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

$(a)\;\normalsize C=\large\frac{\lambda d} {ln(1+K_{0} \lambda d)} \;\normalsize C_{0}\qquad(b)\;\normalsize C=\large\frac{\lambda } {d\;.ln(1+K_{0} \lambda d)} \;\normalsize C_{0}\qquad(c)\;\normalsize C=\large\frac{\lambda d} {ln(1+\large\frac{\lambda d}{K_{0}} )} \;\normalsize C_{0}\qquad(d)\;\normalsize C=\large\frac{\lambda } {d\;.ln(1+\large\frac{K_{0}} {\lambda d})} \;\normalsize C_{0}$

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1 Answer

You need to integrate as the capacitance is varying with x .

Secondly, as you progress through the small dx elements of the dielectric you observe that they are connected in series ,

hence ,(important note i have used k instead of lambda.

integral( 1/dC) = Integral [( ko + kx)eoA] / dx

integrate the above expression , find constant of integration which is -ln[ko]/keo
answered Mar 30, 2015 by anupambisht12
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