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

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Answer & Solution

Correct Answer

(B) \(\frac{{kA}}{2 \ln \left(\frac{2 \varepsilon_{0}+{kd}}{2 \varepsilon_{0}}\right)}\)

Detailed explanation

Taking an element of width \(dx\) at a distance \(x(x < d / 2)\) from left plate

\(dC =\frac{\left(\varepsilon_{0}+ kx \right) A }{ dx }\)

Capacitance of half of the capacitor

\(\frac{1}{C}=\int_{0}^{ d / 2} \frac{1}{ dc }=\frac{1}{ A } \int_{0}^{ d / 2} \frac{ dx }{\varepsilon_{0}+ kx }\)

\(\frac{1}{ C }=\frac{1}{ kA } \ln \left(\frac{\varepsilon_{0}+ kd / 2}{\varepsilon_{0}}\right)\)

Capacitance of second half will be same

\(C _{\text {eq }}=\frac{ C }{2}=\frac{ kA }{2 \ln \left(\frac{2 \varepsilon_{0}+ kd }{2 \varepsilon_{0}}\right)}\)

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