Question

A capacitor is made of two square plates each of side ‘$a$ ’ making a very small angle between them, as shown in figure. The capacitance will be close to

• A. $\frac{{\epsilon }_0{a}^2}{d}\left(1-\frac{\alpha a}{2d}\right)$
• B. $\frac{{\epsilon }_0{a}^2}{d}\left(1-\frac{3\alpha a}{2d}\right)$
• C. $\frac{{\epsilon }_0{a}^2}{d}\left(1+\frac{\alpha a}{d}\right)$
• D. $\frac{{\epsilon }_0{a}^2}{d}\left(1-\frac{\alpha a}{4d}\right)$

Answer 1

The correct option is A

$\frac{{\epsilon }_0{a}^2}{d}\left(1-\frac{\alpha a}{2d}\right)$

Step 1: Given data

Angle between two square plates $=\alpha$

Length of the plate$=a$

Width of the plate $=d$

Step 2: To find the capacitance

In triangle $AEF$

$\frac{EF}{AF}=\tan \alpha$

$\Rightarrow$$\frac{y_1}{x}=\tan \alpha$

$\Rightarrow$ $y_1=x\tan \alpha$ …(1)

$EG=EF+FG$

$\Rightarrow$$EG=x\tan \alpha +d$

Assume a small element $dx$ at a distance $x$ from left end

Now, the capacitance of a small element $dx$ is,

$dC=\frac{A_{EG}.{\epsilon }_0}{d_{EG}}$

$dC=\frac{\left(dx.a\right){\epsilon }_0}{y_1+y_2}$

$dC=\frac{adx{\epsilon }_0}{y_1+y_2}$

$dC=\frac{{\epsilon }_{0}adx}{x\tan \alpha +d}$

Net capacitance $C_{net}={\int }_0^{a}dC$

$$\begin{gathered} ={\int }_0^a\frac{{\epsilon }_{0}adx}{x\tan \alpha +d} \\ ={\epsilon }_{0}a{\int }_0^a\frac{dx}{d+x\tan \alpha } \\ ={\left[\frac{{\epsilon }_{0}a\ln \left(d+x\tan \alpha \right)}{\tan \alpha }\right]}_0^a \\ =\frac{{\epsilon }_{0}a}{\tan \alpha }\left[\ln \left(d+a\tan \alpha \right)-\ln d\right] \end{gathered}$$

For small angles $\tan \alpha \simeq \alpha$

$C_{net}=\frac{a{\epsilon }_0}{\alpha }\ln \left(1+\frac{a}{d}\alpha \right)$ …(1)

$\because \ln (1+x)=x-\frac{x^2}{2}$ …(2) where x<<1

From equations (1) and (2), we get

$\begin{array}{rcl}{C}_{net}& =& \frac{a{\epsilon }_0}{\alpha }\left(\frac{a\alpha }{\alpha }-\frac{{\left({\displaystyle \frac{a\alpha }{d}}\right)}^2}{2}\right)\\ & =& \frac{a{\epsilon }_0}{\alpha }\left(\frac{a\alpha }{d}\right)\left(1-\frac{a\alpha }{2d}\right)\\ & =& \frac{a^2{\epsilon }_0}{d}\left(1-\frac{a\alpha }{2d}\right)\end{array}$

Therefore, the capacitance is $\frac{{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{d}}\left(1-\frac{\alpha a}{2d}\right)$

Hence, the correct answer is option A.