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:

Displacement current exists only when electric field changes with time. No change in electric field → no displacement current.

Answer 1

Concept: Displacement current was introduced by Maxwell to explain continuity of current in circuits containing capacitors. It arises due to time-varying electric field, even where no charge flows physically.
Step 1: Definition of displacement current. Displacement current is defined as: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \] Where:

\( \varepsilon_0 \) = permittivity of free space
\( \Phi_E \) = electric flux

Step 2: Charging capacitor case. When a capacitor is charging:

Conduction current flows in wires
No real charge flows across dielectric gap
But electric field between plates changes with time
Electric flux between plates: \[ \Phi_E = EA \] As voltage increases, electric field changes: \[ E = \frac{V}{d} \Rightarrow \Phi_E \text{ changes with time} \] Thus: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \] This ensures continuity: \[ i_c = i_d \]
Step 3: Conductor with constant voltage. For a conductor with constant applied voltage:

Electric field is constant
Electric flux does not change with time
So: \[ \frac{d\Phi_E}{dt} = 0 \] Hence: \[ i_d = 0 \] Final Answers:

Displacement current: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
For a conductor at constant voltage: \[ i_d = 0 \]

Answer 2

Displacement current is the extra term Maxwell added to Ampere's law so that the law gives a consistent answer no matter what surface is chosen to bound the loop used in the calculation, even when that loop is drawn near a charging capacitor. A good way to see why it is needed, and where its formula comes from, is to look at the contradiction that arises without it.

Step 1: Set up the paradox.
Consider a wire carrying charging current \( i_c \) into one plate of a capacitor. Draw an Amperian loop encircling the wire, away from the plates. Two different surfaces can be bounded by this same loop: a flat disc that cuts straight through the wire, and a bulging surface that passes through the gap between the capacitor plates, never touching the wire.

Step 2: Apply the original (incomplete) Ampere's law to each surface.
Ampere's law states \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 i_{\text{enclosed}} \), where \( i_{\text{enclosed}} \) is the conduction current piercing the chosen surface. For the flat disc, the wire pierces it, so \( i_{\text{enclosed}} = i_c \), giving \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 i_c \). For the bulging surface passing through the gap, where no charge physically crosses, \( i_{\text{enclosed}} = 0 \), giving \( \oint \mathbf{B} \cdot d\mathbf{l} = 0 \). Both surfaces share the same boundary loop, so the result must be the same for both, yet \( \mu_0 i_c \) and \( 0 \) disagree, meaning conduction current alone cannot be the complete source term.

Step 3: Fix the paradox by adding a current-like term through the gap.
Between the plates, while the capacitor is charging, the electric field is changing with time. Maxwell proposed adding a term proportional to the rate of change of electric flux through the bulging surface, the displacement current:\[i_d = \varepsilon_0 \frac{d\Phi_E}{dt}\]so that Ampere's law becomes\[\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 (i_c + i_d)\]

Step 4: Check consistency.
Through the flat disc, only \( i_c \) contributes, giving \( \mu_0 i_c \). Through the bulging surface, only \( i_d \) contributes, and by the charge-field relation for a capacitor, \( i_d \) works out to exactly equal \( i_c \), again giving \( \mu_0 i_c \). The two surfaces now agree, which is exactly the condition needed for Ampere's law to hold for any surface bounded by the loop.

Step 5: Value for a conductor at constant voltage.
If a constant voltage is applied across a conductor, the associated electric field does not change with time, so \( \frac{d\Phi_E}{dt} = 0 \) and therefore \[i_d = 0\] Only steady conduction current flows; there is no time-varying field to contribute a displacement term.