Questions

The space between the plates of a parallel-plate capacitor of capacitance C is filled with three dielectric slabs of identical size as shown in figure. If the dielectric constants of the three slabs are $K_{1}, K_{2} and K_{3},$ then find the new capacitance

Easy

A

$C_{eq} = (K_{1} + K_{2} + K_{3}) \frac{C}{3}$

B

$C_{eq} = (K_{1} + K_{2} + K_{3}) \frac{2 C}{3}$

C

$C_{eq} = (K_{1} + K_{2} + K_{3}) \frac{3 C}{3}$

D

$C_{eq} = (K_{1} + K_{2} + K_{3}) \frac{4 C}{3}$

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