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Electromagnetic Waves - Displacement Current

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Maxwell noticed an inconsistency in Ampere's Circuital Law when applied to a charging capacitor, where the conduction current IcI_c is zero in the gap between the plates, yet a magnetic field is present.

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Displacement Current (IdI_d) is defined as the current that arises due to the time-varying electric field or electric flux ΦE\Phi_E.

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In the gap between the plates of a capacitor during charging or discharging, the conduction current Ic=0I_c = 0 and only displacement current IdI_d exists.

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Outside the capacitor (in the connecting wires), the displacement current Id=0I_d = 0 and only conduction current IcI_c exists.

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The total current I=Ic+IdI = I_c + I_d is continuous across the circuit. During the charging process, the magnitude of IcI_c in the wires equals the magnitude of IdI_d between the plates.

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Ampere-Maxwell Law: The generalized form of Ampere's Law states that the line integral of the magnetic field B\mathbf{B} around any closed loop is μ0\mu_0 times the sum of the conduction current and the displacement current passing through the surface enclosed by the loop.

📐Formulae

Id=ϵ0dΦEdtI_d = \epsilon_0 \frac{d\Phi_E}{dt}

ΦE=∮E⋅dA\Phi_E = \oint \mathbf{E} \cdot d\mathbf{A}

∮B⋅dl=μ0(Ic+Id)\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 (I_c + I_d)

∮B⋅dl=μ0(Ic+ϵ0dΦEdt)\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \left( I_c + \epsilon_0 \frac{d\Phi_E}{dt} \right)

Id=CdVdtI_d = C \frac{dV}{dt}

💡Examples

Problem 1:

A parallel plate capacitor has a capacitance C=2.0 pFC = 2.0 \text{ pF}. It is connected to a V=100 VV = 100 \text{ V} DC source which is then changed such that the potential difference across the plates increases at a rate of 1012 V/s10^{12} \text{ V/s}. Calculate the displacement current IdI_d between the plates.

Solution:

Id=CdVdtI_d = C \frac{dV}{dt} Given C=2.0×10−12 FC = 2.0 \times 10^{-12} \text{ F} and dVdt=1012 V/s\frac{dV}{dt} = 10^{12} \text{ V/s}. Id=(2.0×10−12 F)×(1012 V/s)=2.0 AI_d = (2.0 \times 10^{-12} \text{ F}) \times (10^{12} \text{ V/s}) = 2.0 \text{ A}

Explanation:

Since the capacitance is constant, the displacement current is directly proportional to the rate of change of potential difference across the plates, as derived from Q=CVQ = CV and Id=dQdtI_d = \frac{dQ}{dt}.

Problem 2:

Show that for a parallel plate capacitor of area AA and charge QQ, the displacement current IdI_d is equal to the conduction current IcI_c.

Solution:

Inside the capacitor, the electric field E=σϵ0=QAϵ0E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}. The electric flux ΦE=E⋅A=Qϵ0\Phi_E = E \cdot A = \frac{Q}{\epsilon_0}. Displacement current Id=ϵ0dΦEdt=ϵ0ddt(Qϵ0)=dQdtI_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \frac{d}{dt} \left( \frac{Q}{\epsilon_0} \right) = \frac{dQ}{dt}. Since Ic=dQdtI_c = \frac{dQ}{dt}, therefore Id=IcI_d = I_c.

Explanation:

This derivation proves the continuity of current in a circuit containing a capacitor, bridging the gap between the conduction current in the wires and the displacement current in the vacuum/dielectric.

Displacement Current - Revision Notes & Key Formulas | ICSE Class 12 Physics