Question

Two capacitors of capacitance 10 and 20 micro Farad are connected in series with a 6 V battery. After the capacitors are fully charged , a slab of dielectric constant K is inserted between the plates of the two capcitors. How will the following be affected-
1. Electric field energy stored in the capacitors.
2. the charges on the capacitor
3. The potential difference between the plates of the capacitors.
Justify your answer

Answer 1

It is given that battery remain connected.

The capacitance increases by k times when we insert a slab of dielectric constant K between plates of the capacitor.

C1 = KC.

1. Electric Field Energy is given by CV²/2

Here C - is capacitance and V is voltage.

V remains constant as it is connected to battery and maintained at 6V.

Capacitance increases by K times.

Hence Electric field energy increases by K times for both capacitors.

2. Charge on capacitor is given by equation Q = CV

C - Capacitance V- Voltage.

V is maintained same.

C increases by K times.

Hence Charges on both capacitance increases by K times.

3. Potential difference between the plates remain same as it is connected to battery of 6V.

So potential difference is no change for both capacitors.