Q. 16

# The charge on a parallel plate capacitor varies as .

The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor?

Answer :

Given

The charge stored in the capacitor q = q_{0}cos2πνt.

The displacement current between the plates of a capacitor is given as,

Where, is the electric flux between the plates, which can be determined by using the Gauss’s law

Gauss’s law states that the net electric flux enclosed through a closed surface is equal to the charge enclosed within the surface divided by the permittivity i.e

Therefore, the flux,

Hence, the displacement current is then,

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Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor

HC Verma - Concepts of Physics Part 2Consider the situation of the previous problem. Define displacement resistance R_{d} = V/i_{d} of the space between the plates where V is the potential difference between the plates and i_{d} is the displacement current. Show that R_{d} varies with time as

R_{d} = R (e^{t/τ} – 1)

HC Verma - Concepts of Physics Part 2

You are given a 2μF parallel plate capacitor. How would you establish an instantaneous displacement current of 1mA in the space between its plates?

Physics - ExemplarUsing B = μ_{0} H find the ratio E_{0}/H_{0} for a plane electromagnetic wave propagating through vacuum. Show that is has the dimensions of electric resistance. This ratio is a universal constant called the impedance of free space.

Show that the dimensions of the displacement current are that of an electric current.

HC Verma - Concepts of Physics Part 2