question_answer

A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as \[k(x)=K(1+\alpha x)\] where 'x' is the distance measured from one of the plates. If \[(\alpha d)<<1\], the total capacitance of the system is best given by the expression [JEE MAIN Held on 07-01-2020 Morning]

A) \[\frac{AK{{\varepsilon }_{0}}}{d}(1+\alpha d)\]

B) \[\frac{A{{\varepsilon }_{0}}K}{d}\left( 1+\frac{{{\alpha }^{2}}{{d}^{2}}}{2} \right)\]

C) \[\frac{AK{{\varepsilon }_{0}}}{d}\left( 1+\frac{\alpha d}{2} \right)\]

D) \[\frac{A{{\varepsilon }_{0}}K}{d}\left( 1+{{\left( \frac{\alpha d}{2} \right)}^{2}} \right)\]

Correct Answer: C

Solution :

[c] \[k=K(1+\alpha x)\] \[{{C}_{el}}=\frac{{{\varepsilon }_{0}}K(1+\alpha x)A}{dx}\] \[\therefore \int{d\left( \frac{1}{C} \right)=\frac{1}{{{C}_{el}}}=\int\limits_{0}^{d}{\left( \frac{dx}{{{\varepsilon }_{0}}KA(1+\alpha x)} \right)}}\] \[\Rightarrow \frac{1}{C}=\frac{1}{{{\varepsilon }_{0}}KA\alpha }[In(1+\alpha x)]_{0}^{d}\] \[\Rightarrow \frac{1}{C}=\frac{1}{{{\varepsilon }_{0}}KA\alpha }In(1+\alpha d)\] \[=\frac{1}{{{\varepsilon }_{0}}KA\alpha }\left[ \alpha d-\frac{{{\alpha }^{2}}{{d}^{2}}}{2} \right]\] \[=\frac{1}{{{\varepsilon }_{0}}KA}\left[ 1-\frac{\alpha d}{2} \right]\] \[\therefore C=\frac{{{\varepsilon }_{0}}KA}{d\left( 1-\frac{\alpha d}{2} \right)}\Rightarrow C=\frac{{{\varepsilon }_{0}}KA}{d}\left( 1+\frac{\alpha d}{2} \right)\]