A linear dielectric sphere of relative permittivity ε_r is placed in a uniform field E₀. The field inside the sphere is:

Non-uniform, largest at the poles

For a plane wave in a good conductor (σ ≫ ωε), which statements are correct?

The skin depth is proportional to 1/√f

The magnetic energy density far exceeds the electric

The skin depth is proportional to 1/√f

The skin depth increases with conductivity

The magnetic energy density far exceeds the electric

A wave has E = E₀(x̂ cos(kz − ωt) + ŷ sin(kz − ωt)). It is:

Elliptically polarised with unequal axes

Electromagnetic Theory: Boundary-Value Problems, Images, Media, Multipoles, Potentials and Gauges, Waves, Fresnel Coefficients and the Poynting Vector

Section 5 of the GATE Physics paper. The electromagnetics chapters of the engineering papers start from Maxwell’s equations and head for transmission lines and antennas; this one starts from the boundary-value problems of electrostatics and heads for the field theory. It follows the syllabus: electrostatic and magnetostatic boundary-value problems, the method of images and separation of variables; dielectrics, conductors and magnetic materials; the multipole expansion; Maxwell’s equations; scalar and vector potentials with the Coulomb and Lorenz gauges; electromagnetic waves in free space, in non-conducting media and in conductors; reflection and transmission at normal and oblique incidence; polarisation; and the Poynting vector and theorem, with the energy and momentum a wave carries. SI units throughout, with ε₀ = 8.854 × 10⁻¹² F/m, μ₀ = 4π × 10⁻⁷ H/m and c = 3.00 × 10⁸ m/s.

1. Boundary-value problems: uniqueness, images and separation of variables

In a charge-free region the potential obeys Laplace’s equation ∇²V = 0 (Poisson’s ∇²V = −ρ/ε₀ where there is charge). The uniqueness theorem — V is fixed once it, or the total charge on each conductor, is given on every boundary — licenses any trick that satisfies the equation and the boundary conditions, because whatever works is the answer.

Point charge q at height d above a grounded plane: replace the plane by an image −q at depth d. The induced surface charge totals −q, and the force on q is q²/[4πε₀(2d)²], attractive. For 1 μC at 0.1 m this is 8.988 × 10⁹ × 10⁻¹²/0.04 = 0.225 N.
Point charge q at distance a from the centre of a grounded sphere of radius R: the image is q′ = −qR/a at distance R²/a from the centre. An isolated neutral sphere needs a second image +qR/a at the centre.
Separation of variables: in Cartesian coordinates, V = X(x)Y(y) turns ∇²V = 0 into X″/X = −Y″/Y = k², giving sines in one direction and exponentials in the other; the boundary conditions fix k and a Fourier series fixes the coefficients. In spherical coordinates with azimuthal symmetry, V = Σ(Aₗrˡ + Bₗr^−(l+1))Pₗ(cos θ).

ℹ️ The dielectric sphere in a uniform field

Only the l = 1 term survives, and the field inside is uniform: E_in = 3E₀/(ε_r + 2), weaker than E₀ because the bound surface charge sets up an opposing field. A conducting sphere is the limit ε_r → ∞: E_in = 0, and the sphere acquires a dipole moment 4πε₀R³E₀.

2. Dielectrics, conductors, magnetic materials and the multipole expansion

A dielectric’s polarisation P produces bound charges ρ_b = −∇·P and σ_b = P·n̂. The displacement D = ε₀E + P has ∇·D = ρ_free, and in a linear medium D = εE with ε = ε₀(1 + χ_e). A magnetised material has bound currents J_b = ∇×M and K_b = M×n̂; H = B/μ₀ − M has ∇×H = J_free, and B = μH with μ = μ₀(1 + χ_m). In a conductor in electrostatic equilibrium, E = 0 inside, any charge sits on the surface, and E just outside is σ/ε₀ along the normal.

Boundary conditions at an interface (no free surface charge or current)
Continuous	Discontinuous
E_tangential, D_normal	E_normal jumps by the ratio of permittivities
H_tangential, B_normal	B_tangential jumps by the ratio of permeabilities

Far from a localised charge distribution, the multipole expansion V = (1/4πε₀)Σ (1/r^l+1) ∫ r′ˡ Pₗ(cos α) ρ dτ′ sorts the potential by its fall-off: monopole (total charge) as 1/r, dipole as 1/r², quadrupole as 1/r³. The dipole term is V = p cos θ/(4πε₀r²), and its field falls as 1/r³. The lowest non-vanishing moment is independent of the origin; a neutral pair of charges has zero monopole, so its dipole moment is origin-independent. The magnetic expansion has no monopole term at all, and its dipole term uses m = I × area.

