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, $C_1=\frac{{ϵ}_0{K}_{1}A}{2d}+\frac{{ϵ}_0{K}_{2}A}{2d}$
$=\frac{{ϵ}_{0}KA}{2d}+\frac{{ϵ}_{0}KA}{8d}$ $\left[\begin{array}{cc}\text{Given :}& K_1=K\\ \text{and}& K_2=\frac{K}{4}\end{array}\right]$
$=\frac{{\left(ϵ\right)}_{0}KA}{d}\left(\frac{1}{2}+\frac{1}{8}\right)=\frac{{\left(ϵ\right)}_{0}KA}{d}\left(\frac{4+1}{8}\right)$
$=\frac{5{\left(ϵ\right)}_{0}KA}{8d}...\left(i\right)$
and $C_2=\frac{{ϵ}_{0}A}{\frac{d}{2K_1}+\frac{d}{2K_2}}=\frac{{ϵ}_{0}A}{\frac{d}{2K}+\frac{d}{\frac{2K}{4}}}$
$=\frac{\left(2K\right){\left(ϵ\right)}_{0}A}{5d}...\left(ii\right)$
Dividing eq. $\left(i\right)$ by eq. $\left(ii\right)$
$\frac{C_1}{C_2}=\frac{\frac{5{ϵ}_{0}KA}{8d}}{\frac{2K{ϵ}_{0}A}{5d}}=\frac{5\times 5}{2\times 8}=\frac{25}{16}$