Question 99 of 180

A parallel plate capacitor with plate area '$A$' and distance of separation '$d$' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as:
$\epsilon \left(x\right)={\epsilon }_{0}+kx$, for $\left(0\lt x\leq \dfrac{d}{2}\right)$
$\epsilon \left(x\right)={\epsilon }_{0}+k\left(d-x\right)$, for $\left(\dfrac{d}{2}\leq x\leq d\right)$

A${\left({\epsilon }_{0}+\dfrac{kd}{2}\right)}^{2/kA}$B$\dfrac{kA}{2ln\left(\dfrac{2{\epsilon }_{0}+kd}{2{\epsilon }_{0}}\right)}$C$0$D$\dfrac{kA}{2}\ln \left(\dfrac{2{\epsilon }_{0}}{2{\epsilon }_{0}-kd}\right)$

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Answer: (B) $\dfrac{kA}{2ln\left(\dfrac{2{\epsilon }_{0}+kd}{2{\epsilon }_{0}}\right)}$

Taking an element of width dx at a distance $x\left(x\lt d/2\right)$ from left plate
$dc=\dfrac{({\epsilon }_{0}+kx)A}{dx}$
Capacitance of half of the capacitor
$\dfrac{1}{C}={\int }_{0}^{d/2}\dfrac{1}{dc}=\dfrac{1}{A}{\int }_{0}^{d/2}\dfrac{dx}{{\epsilon }_{0}+kx}$
$\dfrac{1}{C}=\dfrac{1}{kA}ln\left(\dfrac{{\epsilon }_{0}+kd/2}{{\epsilon }_{0}}\right)$
Capacitance of second half will be same
${C}_{eq}=\dfrac{C}{2}=\dfrac{kA}{2ln\left(\dfrac{2{\epsilon }_{0}+kd}{2{\epsilon }_{0}}\right)}$