At a large distance r along the axis of an ideal point electric dipole, how does the electrostatic potential vary with r?

As 1/r², because the two opposite charges cancel the leading 1/r term

As 1/r, in the same way as the potential of a single point charge

As 1/r³, in the same way as the electric field of the dipole itself

As 1/r², because the two opposite charges cancel the leading 1/r term

As ln r, in the same way as the potential of an infinite line of charge

A solid sphere of radius a has uniform density, and M is its total mass. How does the gravitational acceleration vary inside the sphere, for 0 < r < a?

It increases linearly with r, from zero at the centre to GM/a² at the surface

It is zero everywhere inside, exactly as it is inside a thin uniform spherical shell

It stays equal to the surface value GM/a², because the total mass of the sphere is unchanged

It rises as 1/r² towards the centre, because the distance to the centre of mass keeps shrinking

For an ideal point dipole, the electric field magnitude on the dipole axis at distance r is compared with that on the equatorial plane at the same distance r. What is the ratio (axis to equator)?

Two, because E = 2p/(4πε0 r³) on the axis and E = p/(4πε0 r³) on the equatorial plane

One half, because the axis points along the dipole while the equator points across it

One, because both points are at the same distance r from the centre of the dipole

Four, because the field of a dipole falls as the cube of the distance from its centre

Two, because E = 2p/(4πε0 r³) on the axis and E = p/(4πε0 r³) on the equatorial plane

A potential satisfies Laplace's equation in the strip 0 0. It is zero on x = 0, on x = a and as y → ∞. Which form is an acceptable separated solution?

sin(nπx/a) exp(−nπy/a), a sine in x that vanishes at both edges and a decay in y

sinh(nπx/a) sin(nπy/a), which is the product of a hyperbolic function of x and a sine function of y

sin(nπx/a) exp(−nπy/a), a sine in x that vanishes at both edges and a decay in y

cos(nπx/a) exp(−nπy/a), a cosine in x that decays as y increases

sin(nπx/a) exp(+nπy/a), a sine in x that grows exponentially with y

Which of the following functions of the plane coordinates, with r = √(x² + y²), satisfies Laplace's equation in two dimensions away from the origin?

x² − y², whose x and y second derivatives are equal and opposite

x² + y², whose second derivatives with respect to x and y are both positive

r itself, which increases smoothly and symmetrically away from the origin

1/r, which decays smoothly and symmetrically away from the origin

x² − y², whose x and y second derivatives are equal and opposite

A uniform conducting earth is probed at low frequency, so that the displacement current is negligible against the conduction current. For the electric field E with time dependence exp(iωt), which statement is correct?

E obeys a diffusion equation, equivalent to ∇²E + k²E = 0 with k² = −iωμσ, so it is damped

E obeys the wave equation ∇²E = με ∂²E/∂t² and travels without any damping at the speed 1/√(με)

E obeys a diffusion equation, equivalent to ∇²E + k²E = 0 with k² = −iωμσ, so it is damped

E obeys Laplace's equation ∇²E = 0 because the fields are effectively static at such low frequencies

E obeys the Helmholtz equation with a real k = ω√(με), so that only its phase changes with depth

A function φ satisfies Laplace's equation inside a closed region V and φ = 0 on the whole boundary S. What does Green's first identity show about φ inside V?

φ = 0 everywhere in V, since ∫|∇φ|² dV = ∮ φ ∂φ/∂n dS = 0

φ can still be positive inside, because a harmonic function is allowed to have an interior maximum

φ can be non-zero inside, because Laplace's equation alone cannot fix a solution without initial conditions

φ = 0 everywhere in V, since ∫|∇φ|² dV = ∮ φ ∂φ/∂n dS = 0

φ is a constant fixed only up to an additive constant, as for Neumann data on the boundary

A point charge q is at distance d from the centre of a grounded conducting sphere of radius a, with d > a. Which image charge reproduces the zero potential on the sphere?

−qa/d, placed at distance a²/d from the centre on the line joining the centre to q

−q, placed at the mirror point of q in the surface, at distance 2a − d from the centre

−qd/a, placed at distance a²/d from the centre on the line joining the centre to q

+qa/d, placed at distance a²/d from the centre on the line joining the centre to q

Recovering the density distribution from a measured gravity anomaly is described by a first-kind Fredholm equation. Why is this inverse problem called ill-posed?

