The capacitance of parallel-plate capacitor is 4μF. If a dielectric material of dielectric constant 16 is placed between the plates then the new capacitance will be :
A
1/64μF
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B
0.25μF
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C
64μF
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D
40μF
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Solution
The correct option is C64μF If the capacitance of the parallel plate capacitor without dielectric is C, the capacitance of it with dielectric will be C′=kC=16×4=64μF