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 $\left(\frac{d}{2} \leq x \leq d\right)$
JEE MAIN 2021, Diffcult
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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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