91Ó°ÊÓ

The metal slab is inserted between the plates of the capacitor that makes some distance from the plates.

The expression to calculate the capacitance of the capacitor is given as follows.

Here, is the area of the plates, and is the separation between the plates.

Consider the figure after placing the metal slab between the plates of the capacitor.

When the metal slab is placed between the plates of the capacitor, two capacitance is formed in the gaps of the capacitor plates and metal slab, and these two capacitors are in the series, but separation distance is decreased due to the slab which results to increase the value of the capacitance.

The rate of change of the capacitor current is given as follows.

$\frac{dl}{dt}=\frac{I_0}{RC}{e}^{-t/RC}$

Here,$I_0$ is the initial current and$R$ is the resistance.

From the above the capacitance is inversely proportional to the current. As the value of the capacitance increases the current decreases.

Hence this circuit keeps constant more nearly constant than capacitor with the plastic plate of Question 4.