Displacement Current
Displacement current is defined as epsilon_0 times the rate of change of electric flux in vacuum. For a capacitor carrying a charging current, the total displacement current through the full gap equals the conduction current in the wires.
Why this shows up in the exam
Explaining magnetic fields inside capacitor gaps · Completing the Ampere-Maxwell law · Analyzing high-frequency dielectric behavior
Learn the idea
A changing electric flux acts as a current source for magnetic field even where no charge crosses. Between charging capacitor plates, electrons do not cross the dielectric gap, yet the electric field changes. Maxwell's displacement current accounts for the magnetic field there and keeps current continuity consistent.
🧠 Memory hook: No crossing charge is needed; changing electric flux counts.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- I_d = epsilon₀ d(Phi_E)/dt — displacement current in vacuum
- I_d = I_c — equality for the full surface spanning an ideal charging capacitor
- J_d = epsilon dE/dt — displacement-current density in a linear dielectric
How to approach it
- 1Identify the surface and changing electric flux through it
- 2Use I_d = epsilon d(Phi_E)/dt with the correct permittivity
- 3For a full ideal capacitor gap, equate I_d to the wire current
Common slip-ups that cost marks
- •Treating displacement current as electrons crossing the dielectric
- •Using electric field instead of its time rate of change
- •Assuming displacement current exists only in vacuum
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
An electromagnetic wave in vacuum has wavelength 1 m. Take c = 3 x 10^8 m/s. Find its frequency in units of 10^8 Hz.
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