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Q.

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is $2 μ F$ . The separation is reduced to half and it is filled with a dielectric substance of value 2.8.
The final capacity of the capacitor is :

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a

$15.6 μ F$

b

$19.2 μ F$

c

$11.2 μ F$

d

$22.4 μ F$

answer is A.

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Detailed Solution

$\frac{C_{1}}{C_{2}} = \frac{d_{2}}{k d_{1}}$

Initial capacitance $C = 2 uF = \frac{ε_{0} A}{d}$

New capacitance

$C^{'} = \frac{{Kε}_{0} A}{d^{'}} = \frac{2.8 \times ε_{0} A}{(d / 2)} = 5.6 \frac{ε_{0} A}{d} = 5.6 \times 2 μF = 11.2 μF$

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The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2μF . The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is :