Electromagnetic Waves, Antennas, and RF: Ampere–Maxwell Law and Displacement Current - Exercises and Complete Worked Solutions

This companion article extends EM14 with a more demanding self-study problem set on charging capacitors, displacement current, the integral and differential Ampere–Maxwell laws, charge continuity, and the first direct derivation of electromagnetic wave equations. All exercises are stated first. Complete worked solutions follow in Part II.

The central relations are

(1)

(2)

and

(3)

Together with Faraday’s law,

(4)

the vacuum form of the Ampere–Maxwell equation provides the second dynamical curl relation required for electromagnetic waves [1, 2, 3, 4, 5].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. In the capacitor problems, keep three quantities distinct: conduction current in material Conductors, displacement current associated with changing electric flux, and the magnetic circulation around a chosen contour. In the differential-law problems, write the vector-calculus identity or theorem used at each step rather than treating the transition from integral to differential form as a formal symbol change.

Part I: Exercises

Exercise 1: the charging-capacitor surface paradox

A capacitor is charging with a steady instantaneous wire current I. A closed contour C circles one lead wire.

Two different surfaces share the same boundary C:

• S₁ cuts through the wire;
• S₂ bulges outward and passes through the capacitor gap without crossing the wire.

For each surface:

1. state the enclosed conduction current;
2. state what the original magnetostatic Ampere law would predict for
   
3. explain why the two predictions cannot both be correct;
4. identify the additional quantity that restores surface independence.

Your explanation should distinguish conduction current from changing electric flux.

Exercise 2: time-dependent charging current from Q(t)

The charge on a capacitor plate is

where

Find:

1. the conduction current I_cond(t) = dQ∕dt;
2. the displacement current I_d(t) in the ideal capacitor gap;
3. the numerical value of both currents at t = τ;
4. the limiting values as t → 0⁺ and t →∞.

Explain physically why the displacement current decreases as the capacitor approaches its final charge.

Exercise 3: derive displacement current from the capacitor field

A vacuum parallel-plate capacitor has plate area A, separation d, and voltage V (t).

Assume a uniform electric field between the plates and neglect fringing.

Starting from

derive, in sequence,

and finally

where

Then explain why this equals the conduction current in the wires for an ideal charging capacitor.

Exercise 4: displacement current for a nonuniform electric field

A circular surface of radius R lies in a time-varying electric field

Find:

1. the electric flux
   
2. the displacement current
   
3. the peak displacement-current magnitude.

Do not replace the field by an average value before performing the surface integral.

Exercise 5: conduction and displacement current in the integral law

For a chosen surface bounded by contour C, suppose the enclosed conduction current is

while the electric flux changes at the rate

Find:

1. the displacement current I_d;
2. the total effective current
   
3. the magnetic circulation
   

State explicitly why the conduction and displacement-current terms are added before multiplying by μ₀.

Exercise 6: derive the magnetic field in a charging capacitor gap

An ideal circular parallel-plate capacitor has plate radius R and carries charging current I. Assume the displacement-current density is uniform across the plate area and neglect fringing.

Using a circular Amperian contour of radius r centered on the capacitor axis:

1. derive the enclosed displacement current for r < R;
2. derive
   
3. derive
   
4. show that the two expressions agree at r = R;
5. describe how B(r) scales with r on each side of R.

Exercise 7: infer the changing electric field from a measured magnetic field

Inside an ideal circular capacitor gap, an Amperian circle of radius

has measured magnetic-field magnitude

Assume the electric field is spatially uniform over the enclosed area and there is no conduction current through the gap.

Use the Ampere–Maxwell law to find

Then determine the corresponding displacement-current density magnitude

Exercise 8: vector form of displacement-current density

The electric field in a region is

Find:

1. ∂E∕∂t;
2. the displacement-current density
   
3. |J_d| at t = 0;
4. the direction of J_d at t = 0.

Explain why displacement current is a vector field, not merely a scalar circuit current.

