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  4. Two material having the dielectric constants K1 and K2 are filled between two parallel plates of a capacitor. Here area of each plate is A and the distance between the plates is d. <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/0b9973ebbb625317ef58bdc92eb16fc4-.png /> The capacity of the capacitor is

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Q. Two material having the dielectric constants $K_1$ and $K_2$ are filled between two parallel plates of a capacitor.
Here area of each plate is $A$ and the distance between the plates is $d$.
image
The capacity of the capacitor is

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Solution:

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, $C^{\prime}=C_1+C_2$
$=\frac{K_1 \varepsilon_0 A / 2}{d}+\frac{K_2 \varepsilon_0 A / 2}{d}=\frac{\left(K_1+K_2\right) \varepsilon_0 A}{d}$