Question

A parallel-plate capacitor of area $A$, plate separation d and capacitance $C$ is filled with four dielectric materials having dielectric constants $k_1, k_2, k_3$ and $k_4$ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by

• A. $k = k _1 + k_2 + k_3 + 3k_4$
• B. $k = \frac{2}{3} (k_1 + k_2 + k_3) + 2k_4$
• C. $\frac{2}{k} = \frac{3}{k_1 + k_2 + k_3} + \frac{1}{k_4}$
• D. $\frac{1}{k} = \frac{1}{k_1 } + \frac{1}{k_2} + \frac{1}{k_3} + \frac{3}{2 k_4}$

Answer 1

Answer: (C)

$\frac{2}{k} = \frac{3}{k_1 + k_2 + k_3} + \frac{1}{k_4}$

Answer 2

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$
$\frac{d}{kA \in_{0} } = \frac{3d}{2\left(k_{1} + k_{2}+k_{3}\right)A \in_{0}} + \frac{d}{2k_{4}A \in_{0}}$
$\frac{d}{kA \in_{0}} = \frac{d}{A \in_{0}} \left[\frac{3}{3\left(k_{1} + k_{2} +k_{3}\right)} + \frac{1}{2k_{4}}\right]$
$\frac{2}{k} = \frac{3}{k_{1} + k_{2} +k_{3}} = \frac{1}{k_{4} }$