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Question

(B) The given arrangement can be considered as two capacitors connected in parallel,each having an area $A/2$ and plate separation $t$. For the first

Vedclass · Accepted Answer

$\frac{\varepsilon_0 A}{t} \cdot \frac{k_1 + k_2}{2}$

Vedclass · Answer

(B) The given arrangement can be considered as two capacitors connected in parallel,each having an area $A/2$ and plate separation $t$. For the first capacitor with dielectric constant $k_1$,the capacitance is $C_1 = \frac{k_1 \varepsilon_0 (A/2)}{t} = \frac{k_1 \varepsilon_0 A}{2t}$. For the second capacitor with dielectric constant $k_2$,the capacitance is $C_2 = \frac{k_2 \varepsilon_0 (A/2)}{t} = \frac{k_2 \varepsilon_0 A}{2t}$. Since they are in parallel,the equivalent capacitance $C = C_1 + C_2$. $C = \frac{k_1 \varepsilon_0 A}{2t} + \frac{k_2 \varepsilon_0 A}{2t} = \frac{\varepsilon_0 A}{2t}(k_1 + k_2) = \frac{\varepsilon_0 A}{t} \cdot \frac{k_1 + k_2}{2}$. Thus,the correct option is $B$.

$A$ parallel plate capacitor is filled with two dielectrics as shown in the figure. The area of each plate is $A \; m^2$ and the separation between the plates is $t \; m$. The dielectric constants are $k_1$ and $k_2$ respectively. Its capacitance in farad will be:

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