The capacity of a parallel plate condenser is $10 μ F$ , when the distance between its plates is $8$ cm. If the distance between the plates is reduced to $4$ cm, then the capacity of this parallel plate condenser will be :

5 μ F

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10 μ F

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20 μ F

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40 μ F

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Solution

The correct option is B $20 μ F$
The capacitance of parallel plate capacitor is $C = \frac{A ϵ_{0}}{d}$ . where $A =$ area of plate.
Thus, $C \propto \frac{1}{d}$
If d is halved then the capacitance will be doubled.
Hence the capacitance becomes $C^{'} = 2 C = 2 (10) = 20 μ F$

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The capacity of a parallel plate condenser is 10μF, when the distance between its plates is 8cm. If the distance between the plates is reduced to 4 cm, then the capacity of this parallel plate condenser will be :

The correct option is B 20μFThe capacitance of parallel plate capacitor is C=Aϵ0d. where A= area of plate. Thus, C∝1dIf d is halved then the capacitance will be doubled.Hence the capacitance becomes C′=2C=2(10)=20μF

The capacity of a parallel plate condenser is $10 μ F$ , when the distance between its plates is $8$ cm. If the distance between the plates is reduced to $4$ cm, then the capacity of this parallel plate condenser will be :

The correct option is B $20 μ F$
The capacitance of parallel plate capacitor is $C = \frac{A ϵ_{0}}{d}$ . where $A =$ area of plate.
Thus, $C \propto \frac{1}{d}$
If d is halved then the capacitance will be doubled.
Hence the capacitance becomes $C^{'} = 2 C = 2 (10) = 20 μ F$