A parallel plate capacitor of area A, plate separation d and capacitance $C$ is filled with three different dielectric materials having dielectric constants $K_{1},K_{2}\ and\ K_{3}$ as shown. If a single dieletric material is to be used to have the same capacitance $C$ is this capacitors, then its dielectric constant $K$ is given by [image:2_image43.emf]

Question

Dr. Abubacker · Accepted Answer

(b) Capacitance of two capacitors each of area $\frac{A}{2}\ ,$ plate separation $d$ but dielectric constants $K_{1}$ and $K_{2}\ $respectively joined in parallel
$C_{1} = \frac{K_{1}\varepsilon_{0}\left( \frac{A}{2} ight)}{d/2} + \frac{K_{2}\varepsilon_{0}\left( \frac{A}{2} ight)}{d/2} = \frac{\left( K_{1} + K_{2} ight)\varepsilon_{0}A}{d}$
It is in series with a capacitor of plate area $A,$ plate separation $d/2\ $and dielectric constant$K_{3}\ ie,\ C_{2} = \frac{K_{3}\varepsilon_{0}A}{d/2}$.
If resultant capacitance be taken as $C = \frac{K_{\ }\varepsilon_{0}A}{d}$,
Then $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$
$	herefore\frac{d}{K\varepsilon_{0}A} = \frac{d}{\left( K_{1} + K_{2} ight)\varepsilon_{0}A} + \frac{d/2}{\begin{array}{r}
K_{3}\varepsilon_{0}A \
\ 
\end{array}}$
$\Rightarrow \frac{1}{K} = \frac{1}{K_{1}{+ \ K}_{2}} + \frac{1}{2K_{3}}$

Explanation

Reviewed by Dr. Abubacker

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