3. Maxwell’s equations, potentials and gauges

Maxwell’s equations in matter
Differential form	Content
∇·D = ρ_free	Gauss: charge is the source of D
∇·B = 0	no magnetic monopoles
∇×E = −∂B/∂t	Faraday: a changing B induces a curling E
∇×H = J_free + ∂D/∂t	Ampère–Maxwell: the displacement current completes the law

Without the displacement current ∂D/∂t, the divergence of Ampère’s law would force ∇·J = 0, contradicting the continuity equation ∇·J + ∂ρ/∂t = 0 whenever charge accumulates, as on a charging capacitor. Because ∇·B = 0, one can write B = ∇×A; Faraday’s law then allows E = −∇V − ∂A/∂t. The potentials are not unique: the gauge transformation A → A + ∇λ, V → V − ∂λ/∂t leaves E and B unchanged.

Coulomb gauge ∇·A = 0: V obeys Poisson’s equation with the instantaneous charge density — convenient for statics and radiation in source-free regions, and not manifestly relativistic.
Lorenz gauge ∇·A + (1/c²)∂V/∂t = 0: both potentials obey wave equations, □²V = −ρ/ε₀ and □²A = −μ₀J, with the same operator — the Lorentz-covariant choice, and the one whose solutions are the retarded potentials.

4. Waves in free space, dielectrics and conductors

In a source-free linear medium Maxwell’s equations give ∇²E = με ∂²E/∂t², a wave of speed v = 1/√(με) = c/n with n = √(ε_rμ_r). For a plane wave E, B and k are mutually perpendicular, B = E/v, and the intrinsic impedance is η = √(μ/ε): 376.7 Ω in vacuum and 376.7/2 ≈ 188 Ω for ε_r = 4, μ_r = 1.

In a conductor of conductivity σ the wavenumber is complex, k = β + iα. For a good conductor (σ ≫ ωε) α = β = 1/δ with the skin depth δ = √(2/(ωμσ)) = 1/√(πfμσ): the amplitude falls as e^−z/δ, B lags E by 45°, the phase velocity is ωδ, and almost all the field energy is magnetic. For copper (σ = 5.8 × 10⁷ S/m) at 1 MHz, δ = 1/√(π × 10⁶ × 4π × 10⁻⁷ × 5.8 × 10⁷) = 66 μm; at 50 Hz it is about 9 mm, which is why high-frequency current flows only in the surface of a wire.

⚠️ Skin depth falls with frequency and with conductivity

δ ∝ 1/√(fσ). Quadrupling the frequency halves δ; a better conductor has a thinner skin. The intuition that a better conductor lets the wave in further is exactly backwards: a perfect conductor has δ = 0 and reflects everything.

5. Reflection, transmission and polarisation

At normal incidence from medium 1 to medium 2 (non-magnetic), the amplitude coefficients are r = (n₁ − n₂)/(n₁ + n₂) and t = 2n₁/(n₁ + n₂), and the power coefficients are R = r² and T = (n₂/n₁)t², with R + T = 1. Air to glass (n = 1.5) reflects R = (0.5/2.5)² = 4%. The negative r for n₂ > n₁ is the half-wave phase change on reflection from a denser medium.

Oblique incidence splits into s (TE, E perpendicular to the plane of incidence) and p (TM, E in it), each with its own Fresnel coefficients, e.g. r_s = (n₁cos θᵢ − n₂cos θₜ)/(n₁cos θᵢ + n₂cos θₜ).
Brewster angle tan θ_B = n₂/n₁: r_p = 0, so reflected light is purely s-polarised, and the reflected and refracted rays are perpendicular. For n₂/n₁ = √3, θ_B = 60°.
Total internal reflection for n₁ > n₂ beyond sin θ_c = n₂/n₁: |r| = 1, and an evanescent field decays into medium 2 without carrying power on average.
Polarisation is the pattern traced by E: linear when the two transverse components are in phase, circular when they are equal and 90° apart, elliptical otherwise.

