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)

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

$ε_{0} (\frac{{AK}_{1}}{d_{1}} + \frac{{AK}_{2}}{d_{2}} + \frac{{AK}_{3}}{d_{3}})$

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

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

answer is A.

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Detailed Solution

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

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