Question

Two dielectrics of dielectric constants $K_{1}$ and $K_{2}$ are filled in gap of parallel plate capacitor as shown in figure
The capacitance of capacitor will be:

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

Answer 1

Answer: (A)

In each capacitor, the area of the plate will be $\frac{A}{2}.$
In the given combination, the arrangement is equivalent to two capacitors connected in parallel. Also in each capacitor the area of the plate will be $\frac{A}{2}.$
Therefore, equivalent capacitance
$C'=C_{1}+C_{2}$
$=\frac{K_{1} \varepsilon_{0} A / 2}{d}+\frac{K_{2} \varepsilon_{0} A / 2}{d}$
$=\frac{\varepsilon_{0} A}{2 d}\left(K_{1}+K_{2}\right)$