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 figure below

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 figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constants K is given by

TestHub · Accepted Answer

The given capacitor can be considered as a combination of four smaller capacitors.

The top half consists of three capacitors in parallel, each with area $A /3$ and thickness $d /2$ . Their capacitances are:
$C_{1} = \frac{k _{1} ϵ _{0} ( A /3 )}{d /2} = \frac{2 k _{1} ϵ _{0} A}{3 d}$
$C_{2} = \frac{k _{2} ϵ _{0} ( A /3 )}{d /2} = \frac{2 k _{2} ϵ _{0} A}{3 d}$
$C_{3} = \frac{k _{3} ϵ _{0} ( A /3 )}{d /2} = \frac{2 k _{3} ϵ _{0} A}{3 d}$

The equivalent capacitance of this top part is:
$C_{t o p} = C_{1} + C_{2} + C_{3} = \frac{2 ϵ _{0} A}{3 d} (k_{1} + k_{2} + k_{3})$

The bottom half is a single capacitor with area $A$ and thickness $d /2$ . Its capacitance is:
$C_{4} = \frac{k _{4} ϵ _{0} A}{d /2} = \frac{2 k _{4} ϵ _{0} A}{d}$

The top part $C_{t o p}$ and the bottom part $C_{4}$ are in series. The equivalent capacitance $C$ of the entire system is given by:
$\frac{1}{C} = \frac{1}{C _{t o p}} + \frac{1}{C _{4}}$
$\frac{1}{C} = \frac{3 d}{2 ϵ _{0} A ( k _{1} + k _{2} + k _{3} )} + \frac{d}{2 k _{4} ϵ _{0} A}$

If a single dielectric of constant $k$ is used, the capacitance would be:
$C = \frac{k ϵ _{0} A}{d} \Rightarrow \frac{1}{C} = \frac{d}{k ϵ _{0} A}$

Equating the two expressions for $\frac{1}{C}$ :
$\frac{d}{k ϵ _{0} A} = \frac{d}{2 ϵ _{0} A} (\frac{3}{k _{1} + k _{2} + k _{3}} + \frac{1}{k _{4}})$

Canceling $\frac{d}{ϵ _{0} A}$ from both sides and multiplying by 2:
$\frac{2}{k} = \frac{3}{k _{1} + k _{2} + k _{3}} + \frac{1}{k _{4}}$

Answer: $\frac{2}{k} = \frac{3}{k _{1} + k _{2} + k _{3}} + \frac{1}{k _{4}}$

Physics - Capacitor Question with Solution | TestHub

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