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  • 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 : <img alt="\begin{aligned} &\varepsi

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 :
\begin{aligned} &\varepsilon(x)=\varepsilon_{0}+k x, \text { for }\left(0<x \leq \frac{d}{2}\right) \\ &\varepsilon(x)=\varepsilon_{0}+k(d-x), \text { for }\left(\frac{d}{2} \leq x \leq d\right) \end{aligned}
 
Option: 1 \left(\varepsilon_{0}+\frac{\mathrm{kd}}{2}\right)^{2 / \mathrm{kA}}
Option: 2 \frac{\mathrm{kA}}{2 \ln \left(\frac{2 \varepsilon_{0}+\mathrm{kd}}{2 \varepsilon_{0}}\right)}
Option: 3 0
Option: 4 \frac{\mathrm{kA}}{2} \ln \left(\frac{2 \varepsilon_{0}}{2 \varepsilon_{0}-\mathrm{kd}}\right)

Answers (1)

best_answer


Capacitance of element dc= \frac{A\varepsilon }{dx}
For series \frac{1}{c_{eq}}= \int \frac{1}{dc}
\frac{1}{c_{eq}}= \int \frac{1}{dc}\doteq \int \frac{dx}{A\varepsilon }= \int \frac{dx}{A\varepsilon }
\frac{1}{c_{eq}}=\int ^{\frac{d}{2}}_{0}\frac{dx}{A\left ( \varepsilon _{0} +kx\right )}+\int ^{d}_{\frac{d}{2}}\frac{dx}{A\left (\varepsilon _{0}+k\left ( d-x \right ) \right )}
\frac{1}{c_{eq}}=\frac{1}{A}\left [ \frac{In\left ( \varepsilon _{0}+kx \right )}{\frac{d}{dx}\left (\varepsilon _{0}+kx \right )} \right ]^{\frac{d}{2}}_{0}+\frac{1}{A}\left [ \frac{In\left (\,\varepsilon _{0}+k\left ( d-x \right ) \right )}{\frac{d}{dx}\left ( \varepsilon _{0} +k\left ( d-x \right )\right )} \right ]^{d}_{\frac{d}{2}}
\\ \frac{1}{c_{eq}}=\frac{1}{A}\left [ \frac{In\left ( \varepsilon _{0} +k\left ( \frac{d}{2} \right )-In\,\varepsilon _{0}\right )}{k} \right ]+\frac{1}{A}\left [ \frac{In\left [ \left (\varepsilon _{0}+k\left ( 0 \right )-In\left ( \varepsilon _{0} +k\left ( \frac{d}{2} \right )\right ) \right ) \right ]}{-k} \right ] \\= \frac{1}{Ak}\left [ In\left (\varepsilon _{0}+k\left ( \frac{d}{2} \right ) \right ) -In\, \varepsilon _{0}-In\left ( \varepsilon _{0} \right )+In\left ( \varepsilon _{0}+\frac{kd}{2} \right )\right ]
\frac{1}{C_{eq}}= \frac{2}{AK}\left [ In\left (\epsilon _{0}+\frac{kd}{2} \right ) -In\left ( \varepsilon _{0} \right )\right ]
{C_{eq}}= \frac{KA}{2}In\left ( \frac{\varepsilon _{0}+k\frac{d}{2}}{\varepsilon _{0}} \right )
The correct option is (2)

Posted by

vishal kumar

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