Small errors in the data can produce large changes in the recovered density, and many densities fit the data

The kernel of the integral is not symmetric, so no solution of the equation can exist for any gravity data

The equation has an identity term outside the integral, which makes the solution grow without bound

Gravity is a vector field, so the density can be recovered only in the direction of the measured component

Which of the following statements about Maxwell's equations and the Helmholtz equation are correct?

Because ∇·B = 0, the magnetic induction can always be written as B = ∇ × A for some vector potential A

In an insulating uniform medium, for exp(iωt) time dependence, each Cartesian component of E obeys ∇²E + k²E = 0 with k = ω√(με)

Because ∇·B = 0, the magnetic induction can always be written as B = ∇ × A for some vector potential A

In an insulating uniform medium, for exp(iωt) time dependence, each Cartesian component of E obeys ∇²E + k²E = 0 with k = ω√(με)

In a time-varying field E remains curl-free, so it can always be written as −∇V alone, with no vector-potential term

In the quasi-static limit the wavenumber k = √(−iωμσ) is real, so the fields propagate without attenuation

Which of the following statements about Green's identities and boundary-value problems for a source-free interior region are correct?

Green's second identity is ∫(φ∇²ψ − ψ∇²φ) dV = ∮(φ ∂ψ/∂n − ψ ∂φ/∂n) dS

The interior Dirichlet problem for Laplace's equation, with φ given on the boundary, has a unique solution

The interior Neumann problem for Laplace's equation has a solution that is unique only up to an additive constant

Green's second identity is ∫(φ∇²ψ − ψ∇²φ) dV = ∮(φ ∂ψ/∂n − ψ ∂φ/∂n) dS

The interior Dirichlet problem for Laplace's equation, with φ given on the boundary, has a unique solution

The interior Neumann problem for Laplace's equation has a solution that is unique only up to an additive constant

For the interior Neumann problem of Laplace's equation, ∂φ/∂n may be prescribed arbitrarily on the boundary

Which of the following statements about the method of images are correct?

A point charge q above an infinite grounded conducting plane has an image charge −q at the mirror point of q in the plane

The total charge induced on the grounded conducting plane equals −q, the negative of the point charge

For a surface current electrode on a half-space under insulating air, the image has the same sign and doubles the full-space potential

A point charge q above an infinite grounded conducting plane has an image charge −q at the mirror point of q in the plane

The total charge induced on the grounded conducting plane equals −q, the negative of the point charge

For a surface current electrode on a half-space under insulating air, the image has the same sign and doubles the full-space potential

The force on the charge above a grounded plane is q²/(4πε0 h²), where h is the height of the charge above the plane

Potential and Time Varying Fields: Potentials, Laplace, Maxwell and Helmholtz Equations, Green's Theorem, Images and Integral Equations

This chapter answers the Potential and Time Varying Fields heading of Part B2 of the Geology and Geophysics (GG) paper: “Scalar and vector potential fields; Laplace, Maxwell and Helmholtz equations for solution of different types of boundary value problems in Cartesian, cylindrical, and spherical polar coordinates; Green’s theorem; Image theory; integral equations in potential and time-varying field theory.” It teaches scalar and vector potentials, the Maxwell equations and their reduction to the Helmholtz equation, separation of variables for Laplace's equation in the three coordinate systems, the potentials of standard sources, Gauss's and Green's theorems with uniqueness, the method of images, and integral equations in gravity and electromagnetics. The numericals are boundary-value solutions in a strip and a sphere, wavenumber, divergence-theorem fluxes, image charges and forces, and a separable integral equation.

1. Scalar and vector potential fields

A scalar potential φ(r) describes a field by one number at each point, and the field is its gradient: E = −∇V in electrostatics. Geophysics usually writes the gravity potential with the positive sign, U = G ∫ ρ/|r − r′| dV, so that g = +∇U points toward the mass. A field that is the gradient of a potential is conservative: ∇ × F = 0, the line integral between two points is independent of the path and vanishes round a closed loop. Where sources exist, φ obeys Poisson's equation, ∇²φ = −ρ/ε0 for electrostatics and ∇²U = −4πGρ for gravity. In a source-free region it obeys Laplace's equation, ∇²φ = 0.

A vector potential A is defined for a solenoidal field, ∇ · B = 0, by B = ∇ × A. It is not unique, since adding the gradient of any scalar leaves B unchanged (gauge freedom); the Coulomb gauge ∇ · A = 0 is common. By the Helmholtz decomposition, a well-behaved field that vanishes at infinity is the sum of a curl-free part −∇φ and a divergence-free part ∇ × A, and is fixed by its divergence and its curl.

