Question

The figure shows two identical capacitors C1 and C2 each of 2μF capacitance, connected to a battery of 5 V. Initially switch 'S' is closed. After some time 'S' is left open and dielectric slabs of dielectric constant K = 5 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

Answer 1

Answer:

For 1st capacitor charge will become 5 times potential will remain same in 5V

Answer 2

Answer:

This is the answer and hope this helps you