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Figure shows two identical capacitors, C1 and C2 each of 1 μF capacitance connected to a battery of 6 V. Initially switch 'S' is closed. After sometime 'S' is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

Asked by Topperlearning User 24th April 2015, 11:07 AM
Answered by Expert
Answer:

In C2:

Charge QD = CDVD will not change.

Where CD = KC = increases K times

VD = V/K = decreases K times

In C1:

Charge QD = CD V

Potential V remains the same as 6V

Charge QD = KCV = KQ, increases K times.

Answered by Expert 24th April 2015, 1:07 PM
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