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New Capacitance in Dielectric

The capacitan...

Question

The capacitance of parallel-plate capacitor is $4 μ F$ . If a dielectric material of dielectric constant 16 is placed between the plates then the new capacitance will be :

A

1 / 64 μ F

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B

0.25 μ F

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C

64 μ F

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D

40 μ F

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Solution

The correct option is C $64 μ F$
If the capacitance of the parallel plate capacitor without dielectric is $C$ , the capacitance of it with dielectric will be $C^{'} = k C = 16 \times 4 = 64 μ F$

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