### Maxwell's displacement current

#### Problem:

A thin parallel-plate capacitor of plate separation d is filled with a medium of conductivity σ and permittivity ε0. The plates of the capacitor are circular.  A variable voltage V = V0sinωt is applied to the capacitor.  Assuming that the electric field between the plates is homogeneous, find the magnetic field in the capacitor.

Solution:

• Concepts:
Maxwell's equations in quasi-static situations
• Reasoning:
We assume that the electric field between the plates is homogeneous.  E = E(t)k.
E(t) = -Φ - ∂A/∂t, we neglect the ∂A/∂t term, which leads to radiation.
• Details of the calculation:
×B = μ0j + (1/c2)∂E/∂t, ∮Γ B∙dr =  μ0Ajn dA + (1/c2)∫AE/∂t∙n dA
Symmetry simplifies this equation.
2πρB(ρ) = μ02π∫0ρjρ'dρ' + (2π/c2)∫0ρ∂E/∂t ρ'dρ',  B = B(ρ) (φ/φ).
Since j = σE and E = E(t) k, we have
2πρB(ρ) = μ02πσE ρ2/2 + (2π/c2)∂E/∂t ρ2/2.
B(ρ) = (ρ/2)(μ0σE(t) +(1/c2)(∂E/∂t)),  E(t) = (V0/d)sinωt,  ∂E/∂t = ω(V0/d)cosωt.
B(ρ) = (ρ/2)(V0/d)(μ0σsinωt + (ω/c2)cosωt) = (ρ/2)(V0/d)Csin(ωt + φ).
Ccosφ = μ0σ, Csinφ = ω/c2, C2 = (μ0σ)2 + ω2/c4,  tanφ = ω/(μ0σc2).

#### Problem:

A thin wire carries constant current I into one plate of a charging capacitor, and another thin wire carries constant current I out of the other plate.  The capacitor plates are disks of radius a and separation w << a (so edge effects can safely be neglected).  The region between the plates has ε ~ ε0, μ ~ μ0, but does have a non-negligible, constant conductivity σ.
Note: The capacitor is not an ideal capacitor, since the material between the plates is not a perfect insulator.
(a)  Supposing that the charges are uniformly distributed on the plates, find a differential equation for the charge Q(t) on the plates, and solve it for Q(t), taking Q(0) = 0.
(b)  Find the electric and magnetic fields in the gap.  Approximate the electric field as just that due to the charged plates.  When computing the magnetic field, include all sourcing contributions.

Solution:

• Concepts:
Maxwell's equations
• Reasoning:
When the charge on the plates of a capacitor changes, the changing electric field between the plates induces a magnetic field between the plates.
• Details of the calculation:
(a)  Choose coordinates so that I flows in the z-direction in the wire.
With charge Q(t), the electric field inside is E = k Q(t)/(πa2ε0 ).
The current density inside is j = σE = k σQ(t)/(πa2ε0 ).
Integrating this over an arbitrary surface of constant z between the plates gives the current flowing from one plate to another.
Iinside = ∫j∙dA = σQ(t)/(ε0).
Charge conservation implies that the charge on the plate satisfies
dQ(t)/dt = I – Iinside = I - σQ(t)/(ε0),
with I the constant current flowing into the plate from the wire. We can solve this differential equation.
Q(t) = (Iε0/σ)(1 – exp(-σt/ε0)).
(b)  The electric field between the plates is E = k Q(t)/(πa2ε0 ) = k (I/(πa2σ))(1 – exp(-σt/ε0)).
The magnetic field inside  the plates satisfies B = B(r)φ/φ and Ampere's law gives
B(r)2πr = μ0∫dA(jz + ε0(∂Ez/∂t) = (μ00)(Q(t)σ + ε0dQ(t)/dt)(r2/a2) = μ0I(r2/a2).
So inside the plates, when r < a, we have B(r) = μ0Ir/(2πa2).
Outside the plates we have B(r) = μ0I/(2πr), as usual for a wire.