Aparallel plate capacitor with capacitance c is connected to a power supply av, acquiring a charge on its plates. while it is still connected to the power supply, the plate separation is doubled and a dielectric of constant k = 4.20 is inserted into the gap. what is the new charge on the plates in the final state? a) q=0.4769 b) q=2.109 c) q=4.20g d)

b)

Explanation:

Given that

The power source still connects it means that the voltage difference will be same in above both condition

As we know that

Or we can say that

   ----1

Where d is the distance between two plates ,A is the area.

When K=4.2 and distance become double

   ----2

Now from equation 1 and 2

So

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