A parallel-plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant ${K}_{1}$ and ${K}_{2}$ of same area $\dfrac{A}{2}$ and thickness $\dfrac{d}{2}$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :

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A parallel-plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric

Quantrex Academy · Accepted Answer

Correct answer: A. ε _ 0 A d ((1)/(2) + K _ 1 K _ 2 K _ 1 + K _ 2 )

C _ eq = C _ 1 + C _ 2 C _ 3 C _ 2 + C _ 3 C _ eq = A ε _ 0 2d + K _ 1 A ε _ 0 d × K _ 2 A ε _ 0 d K _ 1 A ε _ 0 d + K _ 2 A ε _ 0 d C _ eq = A ε _ 0 2d + A ε _ 0 d ( K _ 1 K _ 2 K _ 1 + K _ 2 ) = A ε _ 0 d ((1)/(2) + K _ 1 K _ 2 K _ 1 + K _ 2 )

A parallel-plate capacitor with plate area A has separation d...

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