Question

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is $A$ metre $^{2}$ and the separation is $t$ metre. The dielectric constants are $K_{1}$ and $K_{2}$, respectively. Its capacitance in farad will be

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

Answer 1

Answer: (C)

The given arrangement is equivalent to two capacitors in parallel. In each capacitor, the area of the plate will be $A / 2$.
$\therefore $ Equivalent capacitance, $C'=C_{1}+C_{2}$
$=\frac{K_{1} \varepsilon_{0} A / 2}{t}+\frac{K_{2} \varepsilon_{0} A / 2}{t}$
$=\frac{\left(K_{1}+K_{2}\right) \varepsilon_{0} A}{2 t}$