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. 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

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

Correct answer · Option 4

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

Combining parallel and series combinations of the dielectric slabs gives the equivalent dielectric constant formula.