$ are filled in the gap as shown in figure. The new capacitance is


$[$ Take $\ln 2=0.7]$

• A. $25\text{nF}$
• B. $10\text{nF}$
• C. $20\text{nF}$
• D. $15\text{nF}$

Answer 1

The correct option is A $25\text{nF}$


Let us consider a small strip of width $dx$ at $x$ distance from the left end.

From figure,

$\tan \varphi =\frac{1\times {10}^{-3}}{1}=1\times {10}^{-3}$

$\therefore t=x\tan \varphi ={10}^{-3}x$

We can see that both small capacitors formed by the small strip are in series.

So, $\frac{1}{d{C}_{eq}}=\frac{1}{d{C}_1}+\frac{1}{d{C}_2}$

$\Rightarrow \frac{1}{d{C}_{eq}}=\frac{d_1}{K_1{ϵ}_0{A}_1}+\frac{d_2}{K_2{ϵ}_0{A}_2}$

$[$ Area of small plate $=1\times dx=dx]$

$\Rightarrow \frac{1}{d{C}_{eq}}=\frac{{10}^{-3}-{10}^{-3}x}{2{ϵ}_{0}dx}+\frac{{10}^{-3}x}{4{ϵ}_{0}dx}$

$\Rightarrow d{C}_{eq}=\frac{4{ϵ}_{0}dx}{2\times {10}^{-3}-{10}^{-3}x}$

$\Rightarrow C_{eq}=\int d{C}_{eq}=\frac{4{ϵ}_0}{{10}^{-3}}\underset{0}{\overset{1}{\int }}\frac{dx}{2-x}$

$\Rightarrow C_{eq}=\frac{4\times 8.85\times {10}^{-12}}{{10}^{-3}}{[-\ln (2-x\left)\right]}_0^1[\because \int \frac{dx}{x}=\ln x+C]$

$\Rightarrow C_{eq}=3.6\times {10}^{-8}[-\ln 1+\ln 2]$

$\Rightarrow C_{eq}=2.5\times {10}^{-8}\text{F}=25\text{nF}$

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

$$\begin{array}{c}\text{Why this question ?}\\ \text{Tip-we use concept of integration for such}\\ \text{kind of problem. This question helps you}\\ \text{to recall your concept of integration.}\end{array}$$