Displacement current in a charging capacitor.

A capacitor charging: E and ΦE growing in the gap, id = ε₀ dΦE/dt equal to i, and the same B on loops round the wire and round the gap.

On an ac source of fixed voltage, find the frequency at which the displacement current is double its 50 Hz value (100 Hz)

Saying no current flows between the plates, so there is no magnetic field there.

• The electric flux through the flat bottom S in the gap is ΦE = EA = Q/ε₀, using Gauss's law.
• As the charge changes, dΦE/dt = (1/ε₀) dQ/dt = i/ε₀, so ε₀ dΦE/dt = i. This is the missing term.
• The current carried by moving charges in conductors is the conduction current ic. The new term, id = ε₀ dΦE/dt, is the displacement current (Maxwell's displacement current), caused by a changing electric field.
• The total current through a surface is i = ic + id. Outside the plates ic = i and id = 0; in the gap ic = 0 and id = i. So the total is the same for every surface on the loop and B at P comes out the same.
• The generalised law is the Ampere-Maxwell law: ∮B·dl = μ₀ic + μ₀ε₀ dΦE/dt.
• In every respect the displacement current acts like a conduction current; above all, it produces a magnetic field just as a real current would. The field measured at a point M between the plates equals the field just outside at P.
• A steady field in a conducting wire does not change, so there id = 0. In the charging capacitor ic and id sit in different regions; in most real media both are present together, because no medium conducts or insulates perfectly.
• There can be large regions of space with no conduction current at all but a displacement current from a changing E. A magnetic field exists there even though no current source is nearby.

Teaching this idea? Put the simulation on your own page or LMS with this code.