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Q.

A parallel plate capacitor having aseparation 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 1/3 A dielectric constant 2K and another with area 2/3 A and dielectric constant K. If the capacitance of this new capacitor is C then C/C₀ is :

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a

1

b

$\frac{2}{3}$

c

$\frac{4}{3}$

d

$\frac{1}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} C_{0} = \frac{{Kε}_{0} A}{d}; C = \frac{(2 K) ε_{0} (\frac{A}{3})}{d} + \frac{{Kε}_{0} (\frac{2 A}{3})}{d} \\ = \frac{{Kε}_{0} A}{d} (\frac{2}{3} + \frac{2}{3}) = \frac{4}{3} \frac{{Kε}_{0} A}{d} = \frac{4}{3} C_{0}; \frac{C}{C_{0}} = \frac{4}{3} \end{array}$

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