Exercise 9: derive the differential Ampere–Maxwell equation

Begin with the integral law for a fixed surface S bounded by C:

Use Stokes’ theorem and the fixed-surface assumption to derive

At each step, identify which integral or differential theorem justifies the transformation.

Exercise 10: local evaluation of the differential law

At one instant and location,

and

Find:

1. the conduction-current contribution μ₀J;
2. the displacement-current contribution
   
3. the total
   
4. the ratio of the displacement contribution to the conduction contribution.

Use the result to comment on why displacement current can become important for rapidly varying fields.

Exercise 11: derive the charge-continuity equation

Starting from

take the divergence of both sides.

Use

and Gauss’s Law,

to derive

Then repeat the argument after deleting Maxwell’s displacement-current term and explain exactly where the inconsistency appears for a time-varying charge density.

Exercise 12: verify charge continuity for a model charge and current distribution

Suppose

and

Calculate:

1. ∂ρ∕∂t;
2. ∇⋅ J;
3. the sum
   

Determine whether this model satisfies local charge conservation.

Exercise 13: bridge to the electromagnetic wave equation for E

This exercise extends the closing discussion of EM14.

In a source-free vacuum region,

Use

and

Take the curl of Faraday’s law and use the Vector Identity

to derive

Compare this with the standard wave-equation form

and identify the propagation speed v.

Exercise 14: magnetic wave equation and electromagnetic speed

Continue the source-free vacuum analysis.

1. take the curl of the Ampere–Maxwell equation;
2. use Faraday’s law and
   

   to derive

   
3. show that both E and B propagate at
   
4. using
   

   find the wavelength of a vacuum electromagnetic wave with frequency

   

Conclude by explaining, in one or two sentences, why Maxwell’s displacement-current term is essential to the existence of the magnetic-field wave equation.

Part II: Complete Worked Solutions

Solution 1: the charging-capacitor surface paradox

For S₁,

(5)

and the magnetostatic law gives

(6)

For S₂, no conduction charge crosses the capacitor gap, so

(7)

The uncorrected law would therefore predict zero circulation, which is impossible because the geometrical contour C is unchanged. During charging, the electric flux changes. Maxwell’s additional source term is

(8)

For an ideal charging capacitor, I_d = I, so either surface gives

(9)

Conduction current is transport of mobile charge through a material cross section; displacement current is the magnetic-source contribution associated with changing electric flux.

Solution 2: time-dependent charging current from Q(t)

Differentiate

(10)

to obtain

(11)

With Q₀ = 24 μC and τ = 2.0 ms,

(12)

For an ideal capacitor,

(13)

At t = τ,

(14)

Thus

(15)

Also,

(16)

As the plate charge approaches its final value, dQ∕dt tends to zero, so the electric field and electric flux stop changing and I_d tends to zero.

Solution 3: derive displacement current from the capacitor field

For a uniform field normal to the plates,

(17)

Differentiate:

(18)

Therefore,

(19)

Since C = 𝜖₀A∕d,

(20)

For fixed C, Q = CV gives I_cond = dQ∕dt = C dV∕dt = I_d.

Solution 4: displacement current for a nonuniform electric field

Here dA = 2πr dr, so

ΦE(t)	= 2πE0 cos(ωt) ∫ 0Rdr	(21)
	= 2πE0 cos(ωt)0R	(22)
	= cos(ωt).	(23)

Therefore,

(24)

The peak magnitude is

(25)

Solution 5: conduction and displacement current in the integral law

Id	= 𝜖0	(26)
	= (8.854 × 10−12)(4.0 × 108)	(27)
	= 3.54 × 10−3 A.	(28)

Thus

(29)

The effective enclosed source is

(30)

Hence

∮ CB ⋅ dℓ	= μ0Ieff	(31)
	≈ (1.256637 × 10−6)(6.54 × 10−3)	(32)
	≈ 8.22 × 10−9 T m.	(33)

So

(34)

Both source terms have units of amperes and enter additively before the common factor μ₀.

