Question

The gap between the plates of a parallel plate capacitor of area $A$ and distance between plates$d$, is filled with a dielectric whose permittivity varies linearly from ${\in }^1$ at one plate to ${\in }^2$ at the other. The capacitance of capacitor is:

• A. ${\in }_0({\in }_1+{\in }_2)A/d$
• B. ${\in }_0({\in }_1+{\in }_2)A/2d$
• C. ${\in }_{0}A/\left[d/n\left({\in }_2/{\in }_1\right)\right]$
• D. ${\in }_0({\in }_1-{\in }_2)A/\left[d/n\left({\in }_2/{\in }_1\right)\right]$

Answer 1

Answer: (D)

$\begin{array}{c}Asthepermittivityofdielectricvarieslinearlyfrom{\epsilon }_{1}atoneplateto{\epsilon }_2\\ attheother,itisgovernedbyequation,k=\left(\frac{{\epsilon }_2-{\epsilon }_1}{d}\right)x+{\epsilon }_1\\ considerasmallelementofthicknessdxatadis\tan cexfromplate,Then,\\ dV=\frac{E_o}{k}dx\\ {\int }_0^{V}dV={\int }_0^d\frac{\sigma }{{\epsilon }_0}\frac{1}{\left(\frac{{\epsilon }_2-{\epsilon }_1}{d}\right)x+{\epsilon }_1}dx\\ V=\frac{d\sigma }{{\epsilon }_0\left({\epsilon }_2-{\epsilon }_1\right)}\ln \left(\frac{{\epsilon }_2}{{\epsilon }_1}\right)\\ Q=CV\Rightarrow C=\frac{Q}{V}=\frac{\sigma A}{\frac{d\sigma }{{\epsilon }_0\left({\epsilon }_2-{\epsilon }_1\right)}\ln \left(\frac{{\epsilon }_2}{{\epsilon }_1}\right)}=\frac{{\epsilon }_0\left({\epsilon }_2-{\epsilon }_1\right)A}{d\ln \left(\frac{{\epsilon }_2}{{\epsilon }_1}\right)}\\ Hence,theoptionDisthecorrectanswer.\end{array}$