Question

From the figure find the capacitance of the capacitor?

• A. $C=\frac{{\epsilon }_{0}A}{d}\left(\frac{K_1+K_2}{2}\right)$
• B. $C=\frac{{\epsilon }_{0}A}{2d}\left(\frac{K_1{K}_2}{K_1+K_2}\right)$
• C. $C=\frac{{\epsilon }_{0}A}{d}\left(\frac{K_1}{K_2}\right)$
• D. $C=\frac{{\epsilon }_{0}A}{d}\left(\frac{2K_1{K}_2}{K_1+K_2}\right)$

Answer 1

Answer: (B)

The given capacitor can be viewed as a series combination of two capacitors $C_1$ and $C_2$ where
$C_1=\frac{{\epsilon }_0{K}_{1}A/2}{d};C_2=\frac{{\epsilon }_0{K}_{2}A/2}{d}$
where d is the separation between the plates,
$\therefore$ The effective capacitance
$C=\frac{C_1{C}_2}{C_1+C_2}$
$=\frac{\frac{{\epsilon }_0{K}_{1}A/2}{d}\times \frac{{\epsilon }_0{K}_{2}A/2}{d}}{\frac{{\epsilon }_0{K}_{1}A/2}{d}+\frac{{\epsilon }_0{K}_{2}A/2}{d}}=\frac{{\left(\frac{{\epsilon }_{0}A}{2d}\right)}^2{K}_1{K}_2}{\left(\frac{{\epsilon }_{0}A}{2d}\right)\left(K_1+K_2\right)}$
$=\frac{{\epsilon }_{0}A}{2d}\left(\frac{K_1{K}_2}{K_1+K_2}\right)$