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 dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by [image:1_image125.emf]

Question

Dr. Abubacker · Accepted Answer

(b) $C_{1} = \frac{K_{1}\varepsilon_{0}\frac{A}{2}}{\left( \frac{d}{2} ight)} = \frac{K_{1}\varepsilon_{0}A}{d}$
$C_{2} = \frac{K_{2}\varepsilon_{0}\frac{A}{2}}{\left( \frac{d}{2} ight)} = \frac{K_{2}\varepsilon_{0}A}{d}$ and $C_{3} = \frac{K_{3}\varepsilon_{0}A}{\left( \frac{d}{2} ight)} = \frac{2K_{3}\varepsilon_{0}A}{d}$
$\frac{1}{C_{eq}} = \frac{1}{C_{1} + C_{2}} + \frac{1}{C_{3}} = \frac{1}{\frac{\varepsilon_{0}A}{d}\left( K_{1} + K_{2} ight)} + \frac{1}{\frac{\varepsilon_{0}A}{d} 	imes 2K_{3}}$
$\frac{1}{C_{eq}} = \frac{d}{\varepsilon_{0}A}\left( \frac{1}{K_{1} + K_{2}} + \frac{1}{2K_{3}} ight)$
$C_{eq} = \left( \frac{1}{K_{1} + K_{2}} + \frac{1}{2K_{3}} ight)^{- 1}.\frac{\varepsilon_{0}A}{d}$
So $K_{eq} = \left( \frac{1}{K_{1} + K_{2}} + \frac{1}{2K_{3}} ight)^{- 1}$

Explanation

Reviewed by Dr. Abubacker

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