Question

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 :
$\epsilon (x)={\epsilon }_0+kx$, for $\left(0<x\le \frac{d}{2}\right)$
$\epsilon (x)={\epsilon }_0+k(d-x)$, for $\left(\frac{d}{2}\le x\le d\right)$

• A. ${\left({\epsilon }_0+\frac{kd}{2}\right)}^{2/kA}$
• B. $\frac{kA}{2\ln \left(\frac{2{\epsilon }_0+kd}{2{\epsilon }_0}\right)}$
• C. $0$
• D. $\frac{kA}{2}\ln \left(\frac{2{\epsilon }_0}{2{\epsilon }_0-kd}\right)$

Answer 1

Answer: (B)

Taking an element of width $dx$ at a distance $x(x<d/2)$ from left plate $d_c=\frac{\left({\epsilon }_0+kx\right)A}{dx}$
Capacitance of half of the capacitor
$\frac{1}{C}=\underset{0}{\overset{d/2}{\int }}\frac{1}{dc}=\frac{1}{A}\underset{0}{\overset{d/2}{\int }}\frac{dx}{{\epsilon }_0+kx}$
$\frac{1}{C}=\frac{1}{kA}\ln \left(\frac{{\epsilon }_0+kd/2}{{\epsilon }_0}\right)$
Capacitance of second half will be same
$C_{eq}=\frac{C}{2}=\frac{kA}{2\ln \left(\frac{2{\epsilon }_0+kd}{2{\epsilon }_0}\right)}$