Question

A parallel plate capacitor is made of square plates, each of length $l$ separated by a distance $t$ as shown. Capacity when the plates are making small angle $\theta$ with each other as shown in the diagram is $C$, if the square plates are of length $2l$ and separated by distance $2t$ and make the same angle $\theta$, then capacity is $C^{\prime }$ which of the following correctly represent the relation between $C$ and $C^{\prime }$

• A. $C^{\prime }=C$
• B. $C^{\prime }=2C$
• C. $C^{\prime }=3C$
• D. $C^{\prime }=4C$

Answer 1

The correct option is B $C^{\prime }=2C$

Consider a small strip $dx$ of medium between the plates, as shown.



So, capacitance of the capacitor of this strip is given by

$dC=\frac{A{ϵ}_0}{d}=\frac{\left(ldx\right){ϵ}_0}{(t+x\theta )}$

Thus, the capacity of the entire capacitor will be

$C=\int dC={\int }_0^l\frac{\left(ldx\right){ϵ}_0}{t+x\theta }$

$C=\frac{{ϵ}_{0}l}{t}{\int }_0^l{(1+\frac{x\theta }{t})}^{-1}dx$

Applying binomial expansion for small angle $\theta$, we get

${(1+\frac{x\theta }{t})}^{-1}\approx (1-\frac{x\theta }{t})$

$\therefore C=\frac{l{ϵ}_0}{t}{\int }_0^l(1-\frac{x\theta }{t})dx$

$C=\frac{l{ϵ}_0}{t}\left[{\int }_0^{l}dx-\frac{\theta }{t}{\int }_0^{l}xdx\right]=\frac{l{ϵ}_0}{t}(l-\frac{\theta l^2}{2t})$

$\therefore C=\frac{l^2{ϵ}_0}{t}(1-\frac{l\theta }{2t})$

Similarly for second case, $C^{\prime }=\frac{2l^2{ϵ}_0}{t}(1-\frac{l\theta }{2t})$

Thus, $C^{\prime }=2C$

Hence, option (b) is the correct answer.