First slide

Capacitance

Question

A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k₁, k₂,k₃ and k₄ 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

Difficult

A

$k = k_{1} + k_{2} + k_{3} + 3 k_{4}$
B

$k = \frac{2}{3} (k_{1} + k_{2} + k_{3}) + 2 k_{4}$
C

$\frac{2}{k} = \frac{3}{(k_{1} + k_{2} + k_{3})} + \frac{1}{k_{4}}$
D

$k = \frac{2 K_{4}}{3} (\frac{k_{1}}{k_{1} + k_{4}} + \frac{k_{2}}{k_{2} + k_{4}} + \frac{k_{3}}{k_{3} + k_{4}})$

Solution

$c_{1} = \frac{2 k_{1} ε_{0} A}{3 d}, c_{2} = \frac{2 k_{2} ε_{0} A}{3 d}, c_{3} = \frac{2 k_{3} ε_{0} A}{3 d}, if c = \frac{ε_{0} A}{d} C_{1} = \frac{2 k_{1} c}{3}, C_{2} = \frac{2 k_{2} c}{3}, C_{3} = \frac{2 k_{3} c}{3}$
Since c₁,c₂ ,c₃ are parallel ; c_p = c₁ + c₂ + c₃
$= \frac{2 c}{3} (k_{1} + k_{2} + k_{3}); c_{4} = \frac{2 k_{4} ε_{0} A}{d} = 2 k_{4} c$
Since c_p and c₄ are series $\frac{1}{c_{s}} = \frac{1}{c_{p}} + \frac{1}{c_{4}}$
$\frac{1}{kc} = \frac{3}{2 c (k_{1} + k_{2} + k_{3})} + \frac{1}{2 k_{4} c} \frac{2}{k} = \frac{3}{(k_{1} + k_{2} + k_{3})} + \frac{1}{k_{4}}$

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