Question bank/JEE Main/Physics

[IMAGE] A parallel plate capacitor has plates of area A separated by distance
between them. It is filled with adielectric which has a
dielectric constant that varies as k(x) = K(1 + $\alpha$x) where
is the distancemeasured from one of the plates. If ($\alpha$d) << 1,
the total capacitance of the system is best given by theexpression

A$\frac{A \in_{0}K}{d}\left( 1 + \frac{\alpha^{2}d^{2}}{2} \right)b$B$\frac{AK \in_{0}}{d}\left( 1 + \frac{\alpha d}{2} \right)$C$\frac{A \in_{0}K}{d}\left( 1 + \left( \frac{\alpha d}{2} \right)^{2} \right)$D$\frac{AK\epsilon_{0}}{d}(1 + \alpha d)$

Show answer and solution▾

Capacitance of element$= \frac{k\varepsilon_{0}A}{dx}$
[IMAGE] Capacitance of
element$C^{'} = \frac{K(1 + \alpha x)\varepsilon_{0}A}{dx}$
$\sum\frac{1}{C^{'}} = \int_{0}^{d}\mspace{2mu}\frac{dx}{K\varepsilon_{0}A(1 + \alpha x)}$
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}\mathcal{l}n(1 + \alpha d)$
 Given $- \ \alpha d < < 1$
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}\left( \alpha d - \frac{\alpha^{2}d^{2}}{2} \right)$
$\frac{1}{C} = \frac{d}{K\varepsilon_{0}A}\left( 1 - \frac{\alpha d}{2} \right)$
$C = \frac{K\varepsilon_{0}A}{d}\left( 1 + \frac{\alpha d}{2} \right)$

Source: Grass-G / P-12 Physics paper 2 Nov.docx