Question

What is displacement current \((i_d)\)? Considering the case of charging of a capacitor, show that \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}. \] What is the value of \(i_d\) for a conductor across which a constant voltage is applied?

Hint:

Maxwell introduced displacement current to make Ampere's law valid for charging capacitors. \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}. \] A constant electric field gives constant electric flux, therefore displacement current becomes zero.

Answer 1

Concept: While studying Ampere's circuital law, Maxwell discovered an inconsistency in the case of a charging capacitor. Although conduction current flows through the connecting wires, no conduction current passes through the dielectric gap between the capacitor plates. To remove this inconsistency, Maxwell introduced the concept of displacement current. Displacement current is not associated with actual motion of charges through the dielectric. It arises due to a time-varying electric field. The displacement current is defined as \[ i_d=\varepsilon_0\frac{d\Phi_E}{dt}, \] where \[ \Phi_E \] is the electric flux through the region.

Step 1: Consider a charging capacitor. Let a capacitor be connected to a battery through a resistor. During charging,
• conduction current \(i\) flows through the wires,
• equal and opposite charges accumulate on the capacitor plates,
• electric field between the plates increases with time. Thus a time-varying electric field is produced between the plates.

Step 2: Find the electric field between the capacitor plates. Surface charge density on the plates is \[ \sigma=\frac{q}{A}, \] where \(A\) is the plate area. Electric field between the plates is \[ E=\frac{\sigma}{\varepsilon_0}. \] Therefore, \[ E=\frac{q}{\varepsilon_0 A}. \]

Step 3: Calculate electric flux between the plates. Electric flux is \[ \Phi_E=EA. \] Substituting the value of \(E\), \[ \Phi_E = \left( \frac{q}{\varepsilon_0 A} \right)A. \] Hence, \[ \Phi_E = \frac{q}{\varepsilon_0}. \] \[ \boxed{ \Phi_E=\frac{q}{\varepsilon_0} } \]

Step 4: Differentiate with respect to time. Differentiating, \[ \frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dq}{dt}. \] Since \[ \frac{dq}{dt}=i, \] we get \[ \frac{d\Phi_E}{dt} = \frac{i}{\varepsilon_0}. \] Multiplying both sides by \(\varepsilon_0\), \[ i = \varepsilon_0 \frac{d\Phi_E}{dt}. \] The current represented by this changing electric field is called displacement current. Therefore, \[ \boxed{ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} }. \]

Step 5: Value of displacement current for a conductor connected to a constant voltage source. When a constant voltage is applied across a conductor,
• the electric field remains constant,
• electric flux does not change with time. Hence, \[ \frac{d\Phi_E}{dt}=0. \] Therefore, \[ i_d = \varepsilon_0 \left( \frac{d\Phi_E}{dt} \right) = 0. \] Thus, \[ \boxed{i_d=0}. \] Final Answer: Displacement current is the current associated with a time-varying electric field and is given by \[ \boxed{ i_d=\varepsilon_0\frac{d\Phi_E}{dt} }. \] For a conductor across which a constant voltage is applied, \[ \boxed{ i_d=0. } \]