Capacitors

JEE Main level, single correct, medium

105 of 180

Question 105 of 180


A parallel plate capacitor has plates of area A separated by distance dd between them. It is filled with a dielectric which has a dielectric constant that varies as K(x)=K0(1+αx)K\left(x\right)={K}_{0}\left(1+\alpha x\right) where xx is the distance measured from one of the plates. If (αd)<<1,\left(\alpha d\right)\lt \lt 1, the total capacitance of the system is best given by the expression:

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Answer: (A) AK0ϵ0d(1+αd2)\dfrac{A{K}_{0}{\epsilon }_{0}}{d}\left(1+\dfrac{\alpha d}{2}\right)

Capacitance of element, dC=K0(1+αx)ϵ0AdxdC=\dfrac{{K}_{0}\left(1+\alpha x\right){\epsilon }_{0}A}{dx}
∫1dC=∫0ddxK0ϵ0A(1+αx)\int \dfrac{1}{dC}={\int }_{0}^{d}\dfrac{dx}{{K}_{0}{\epsilon }_{0}A\left(1+\alpha x\right)}
⇒1C=1K0ϵ0Aαln⁡(1+αd)\Rightarrow \dfrac{1}{C}=\dfrac{1}{{K}_{0}{\epsilon }_{0}A\alpha }\ln \left(1+\alpha d\right)
Given αd≪1\alpha d≪1
⇒1C=1K0ϵ0Aα(αd−α2d22)\Rightarrow \dfrac{1}{C}=\dfrac{1}{{K}_{0}{\epsilon }_{0}A\alpha }\left(\alpha d-\dfrac{{\alpha }^{2}{d}^{2}}{2}\right)
⇒1C=dK0ϵ0A(1−αd2)\Rightarrow \dfrac{1}{C}=\dfrac{d}{{K}_{0}{\epsilon }_{0}A}\left(1-\dfrac{\alpha d}{2}\right)
⇒C=K0ϵ0Ad(1+αd2)\Rightarrow C=\dfrac{{K}_{0}{\epsilon }_{0}A}{d}\left(1+\dfrac{\alpha d}{2}\right)

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