JEE MainPhysicsDifficultElectrostatics

[IMAGE] A and d are the plate area and distance between plates of a parallel
plate capacitor repectively. The dielectric constant of dielectric
filled inside a parallel plate capacitor is variable and given as
$k(x) = K(1 + \gamma x$ ) where is the distancemeasured from the
left plate as shown in the figure. If $(\gamma d) < < 1$ , then find the
capacitance of capacitor with dielectric.

Show answer and solution

Capacitance of element

= \frac{k\varepsilon_{0}A}{dx}

\sum\frac{1}{C^{'}} = \int_{0}^{d}\mspace{2mu}\frac{dx}{K\varepsilon_{0}A(1 + \gamma x)}

\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}ln(1 + \gamma d)

{\ Given\ \ \gamma d < < 1

C = \right)$$

Source: Grass-G / P-20 Physcis Paper 12 Dec.docx

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