Solution 6: derive the magnetic field in a charging capacitor gap

Uniform displacement-current density gives

(35)

For r < R,

(36)

Ampere–Maxwell then gives

(37)

so

(38)

For r > R, the full displacement current is enclosed:

(39)

At r = R, both expressions give μ₀I∕(2πR), so B is continuous. Thus B ∝ r inside and B ∝ 1∕r outside.

Solution 7: infer the changing electric field from a measured magnetic field

Inside the uniform changing-field region,

(40)

so

(41)

With B = 8.0 × 10⁻⁸ T, r = 2.5 × 10⁻² m, and μ ₀𝜖₀ ≈ 1.113 × 10⁻¹⁷ s²∕m²,

(42)

Then

(43)

so

(44)

Solution 8: vector form of displacement-current density

Differentiate componentwise:

(45)

Therefore,

(46)

At t = 0,

(47)

so

(48)

The local displacement-current density is vector-valued because it inherits direction from ∂E∕∂t; scalar current follows only after a surface integral.

Solution 9: derive the differential Ampere–Maxwell equation

Apply Stokes’ theorem:

(49)

For a fixed surface,

(50)

Therefore,

(51)

Because this holds for arbitrary fixed surfaces,

(52)

Solution 10: local evaluation of the differential law

The conduction contribution is

(53)

The displacement contribution is

(54)

Hence

(55)

The ratio is

(56)

So here the displacement contribution is about 22.1% of the conduction contribution; sufficiently rapid field variation can make it comparable or dominant.

Solution 11: derive the charge-continuity equation

Take the divergence:

(57)

Using Gauss’s law ∇⋅ E = ρ∕𝜖₀,

(58)

Therefore,

(59)

If Maxwell’s term is omitted, taking the divergence of ∇× B = μ₀J forces ∇⋅ J = 0. Continuity would then require ∂ρ∕∂t = 0, so the uncorrected law fails whenever charge accumulates or depletes locally.

Solution 12: verify charge continuity for a model charge and current distribution

For

(60)

(61)

Also,

(62)

so

(63)

Therefore,

(64)

and the model satisfies local charge conservation exactly.

Solution 13: bridge to the electromagnetic wave equation for E

In source-free vacuum, ∇⋅ E = 0 and J = 0. Take the curl of Faraday’s law:

(65)

Use the vacuum Ampere–Maxwell law:

(66)

The identity

(67)

reduces to −∇²E because ∇⋅ E = 0. Hence

(68)

Comparing with ∇²E = (1∕v²)∂²E∕∂t² gives

(69)

Solution 14: magnetic wave equation and electromagnetic speed

Take the curl of the vacuum Ampere–Maxwell law:

(70)

Faraday’s law gives

(71)

Using

(72)

and ∇⋅ B = 0 yields

(73)

Thus both fields propagate at

(74)

For f = 100 MHz = 1.00 × 10⁸ Hz,

(75)

Hence

(76)

Without Maxwell’s displacement-current term, the vacuum Ampere law would reduce to ∇× B = 0 rather than coupling B to ∂E∕∂t, so the curl–curl derivation would lose the time-derivative term needed for the magnetic wave equation.

Common mistakes

• Treating displacement current as ordinary charges crossing the vacuum gap.
• Applying the steady-current Ampere law to a charging capacitor without the electric-flux term.
• Forgetting that the same closed contour may be spanned by many surfaces but must have one unique line integral.
• Replacing a nonuniform electric field by its center value instead of integrating E ⋅ dA.
• Confusing the scalar displacement current I_d with the vector displacement-current density J_d.
• Forgetting the fixed-surface assumption when moving d∕dt through a surface integral.
• Forgetting that ∇⋅ (∇× B) = 0 identically.
• Dropping ∇⋅ E = 0 or ∇⋅ B = 0 without first stating that the region is source-free vacuum.
• Losing a minus sign in a curl–curl wave-equation derivation.

Reinforcement summary

The central Maxwell correction is

(77)

It upgrades Ampere’s magnetostatic law to

(78)

Together with Gauss’s law, this is consistent with

(79)

In vacuum it also couples the magnetic field to a changing electric field, and together with Faraday’s law produces wave equations propagating at

(80)

References