Question

A capacitor is made of a flat plate of area $A$, and a second plate having a stair-like structure as shown in the figure. The area of each stair is $\frac{A}{5}$ and height is $a$. Find the capacitance of the assembly if $C=\frac{{\epsilon }_{0}A}{a}$.

• A. $\frac{170}{165}C$
  
• B. $\frac{78}{150}C$
  
• C. $\frac{160}{300}C$
  
• D. $\frac{137}{300}C$
  

Answer 1

The correct option is D $\frac{137}{300}C$


From the given figure, it is clear that the given arrangement behaves as parallel combination of the five individual capacitors.

Since, area and separation between each plate is given, we can get individual capacitance of all five plates using $C=\frac{{\epsilon }_{0}A}{d}$.

$\Rightarrow C_1=\frac{{\epsilon }_0\left(A/5\right)}{a}=\frac{{\epsilon }_{0}A}{5a}$

Similarly, $C_2=\frac{{\epsilon }_0\left(A/5\right)}{2a}=\frac{{\epsilon }_{0}A}{10a}$

$C_3=\frac{{\epsilon }_0\left(A/5\right)}{3a}=\frac{{\epsilon }_{0}A}{15a}$

$C_4=\frac{{\epsilon }_0\left(A/5\right)}{4a}=\frac{{\epsilon }_{0}A}{20a}$

$C_5=\frac{{\epsilon }_0\left(A/5\right)}{5a}=\frac{{\epsilon }_{0}A}{25a}$

Therefore, the capacitance of the combination,

$C_{eqv}=C_1+C_2+C_3+C_4+C_5$

$\Rightarrow C_{eqv}=\frac{{\epsilon }_{0}A}{5a}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})$

$\Rightarrow C_{eqv}=\frac{137}{300}\frac{{\epsilon }_{0}A}{a}$

$\therefore C_{eqv}=\frac{137}{300}C$ $[\because \frac{{\epsilon }_{0}A}{a}=C]$

Hence, option (d) is the correct answer.