When a parallel plate capacitor is being charged by a dc battery, why
does a galvanometer show a momentary deflection? Obtain the expression for the displacement current and show how it is related to conduction current in a steady state.

 

 
 
 
 
 

Answers (1)
S safeer

In between the plates of the capacitor due to the time-varying electric field, there is a current. This current is known as displacement current. While charging a capacitor total current is the sum of displacement and conduction current. Due to this current galvanometer deflects.

The electric field between the plates of the capacitor 

E=\frac{Q}{\epsilon_0A}

Electric flux

\phi_E=\frac{Q}{\epsilon_0A}\times A=\frac{Q}{\epsilon_0}

During charging the charge is varying with time

Therefore change in flux with respect to time

\frac{d\phi_E}{dt}=\frac{dQ}{dt}\frac{1}{\epsilon_0}

\frac{d\phi_E}{dt}=I_d\frac{1}{\epsilon_0}

Displacement current 

I_d=\epsilon_0 \frac{d\phi_E}{dt}

At steady-state, there is no change in the electric field there for displacement current is zero.

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