6. The Poynting theorem, energy and momentum of waves

The Poynting theorem, −∂u/∂t = ∇·S + J·E, is energy conservation for fields: the field energy u = ½(ε₀E² + B²/μ₀) in a volume falls by the flux of S = E × H out through its surface plus the work done on charges inside. For a plane wave in vacuum the electric and magnetic energies are equal, and the time-averaged intensity is I = ⟨S⟩ = ½cε₀E₀². A 1 mW beam spread over 1 mm² has I = 1000 W/m² and E₀ = √(2I/(cε₀)) ≈ 868 V/m.

A wave also carries momentum, with density S/c², so it exerts a radiation pressure: I/c on a perfect absorber and 2I/c on a perfect reflector at normal incidence. Sunlight at 1360 W/m² presses on an absorbing surface with 1360/(3 × 10⁸) = 4.53 μPa. A circularly polarised wave also carries angular momentum, ±ħ per photon.

Key takeaways

Uniqueness licenses images: −q at depth d for a grounded plane, −qR/a at R²/a for a grounded sphere; a dielectric sphere in E₀ has E_in = 3E₀/(ε_r + 2).
Bound charges −∇·P and P·n̂, bound currents ∇×M and M×n̂; multipoles fall as 1/r, 1/r², 1/r³, and there is no magnetic monopole term.
E = −∇V − ∂A/∂t, B = ∇×A; Coulomb gauge ∇·A = 0, Lorenz gauge ∇·A + (1/c²)∂V/∂t = 0 gives wave equations for both potentials.
Good conductor: δ = √(2/(ωμσ)), falling with f and σ; B lags E by 45°.
R = ((n₁ − n₂)/(n₁ + n₂))² at normal incidence, tan θ_B = n₂/n₁; I = ½cε₀E₀², pressure I/c absorbed and 2I/c reflected.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A point charge of 1 μC is held 0.1 m above an infinite grounded conducting plane. Taking 1/(4πε₀) = 8.988 × 10⁹ N m² C⁻², the magnitude of the force on the charge, in N, to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.225
The image −q sits 0.1 m below the plane, 0.2 m from q, so F = 8.988 × 10⁹ × (10⁻⁶)²/(0.2)² = 0.225 N, attractive. Using the height 0.1 m as the separation gives 0.899 N, four times too large.
A point charge q is at distance a from the centre of a grounded conducting sphere of radius R (a > R). The image charge and its distance from the centre are:
1. −qR/a at R²/a
2. −q at R²/a
3. −qR/a at a − R
4. −qa/R at R²/a
Show answer
Answer: A — −qR/a at R²/a
Requiring V = 0 on the whole sphere forces q′ = −qR/a placed at the inverse point R²/a. The full −q is the plane’s image (the R → ∞ limit), and −qa/R would exceed q in magnitude.
A linear dielectric sphere of relative permittivity ε_r is placed in a uniform field E₀. The field inside the sphere is:
1. Uniform, 3E₀/(ε_r + 2)
2. Uniform, E₀/ε_r
3. Zero
4. Non-uniform, largest at the poles
Show answer
Answer: A — Uniform, 3E₀/(ε_r + 2)
Separation of variables with only the l = 1 Legendre term gives a uniform interior field 3E₀/(ε_r + 2). E₀/ε_r is the result for a slab perpendicular to the field, not a sphere, and zero is the conducting limit ε_r → ∞.
Far from a neutral charge distribution whose dipole moment vanishes but whose quadrupole moment does not, the potential falls off as:
1. 1/r³
2. 1/r²
3. 1/r
4. 1/r⁴
Show answer
Answer: A — 1/r³
The l-th multipole contributes to V as 1/r^l+1: monopole 1/r, dipole 1/r², quadrupole (l = 2) 1/r³. 1/r⁴ is the quadrupole’s field, one power faster than its potential.
Which of the following hold for time-dependent fields in vacuum?
1. ∇·B = 0
2. ∇×E = −∂B/∂t
3. ∇·E = ρ/ε₀
4. ∇×B = μ₀J
Show answer
Answer: A — ∇·B = 0; B — ∇×E = −∂B/∂t; C — ∇·E = ρ/ε₀
The first three are Maxwell’s equations as they stand. ∇×B = μ₀J is Ampère’s law for steady currents only; with time-dependent fields it lacks the displacement term μ₀ε₀∂E/∂t and would contradict charge conservation.
