From the figure find the capacitance of the capacitor? | AIIMS 2001 | Physics

Options

B. $C=\frac{\varepsilon_0 A}{2 d}\left(\frac{K_1 K_2}{K_1+K_2}\right)$

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}

; \quad C_2=

\frac{\varepsilon_0 K_2 A / 2}{d}

where

d

is the separation between the plates,

\therefore \quad

The effective capacitance

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\begin{aligned}& 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}{2 d}\right)^2 K_1 K_2}{\left(\frac{\varepsilon_0 A}{2 d}\right)\left(K_1+K_2\right)}\\ & =\frac{\varepsilon_0 A}{2 d}\left(\frac{K_1 K_2}{K_1+K_2}\right) \end{aligned}