if a capacitor blocks dc, then why do we consider that the circuit is completed when a bulb is connected in series with a capacitor while finding the expression for displacement current?

The right way of saying it is: capacitor blocks DC in STEADY-STATE

Hi Deeksha,

This is a good question, where you can get your concepts cleared about electric currents.

In electromagnetism, displacement current is a quantity that is defined in terms of the rate of change of electric displacement field. Displacement current has the units of electric current density, and it has an associated magnetic field just as actual currents do. However it is not an electric current of moving charges, but a time-varying electric field.

So even when the current in the circuit is zero (ie when there are no moving charges, a displacement current can exist betwwen the plates of the capacitor.

And hence note that, if the capacitor is fully charged, the electric field between the plates would become a constant and wouldn't vary with time, and hence I_d = 0.

But Note, on a fully charged capacitor, if the distance betwen the plates keep changing with time, the Electric field will vary between the paltes with time, and hence I_d ≠ 0, whereas the current in the circuit would be 0.

Hope that makes the concept clear.

All the best.

Regards,

Ashwin (IIT Madras).