The Lorenz gauge condition is:
1. ∇·A + (1/c²)∂V/∂t = 0
2. ∇·A = 0
3. ∇×A = 0
4. ∇·A − ∂V/∂t = 0
Show answer
Answer: A — ∇·A + (1/c²)∂V/∂t = 0
The Lorenz condition couples A and V so that both obey the same wave equation and the condition is Lorentz invariant. ∇·A = 0 is the Coulomb gauge; ∇×A = 0 would make B vanish and is not a gauge choice at all.
Copper has σ = 5.8 × 10⁷ S/m and μ = μ₀ = 4π × 10⁻⁷ H/m. The skin depth at 1 MHz, in μm, to the nearest integer, is ____.
Numerical answer — type the value.
Show answer
Answer: 66
δ = 1/√(πfμσ) = 1/√(3.1416 × 10⁶ × 1.2566 × 10⁻⁶ × 5.8 × 10⁷) = 1/√(2.29 × 10⁸) = 6.61 × 10⁻⁵ m = 66 μm. Dropping the π, 1/√(fμσ), gives 117 μm; using ω in place of πf in 1/√(πfμσ) gives 47 μm.
For a plane wave in a good conductor (σ ≫ ωε), which statements are correct?
1. B lags E by 45°
2. The skin depth is proportional to 1/√f
3. The skin depth increases with conductivity
4. The magnetic energy density far exceeds the electric
Show answer
Answer: A — B lags E by 45°; B — The skin depth is proportional to 1/√f; D — The magnetic energy density far exceeds the electric
k = (1 + i)/δ has phase 45°, and B = kE/ω lags E by that angle. δ = √(2/(ωμσ)) falls as 1/√f and as 1/√σ, so a better conductor has a thinner skin. |B|/|E| = |k|/ω is large, so B²/μ ≫ εE²: the energy is mostly magnetic.
The intrinsic impedance of free space is 376.7 Ω. For a non-magnetic lossless dielectric with ε_r = 4, the intrinsic impedance in Ω, to the nearest integer, is ____.
Numerical answer — type the value.
Show answer
Answer: 188
η = √(μ/ε) = η₀/√ε_r = 376.7/2 = 188 Ω. Dividing by ε_r instead of its square root gives 94 Ω.
Light falls normally from air onto glass of refractive index 1.5. The fraction of the incident power reflected is ____.
Numerical answer — type the value.
Show answer
Answer: 0.04
R = ((n₁ − n₂)/(n₁ + n₂))² = (−0.5/2.5)² = 0.04. The amplitude coefficient is −0.2; reporting it, or its magnitude, instead of its square is the usual error.
Light in air is incident on a medium of refractive index 1.732. The Brewster angle, in degrees, to the nearest integer, is ____.
Numerical answer — type the value.
Show answer
Answer: 60
tan θ_B = n₂/n₁ = 1.732 = √3, so θ_B = 60°. At this angle r_p = 0 and the reflected light is s-polarised. Using sin θ = 1/n (a critical-angle formula) gives 35°.
A wave has E = E₀(x̂ cos(kz − ωt) + ŷ sin(kz − ωt)). It is:
1. Circularly polarised
2. Linearly polarised at 45°
3. Elliptically polarised with unequal axes
4. Unpolarised
Show answer
Answer: A — Circularly polarised
The two components have equal amplitude and are 90° out of phase, so the tip of E traces a circle of radius E₀ at fixed z. Linear polarisation at 45° needs the components in phase; unequal amplitudes would make it elliptical.
A 1 mW laser beam of uniform intensity has a cross-section of 1 mm². With c = 3.00 × 10⁸ m/s and ε₀ = 8.854 × 10⁻¹² F/m, the amplitude of its electric field, in V/m, to the nearest integer, is ____.
Numerical answer — type the value.
Show answer
Answer: 868
I = 10⁻³ W/10⁻⁶ m² = 1000 W/m², and I = ½cε₀E₀² gives E₀ = √(2 × 1000/(3 × 10⁸ × 8.854 × 10⁻¹²)) = √(7.53 × 10⁵) = 868 V/m. Leaving out the ½ gives the rms-like value 614 V/m.
Sunlight of intensity 1360 W/m² falls normally on a perfectly absorbing surface. With c = 3.00 × 10⁸ m/s, the radiation pressure in μPa, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 4.53
P = I/c = 1360/(3 × 10⁸) = 4.53 × 10⁻⁶ Pa = 4.53 μPa. A perfect reflector would feel twice this, 9.07 μPa, because the momentum is reversed rather than absorbed.