Question

Calculate the ratio of the capacitance of two capacitors filled with dielectrics of the same dimensions but of different values $K$ and $K/4$ arranged in two ways as shown in figure (i) and (ii)

• A. $5:2$
• B. $25:16$
• C. $5:4$
• D. $2:5$

Answer 1

Answer: (B)

Here, $C_{1}=\frac{\epsilon _{0} K_{1} A}{2 d}+\frac{\epsilon _{0} K_{2} A}{2 d}$
$=\frac{\epsilon _{0} K A}{2 d}+\frac{\epsilon _{0} K A}{8 d}$ $\begin{bmatrix} \text{Given :} & K_{1}=K \\ \text{and} & K_{2}=\frac{K}{4} \end{bmatrix}$
$=\frac{\left(\epsilon \right)_{0} K A}{d}\left(\frac{1}{2} + \frac{1}{8}\right)=\frac{\left(\epsilon \right)_{0} K A}{d}\left(\frac{4 + 1}{8}\right)$
$=\frac{5 \left(\epsilon \right)_{0} K A}{8 d}...\left(i\right)$
and $C_{2}=\frac{\epsilon _{0} A}{\frac{d}{2 K_{1}} + \frac{d}{2 K_{2}}}=\frac{\epsilon _{0} A}{\frac{d}{2 K} + \frac{d}{\frac{2 K}{4}}}$
$=\frac{\left(2 K\right) \left(\epsilon \right)_{0} A}{5 d}...\left(ii\right)$
Dividing eq. $\left(i\right)$ by eq. $\left(ii\right)$
$\frac{C_{1}}{C_{2}}=\frac{\frac{5 \epsilon _{0} K A}{8 d}}{\frac{2 K \epsilon _{0} A}{5 d}}=\frac{5 \times 5}{2 \times 8}=\frac{25}{16}$