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 :
$\varepsilon(x)=\varepsilon_{0}+k x$, for $\left(0 < x \leq \frac{d}{2}\right)$
$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for $\left(\frac{d}{2} \leq x \leq d\right)$

• A. $\left(\varepsilon_{0}+\frac{k d}{2}\right)^{2 / k A}$
• B. $\frac{k A}{2 \ln \left(\frac{2 \varepsilon_{0}+k d}{2 \varepsilon_{0}}\right)}$
• C. $0$
• D. $\frac{k A}{2} \ln \left(\frac{2 \varepsilon_{0}}{2 \varepsilon_{0}-k d}\right)$

Answer 1

Answer: (B)

Taking an element of width $d x$ at a distance $x(x Capacitance of half of the capacitor
$\frac{1}{C}=\int\limits_{0}^{d / 2} \frac{1}{d c}=\frac{1}{A} \int\limits_{0}^{d / 2} \frac{d x}{\varepsilon_{0}+k x}$
$\frac{1}{C}=\frac{1}{k A} \ln \left(\frac{\varepsilon_{0}+k d / 2}{\varepsilon_{0}}\right)$
Capacitance of second half will be same
$C_{e q}=\frac{C}{2}=\frac{k A}{2 \ln \left(\frac{2 \varepsilon_{0}+k d}{2 \varepsilon_{0}}\right)}$