When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge
A
Is less
B
Is same
C
Is more
D
May be less or more depending on the nature of the material introduced
Easy
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C
Is more
c (c) When dielectric is introduced, the capacitance will increase and as the battery remains connected, so the voltage will remain constant. Hence according to $Q = CV,$ the charge will increase.
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