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