The capacitance of a parallel plate condenser, with plate area A and distance between plates d, when filled with a dielectric whose dielectric constant varies as, is $ε (x) = ε_{0} + β . x, for 0 < x < \frac{d}{2}$ $ε (x) = ε_{0} + β (d - x) for \frac{d}{2} < x < d$

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{β \frac{A}{2} ϵ_{0}}{\log_{e} (\frac{ϵ_{0} + \frac{βd}{2}}{ϵ_{0}})}$

$\frac{β \frac{A}{2}}{\log_{e} (\frac{ϵ_{0}}{βd})}$

$\frac{A}{2} \log_{e} (\frac{ϵ_{0}}{\frac{βd}{2}})$

$\frac{A}{2} \log_{e} (\frac{ϵ_{0}}{ϵ_{0} - \frac{βd}{2}})$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Considering an element of thickness dx at a distance x :

${dC}_{1} = \frac{| ε | A}{dx}$

$\frac{1}{C} = \int \frac{1}{dC} = \int_{0}^{d / 2} \frac{dx}{(ϵ_{0} + βx) A} + \int_{d / 2}^{d} \frac{dx}{(ϵ_{0} + β (d - x)) A}$

$= [\frac{1}{A β} (\ln (ϵ_{0} + \frac{β d}{2}) - \ln (ϵ_{0}))] + [- \frac{1}{A β} (\ln (ϵ_{0}) - \ln (ϵ_{0} + \frac{βd}{2}))]$

$\Rightarrow C = \frac{β \frac{A}{2} ϵ_{0}}{\ln (\frac{ϵ_{0} + \frac{βd}{2}}{ϵ_{0}})}$

Watch 3-min video & get full concept clarity