91Ó°ÊÓ

Capacitor has nothing between the plates.

Capacitor has layer of plastic between the plates.

The circuit has the same battery and bulb in the series with the capacitor.

The expression to calculate the capacitor when it is filled with the dielectric material is given as follows.

$C=\frac{E_{0}AK}{d}$

Here, $A$is the area of the plates, $K$is the dielectric constant and $d$is the separation between the plates.

The expression to calculate the charge on the plates is given as follows.

$Q=C△V$…… (i)

Here, $â–³V$is the potential difference between the plates of capacitor.

Calculate the charge on the plates of the capacitors.

Substitute $\frac{E_{0}AK}{d}$for$C$ into equation (i).

From the above equation, it is clear that the charge on the capacitor plates depends on the dielectric material placed between them.

The dielectric field reduces the strength of the electric field, and by keeping the total charge constant on the plates, the potential difference is reduced. The capacitance of the capacitor increases.

Therefore, the capacitance of capacitor and the charge on capacitor is more than capacitor . Due to more charge, equilibrium is reached, and the magnitude of the fringe field is enough to cancel the other fields.

Hence the current stays more nearly constant in the circuit with capacitor .