Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 :

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

Can you answer this question?

1 Answer

0 votes
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
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App