Question

5. A capacitor of 200Pf is charged to potential difference of 100volts. Its plates are then connected in parallel with one another and it is found that P.D. falls to 60V. What is the capacitance of the second capacitor?​

Answer 1

Answer:

Sorry but I don't no this question answer

Answer 2

Answer:

Current flowing will be constant…so charge accumulated before and after will be same.

Q=CV=C1V1+C2V2

200×100=60(200+C2)

As capacitors are connected in parallel voltage across them will be same. So V1=V2

On simplification, C2=133.33 pf

So capacitors capacitance will be 133.33pf