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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
$\frac{2}{k} = \frac{3}{k_1 + k_2 + k_3} + \frac{1}{k_4}$
B
$\frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} + \frac{3}{2k_4}$
C
$k = k_1 + k_2 + k_3 + 3k_4$
D
$k = \frac{2}{3}(k_1 + k_2 + k_3) + 2k_4$
Explanation
Three side-by-side dielectrics add in parallel; that block is in series with $k_4$.
Detailed Solution
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$
$\frac{d}{A\varepsilon_0k} = \frac{1}{\frac{A}{3}\varepsilon_0\frac{(k_1 + k_2 + k_3)}{d/2}} + \frac{1}{\frac{A\varepsilon_0k_4}{d/2}}$
$\Rightarrow \frac{2}{k} = \frac{3}{k_1 + k_2 + k_3} + \frac{1}{k_4}$