Question

A capacitor has rectangular plates of length $a=1\text{m}$ and width $b=0.5\text{m}$. The top plate is inclined at a small angle , as shown in figure. The plate separation varies from $p=y_0$ at the left to $p=2y_0$ at the right, where $y_0$ is much less than $a$ or $b$. Calculate the capacitance of the system.

• A. $\frac{2{\epsilon }_0}{y_0}\ln 2$
  
• B. $\frac{{\epsilon }_0}{2y_0}\ln 2$
  
• C. $\frac{{\epsilon }_0}{2y_0}\ln 3$
  
• D. $\frac{{\epsilon }_0}{y_0}\ln 2$
  

Answer 1

The correct option is B $\frac{{\epsilon }_0}{2y_0}\ln 2$


We consider a differential strip of width $dx$ and length $b$ to approximate a differential capacitor of area $bdx$.


From figure,

$p=y_0+\left(\frac{2y_0-y_0}{a}\right)x$

All such differential capacitors are in parallel arrangement. Thus

$dC=\frac{{\epsilon }_0\left(bdx\right)}{y_0+\left(\frac{y_0}{a}\right)x}$

$\Rightarrow C={\epsilon }_{0}b{\int }_0^a\frac{dx}{y_0+\left(\frac{y_0}{a}\right)x}$

$=\frac{{\epsilon }_{0}b}{\left(y_0/a\right)}\left[\ln \left(\frac{y_0+\frac{y_0}{a}\times a}{y_0}\right)\right]$

$=\frac{{\epsilon }_{0}ab}{y_0}\ln 2$

Substituting the data given in the question,

$C=\frac{{\epsilon }_0\times 0.5\times 1\times \ln 2}{y_0}$

$\Rightarrow C=\frac{{\epsilon }_0\ln 2}{2y_0}$

Hence, option (b) is the correct answer.