Harmonic functions (solutions of Laplace's equation) have no local maximum or minimum inside a source-free region, so extremes lie on the boundary.
The value at a point equals the average over any sphere centred on it (mean-value property), so potential fields are smooth.

⚠️ Check the sign convention

Physics writes V = −GM/r with g = −∇V; geophysics writes U = +GM/r with g = +∇U. The force is the same, the sign of the potential is not, so a stem giving a potential value must state its convention.

2. Maxwell and Helmholtz equations

In a medium Maxwell's equations are ∇·D = ρ_f, ∇·B = 0, ∇×E = −∂B/∂t and ∇×H = J + ∂D/∂t, with D = εE, B = μH and J = σE. Because ∇·B = 0, B = ∇×A always exists. Because of the third equation E is no longer curl-free in a time-varying field: E = −∇V − ∂A/∂t.

Eliminating H in a source-free uniform medium gives ∇²E = με ∂²E/∂t² + μσ ∂E/∂t. For time dependence exp(iωt) this is the Helmholtz equation ∇²E + k²E = 0 with k² = ω²με − iωμσ. In an insulator (σ = 0), k = ω√(με) = ω/v = 2π/λ is real: undamped waves of speed v = 1/√(με). In a conducting earth at low frequency (σ ≫ ωε, the quasi-static limit, displacement current neglected), k² = −iωμσ and k = √(ωμσ/2)(1 − i): the field diffuses and decays as exp(−z/δ) with skin depth δ = √(2/(ωμσ)) ≈ 503 √(ρ/f) m (ρ in Ω·m, f in Hz). The scalar form ∇²p + k²p = 0 with k = ω/c also describes acoustic and seismic waves in a medium of constant speed c.

Field equations and where each is used
Equation	Form	Used for
Laplace	∇²φ = 0	Source-free static potentials (gravity, magnetic, DC electric)
Poisson	∇²φ = −ρ/ε0	Static potentials with sources
Helmholtz	∇²ψ + k²ψ = 0	One frequency: waves (k real) or EM induction (k complex)
Diffusion	∇²ψ = μσ ∂ψ/∂t	Quasi-static EM fields in the earth
Wave	∇²ψ = (1/v²) ∂²ψ/∂t²	Seismic and radar waves in time domain

⚠️ Which k?

k = ω/v = 2π/λ belongs to an insulating medium only. In the quasi-static conductor |k| = √(ωμσ) = √2/δ and k has equal real and imaginary parts, so the field is damped, not propagating. The outgoing point-source solution of Helmholtz's equation is exp(−ikr)/(4πr) for exp(iωt) time dependence.

3. Laplace's equation: separation of variables in three coordinate systems

To solve ∇²φ = 0, write φ as a product of functions of one coordinate each. The equation splits into ordinary differential equations joined by separation constants, and the boundary conditions select the constants.

Cartesian, φ = X(x)Y(y)Z(z): X″/X + Y″/Y + Z″/Z = 0. With X″ = −α²X and Y″ = −β²Y, Z″ = (α² + β²)Z: oscillatory in x and y, exponential in z.
Cylindrical (ρ, θ, z): ∇²φ = (1/ρ)∂ρ(ρ∂ρφ) + (1/ρ²)∂θ²φ + ∂z²φ. Without z: ρⁿ cos nθ, ρ⁻ⁿ cos nθ (and sine forms) and ln ρ for n = 0. With z dependence: Jₙ(kρ) cos nθ exp(±kz), Bessel functions.
Spherical (r, θ, φ) with azimuthal symmetry: (1/r²)∂r(r²∂rφ) + (1/(r² sin θ))∂θ(sin θ ∂θφ) = 0, giving φ = Σ (A_l r^l + B_l r^−(l+1)) P_l(cos θ), with P0 = 1, P1 = cos θ, P2 = (3cos²θ − 1)/2. The l = 0 term is the monopole 1/r and l = 1 gives the dipole cos θ/r².

Example (Cartesian). Strip 0 < x < a, y > 0, with φ = 0 on x = 0, x = a and as y → ∞, and φ = V0 on y = 0. Zero at both edges forces X = sin(nπx/a), so Y decays as exp(−nπy/a), and the Fourier series of the constant V0 on the base gives φ = (4V0/π) Σ over odd n of (1/n) sin(nπx/a) exp(−nπy/a). Higher terms die faster with y, so away from the base the n = 1 term dominates: at x = a/2, y = a, φ ≈ (4/π)e^(−π) V0 = 0.055 V0.

