The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate = A)

\frac{ε_{0} A}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}} + \frac{d_{3}}{K_{3}}}

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\frac{ε_{0} A}{(\frac{d_{1} + d_{2} + d_{3}}{K_{1} + K_{2} + K_{3}})}

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\frac{ε_{0} A (K_{1} K_{2} K_{3})}{d_{1} d_{2} d_{3}}

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ε [\frac{A K_{1}}{d_{1}} + \frac{A K_{2}}{d_{2}} + \frac{A K_{3}}{d_{3}}]

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Solution

The correct option is A $\frac{ε_{0} A}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}} + \frac{d_{3}}{K_{3}}}$
As the given combination is in series
$\frac{1}{C_{e q}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$= \frac{d_{1}}{K_{1} ϵ_{o} A} + \frac{d_{2}}{K_{2} ϵ_{o} A} + \frac{d_{3}}{K_{3} ϵ_{o} A}$
$\frac{1}{C_{e q}} = \frac{1}{ϵ_{o} A} [\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}} + \frac{d_{3}}{K_{3}}]$
$C_{e q} = \frac{ϵ_{o} A}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}} + \frac{d_{3}}{K_{3}}}$

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