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A parallel plate condenser is charged by connecting it to a battery. Without disconnecting the battery, the space between the plates is completely filled with a medium of dielectric constant k. Then:

potential becomes 1/k times
charge becomes k times
energy becomes 1/k times
electric intensity becomes k times

A

potential becomes 1/k times

B

energy becomes 1/k times

C

charge becomes k times

D

electric intensity becomes k times

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Solution

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When the battery is not disconnected the potential across the capacitor must be same as voltage of battery (kirchoffs voltage law)
Now when dielectric slab is added capacitance increases by k.
we know that $Q = C V$
V is constant, C increases by k $∴$
Q also increases by k.

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