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(D) The capacitor can be modeled as two capacitors in series,each with plate area $A$ and dielectric thicknesses $d_1 = \frac{d}{4}$ and $d_2 = \frac{

Vedclass · Accepted Answer

$\frac{4 A \varepsilon_0}{d} \left[\frac{K_1 K_2}{3 K_1+K_2}\right]$

Vedclass · Answer

(D) The capacitor can be modeled as two capacitors in series,each with plate area $A$ and dielectric thicknesses $d_1 = \frac{d}{4}$ and $d_2 = \frac{3d}{4}$.Capacity of the $1^{	ext{st}}$ capacitor:$C_1 = \frac{K_1 \varepsilon_0 A}{d/4} = \frac{4 K_1 \varepsilon_0 A}{d}$Capacity of the $2^{	ext{nd}}$ capacitor:$C_2 = \frac{K_2 \varepsilon_0 A}{3d/4} = \frac{4 K_2 \varepsilon_0 A}{3d}$Since they are in series,the equivalent capacitance $C$ is given by:$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$$\frac{1}{C} = \frac{d}{4 K_1 \varepsilon_0 A} + \frac{3d}{4 K_2 \varepsilon_0 A}$$\frac{1}{C} = \frac{d}{4 \varepsilon_0 A} \left[\frac{1}{K_1} + \frac{3}{K_2}ight]$$\frac{1}{C} = \frac{d}{4 \varepsilon_0 A} \left[\frac{K_2 + 3 K_1}{K_1 K_2}ight]$Therefore,$C = \frac{4 \varepsilon_0 A}{d} \left[\frac{K_1 K_2}{3 K_1 + K_2}ight]$.

Two dielectric slabs having dielectric constants $K_1$ and $K_2$ of thicknesses $\frac{d}{4}$ and $\frac{3d}{4}$ are inserted between the plates of a capacitor as shown in the figure. The net capacitance between $A$ and $B$ is [where $\varepsilon_0$ is the permittivity of free space].

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