Question

A capacitor, as shown in figure has square plates of length l and are inclined at an angle $\theta$ with one another. For small value of $\theta$ , capacitance is given by:

• A. $\frac{{ϵ}_0{l}^2}{d}(1-\frac{\theta l}{2d})$
• B. $\frac{{ϵ}_0{l}^2}{2d}(1-\frac{\theta l}{d})$
• C. $\frac{{ϵ}_0{l}^2}{d}(1+\frac{\theta l}{d})$
• D. $\frac{{ϵ}_0{l}^2}{2d}(1+\frac{\theta l}{d})$

Answer 1

Answer: (A)

$\frac{{ϵ}_0{l}^2}{d}(1-\frac{\theta l}{2d})$

At a distance $x$ on the inclined plate, let us consider a small segment of length $dx$.

The small segment is at a distance of $D$ from lower plate.
So, capacitance due to the small segment:

$dC=\frac{{ϵ}_{0}dA}{D}=\frac{{ϵ}_{0}dxl}{(d+x\theta )}$ as $D=d+x\sin \theta$ and $\sin \theta \approx \theta$ for small $\theta$

Integrating above, we get:

$C_{net}={\int }_0^l\frac{{ϵ}_{0}dA}{D}$$=\frac{{ϵ}_{0}l}{\theta }{\int }_0^l\frac{\theta dx}{(d+x\theta )}$$=\frac{{ϵ}_{0}l}{\theta }\times ln\frac{(d+l\theta )}{d}$
$=\frac{{ϵ}_{0}l}{\theta }\times ln(1+\frac{l\theta }{d})$
Using logarithm expansions,
$C_{net}=\frac{{ϵ}_{0}l}{\theta }(\frac{l\theta }{d}-\frac{1}{2}\times \frac{(l\theta {)}^2}{d^2})$
$=\frac{{ϵ}_0{l}^2}{d}(1-\frac{\theta l}{2d})$