Example (spherical). A sphere of radius a has φ(a, θ) = V0 cos θ, with φ finite at r = 0 and φ → 0 as r → ∞. Only l = 1 matches cos θ. Inside, φ = V0 (r/a) cos θ (the r⁻² term is excluded at the centre); outside, φ = V0 (a/r)² cos θ (the r term is excluded at infinity). The two agree at r = a. At r = 2a and θ = 60°, φ = V0 × (1/4) × 0.5 = 0.125 V0.

⚠️ Boundary conditions decide the solution

The general solution has two pieces per mode, such as r^l and r^−(l+1), exp(+kz) and exp(−kz). The physical conditions choose between them. Forgetting the condition at infinity, or at the centre, is the commonest error in these problems.

4. Potentials of point, line, dipole and sphere sources

In the positive geophysical convention a point mass gives U = GM/r and g = GM/r² toward the mass. An infinite line source of mass λ per unit length gives U = −2Gλ ln ρ + const and g = 2Gλ/ρ. An infinite uniform plane sheet of surface density σ gives a constant field g = 2πGσ, independent of distance, which is the origin of the Bouguer slab. A dipole (charges +q and −q a distance d apart, moment p = qd) has far potential V = p cos θ/(4πε0 r²), with E_r = 2p cos θ/(4πε0 r³) and E_θ = p sin θ/(4πε0 r³): the potential falls as r⁻² and the field as r⁻³, and the field on the axis is twice that on the equatorial plane at the same r.

How potential and field fall with distance
Source	Potential	Field
Point (monopole)	1/r	1/r²
Dipole	1/r²	1/r³
Quadrupole	1/r³	1/r⁴
Infinite line (2-D)	ln ρ	1/ρ
Infinite plane sheet	Linear in distance	Constant

A uniform sphere of radius a, density ρ and mass M = 4πa³ρ/3 acts, outside, exactly as a point mass at its centre (the shell theorem): U = GM/r. Inside, only the mass within radius r attracts, so g = GMr/a³ = (4/3)πGρ r, increasing linearly from zero at the centre to GM/a² at the surface; the potential inside is U = 2πGρ(a² − r²/3). Any finite source has a multipole expansion U = GM/r + (dipole term)/r² + (quadrupole term)/r³ + …, so far away the lowest non-vanishing term dominates; taking the origin at the centre of mass removes the dipole term.

🎯 Why the slab field is constant

The pull of a patch falls as 1/r², but a patch inside a given cone of directions has area growing as r²/cos θ and its vertical part carries cos θ, so each element of solid angle adds Gσ dΩ. The half-space of directions spans 2π, giving 2πGσ at any height: the Bouguer slab term.

5. Gauss's and Green's theorems, and uniqueness

The divergence (Gauss) theorem is ∮ F·n dS = ∫ ∇·F dV. For gravity it gives ∮ g·n dS = −4πG M_enclosed: the flux depends only on the enclosed mass, whatever its shape. Example: F = (x², y², z²) on the cube 0 ≤ x, y, z ≤ 2. Then ∇·F = 2(x + y + z) and the volume integral is 2 × 3 × (∫x dV) = 2 × 3 × 8 = 48. Check: the faces x = 2, y = 2 and z = 2 each give 4 × 4 = 16, and the faces at 0 give nothing.

Green's identities follow by putting F = φ∇ψ in the divergence theorem. The first is ∫(φ∇²ψ + ∇φ·∇ψ) dV = ∮ φ ∂ψ/∂n dS; swapping φ and ψ and subtracting gives the second, ∫(φ∇²ψ − ψ∇²φ) dV = ∮(φ ∂ψ/∂n − ψ ∂φ/∂n) dS. Take ψ = G(r, r′) = 1/(4π|r − r′|), which satisfies ∇²G = −δ(r − r′). For ∇²φ = −f the second identity gives the Green's representation φ(r) = ∫ G f dV′ + ∮ (G ∂φ/∂n′ − φ ∂G/∂n′) dS′: the potential inside is fixed by the sources inside and by φ and ∂φ/∂n on the boundary. A Green's function that vanishes on S leaves only φ on S to be given.

