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A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance C₀ . Now one-third of the material is replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area $\frac{1}{3}$ A, dielectric constant 2K and another with area $\frac{2}{3}$ A and dielectric constant K. If the capacitance of this new capacitor is C then $\frac{C}{C_{0}}$ is :
Option: 1
$1$

Option: 2
$\frac{4}{3}$

Option: 3
$\frac{2}{3}$

Option: 4
$\frac{1}{3}$

Answers (1)

best_answer

$C_{0}=\frac{k \in_{0} A}{d}$

and

$C=\frac{k \in_{0} 2}{3 d}+\frac{2 k \in_{0} A}{3 d}=\frac{4}{3} \frac{k \in_{0} A}{d}$

So

$\frac{C}{C_{0}}=\frac{\frac{4}{3} \frac{k \in_{0} A}{d}}{\frac{k \epsilon_{0} A}{d}}=\frac{4}{3}$

Posted by

Suraj Bhandari

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