**CONCEPT OF DISPLACEMENT CURRENT (DIFFERENCE BETWEEN DISPLACEMENT CURRENT AND CONDUCTION CURRENT)**

Let there be a parallel R-C network with a voltage V as shown in fig .Let the current through resistor R is I_{c} and by Ohm’s law it is given by

I_{c}=V/R

And current through capacitor C is I_{d} and is given by

I_{d}=dQ/dt

I_{d}=Cdv/dt ( dQ=Cdv)(1)

In practice ,the current does not flow through the capacitor . But ,the current that flows out of one electrode of capacitor equals the current that flows in to the other electrode. The net effect is as if there is a current flowing through the path containing the capacitor. But current, I_{c}actually flows through the resistor.

Hence ,from the above result ,current flowing through the resistor is known as **conduction current **and it obeys Ohm’s law,while the current flowing through the capacitor is commonly known as **Displacement current.**

**Mathematical Proof.**As the electric field inside each element equals the voltage V across the element divided by its length d

That is E=V/d or V=ED (2)

Now the current density in resistor is given by

J_{c}=I_{c}/A=σE (3)

Where A= cross-sectional area

σ=conductivity of resistance element

Also capacitance of a parallel plate capacitor is given by

C=ε_{0}A/d (4)

Now rewrite equation(1)

I_{d}=C dV/dt

By substituting the values of V and C from equations (2 and 4) in above equation,we get

I_{d}= ε_{ 0}A/d( E/t)

I_{d}= ε_{0}A E/t (5)

Therefore current density J_{d} inside capacitor is

J_{d}=I_{d}/A

Substititing value of I_{d} from equation (5) in above equation,we get

J_{d}= ε_{ 0}/A E/t

Or J_{d}= ε_{0} E/t (6a)

Or J_{d}= D/t (6b)

Where D= ε_{ 0}E=Electric displacement vector

And J_{d}=Displacement current density

Equation (6a) proves that displacement current density arises whenever there will be change in electric field E that is (E/t≠0)