Question

The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation :
$K\left(x\right)=K_{0}+\lambda x\left(\lambda=a\,\text{constant}\right)$
The capacitance $C$, of this capacitor, would be related to its 'vacuum' capacitance $C_o$ as per the relation :

• A. $C=\frac{\lambda d}{\ell n\left(1+K_{o}\lambda d\right)}C_{o}$
• B. $C=\frac{\lambda }{d. \ell n\left(1+K_{o}\lambda d\right)}C_{o}$
• C. $C=\frac{\lambda d}{\ell n\left(1+\lambda d/K_{o}\right)}C_{o}$
• D. $C=\frac{\lambda d}{d. \ell n\left(1+K_{o}/\lambda d\right)}C_{o}$

Answer 1

Answer: (C)

$d _{ C }=\frac{\left( K _{0}+\lambda_{ x }\right) A }{ dx }$
$\int \frac{1}{ d _{ C }}=\int\limits_{0}^{\phi} \frac{ dx }{ A \cdot\left( K _{0}+\lambda x \right)}$
$C _{0}=\frac{\varepsilon_{0} A }{ d } \frac{1}{\lambda} \int\limits_{ K _{0}}^{ K _{0}+\lambda d } \frac{ dt }{ At }$
$K _{0}+\lambda x = t\,\,\,\, \lambda d x = dt$
$C=\frac{1}{\lambda A } \ln \left[\frac{ K _{0}+\lambda d }{ K _{0}}\right]$