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 :
0.25μF
1/64μF
64μF
40μF
A
64μF
B
40μF
C
0.25μF
D
1/64μF
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Solution
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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
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