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

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charge becomes k times

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energy becomes 1/k times

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electric intensity becomes k times

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Solution

The correct option is B charge becomes k times
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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