A parallel plate capacitor with plate area A and plate separation d=2 m has a capacitance of 4 µF. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K=3 (as shown in figure) will be :

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8µF

2µF

6µF

32µF

answer is C.

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Detailed Solution

$C_{original} = \frac{{Aε}_{0}}{d}$

$C_{1} = \frac{{Aε}_{0}}{d / 2} = \frac{2 {Aε}_{0}}{d} = C$
$C_{2} = \frac{{KAε}_{0}}{d / 2} = \frac{2 {KAε}_{0}}{d} = \frac{6 {Aε}_{0}}{d} = 3 C$
C₁ & C₂ are in series
$C_{new} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{C \times 3 C}{C + 3 C} = \frac{3 C}{4}$
$= \frac{3}{4} \times \frac{2 {Aε}_{0}}{d} = \frac{3}{2} \times \frac{{Aε}_{0}}{d}$
$\begin{array}{l} C_{new} = \frac{3}{2} C_{original} \\ = \frac{3}{2} \times 4 = 6 μF \end{array}$

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