Question

From the figure find the capacitance of the capacitor?

• A. $C=\frac{\varepsilon_{0}A}{d}\left(\frac{K_{1}+K_{2}}{2} \right)$
• B. $C=\frac{\varepsilon_{0}A}{2d}\left(\frac{K_{1}K_{2}}{K_{1}+K_{2}}\right)$
• C. $C=\frac{\varepsilon_{0}A}{d}\left(\frac{K_{1}}{K_{2}} \right)$
• D. $C=\frac{\varepsilon_{0}A}{d}\left(\frac{2K_{1}K_{2}}{K_{1}+K_{2}} \right)$

Answer 1

Answer: (B)

The given capacitor can be viewed as a series combination of two capacitors $C_{1}$ and $C_{2}$ where
$C_{1} = \frac{\varepsilon_{0}K_{1}A/2}{d}; C_{2} = \frac{\varepsilon_{0}K_{2}A/2}{d}$
where d is the separation between the plates,
$\therefore $ The effective capacitance
$C = \frac{C_{1}C_{2}}{C_{1}+C_{2}}$
$= \frac{\frac{\varepsilon_{0}K_{1}A/2}{d}\times \frac{\varepsilon_{0}K_{2}A/2}{d}}{\frac{\varepsilon_{0}K_{1}A/2 }{d}+\frac{\varepsilon_{0}K_{2}A/2}{d}} = \frac{\left(\frac{\varepsilon_{0}A}{2d}\right)^{2}K_{1}K_{2}}{\left(\frac{\varepsilon_{0}A}{2d}\right)\left(K_{1}+K_{2}\right)}$
$= \frac{\varepsilon_{0}A}{2d}\left(\frac{K_{1}K_{2}}{K_{1}+K_{2}}\right)$