Uniqueness. If φ1 and φ2 both solve Laplace's equation in V with the same φ on S (Dirichlet data), then w = φ1 − φ2 is harmonic and zero on S. The first identity with φ = ψ = w gives ∫|∇w|² dV = 0, so w = 0. For Neumann data (∂φ/∂n given) the solution is unique up to an additive constant. Consequently any solution found by any method, whether separation of variables or images, is the solution.

⚠️ Neumann data cannot be chosen freely

For Laplace's equation inside a closed surface the net outward flux ∮ ∂φ/∂n dS must be zero, because there is no source inside. Dirichlet data can be any continuous function, but Neumann data must satisfy this compatibility condition before a solution exists.

6. Image theory

The method of images replaces a boundary by fictitious sources outside the region so that the boundary condition holds. By the uniqueness theorem the result is the true potential inside the region. A point charge q at height h above an infinite grounded conducting plane has an image −q at depth h. The potential above the plane is that of the two charges, the induced surface charge totals −q, and the force on q is q²/(4πε0 (2h)²) toward the plane, because the distance to the image is 2h. Example: q = 5 nC at h = 0.30 m gives F = 9 × 10⁹ × (5 × 10⁻⁹)²/0.6² = 6.25 × 10⁻⁷ N.

Boundaries and their images
Boundary	Image of a source q
Grounded conducting plane (φ = 0)	−q at the mirror point
Insulating plane (no normal gradient), e.g. ground surface for current	+q at the mirror point: the potential doubles
Grounded sphere of radius a, source at distance d > a from the centre	−qa/d at distance a²/d from the centre, on the same line
Interface between resistivities ρ1 and ρ2	kq at the mirror point, k = (ρ2 − ρ1)/(ρ2 + ρ1)

The interface row contains the two limits: ρ2 → ∞ gives k = +1 (insulating) and ρ2 → 0 gives k = −1 (grounded conductor). For a current electrode I on the surface of a layer of resistivity ρ1 and thickness h over a half-space, the images between the surface and the interface form an infinite series, V = (Iρ1/2π)[1/r + 2 Σ kⁿ/√(r² + (2nh)²)], n = 1, 2, …. For example, with a sphere of a = 0.20 m and a 10 nC charge at d = 0.50 m the image is −4 nC at 0.08 m from the centre. In time-varying fields a perfect conductor mirrors an electromagnetic dipole; for a finitely conducting earth the complex image method uses a complex depth set by the skin depth.

⚠️ The distance to the image, not to the plane

The force on a charge above a grounded plane uses the distance to the image, 2h; using h overestimates it by a factor of four. The image lies outside the region and is only a device: the field below the plane is zero.

7. Integral equations in potential and time-varying field theory

An integral equation has the unknown inside an integral. In a Fredholm equation of the first kind f(x) = ∫ K(x, s) φ(s) ds the unknown appears only under the integral; in the second kind φ(x) = f(x) + λ ∫ K(x, s) φ(s) ds it also appears outside. A kernel that separates as a(x)b(s) reduces the second kind to algebra. Example: φ(x) = x + ∫₀¹ x s φ(s) ds. The integral is a number C = ∫₀¹ s φ(s) ds, so φ(x) = x(1 + C). Then C = (1 + C) ∫₀¹ s² ds = (1 + C)/3, which gives C = 1/2 and φ(x) = 1.5x.

In potential theory the forward problems are integrals: the gravity anomaly g(r) = G ∫ Δρ(r′)(z′ − z)/|r − r′|³ dV′, or the field of a surface layer. Recovering Δρ from g is a first-kind equation and is ill-posed: the kernel smooths, so small data errors produce large changes in the model, and the solution is non-unique, since many densities give the same external field. It needs regularisation and constraints. After discretising the model into n cells the equation becomes the matrix equation A m = d, and a first-kind kernel gives a badly conditioned A.

In time-varying fields the volume integral equation for EM scattering is E(r) = E_p(r) + ∫ Ĝ(r, r′) Δσ(r′) E(r′) dV′, with E_p the primary field of the background, Δσ the anomalous conductivity and Ĝ the background Green's function. The unknown E sits inside the integral over the anomalous body, so it is a second-kind equation, and cells are needed only there. Boundary integral equations reduce a 3-D problem to its surfaces, for example the unknown surface charge q_s of a conductor in V(r) = ∮ q_s/(4πε0|r − r′|) dS′. Both are solved by cell discretisation (the method of moments).

ℹ️ First kind versus second kind

A second-kind equation has an identity term outside the integral and is well-posed away from special values of λ (the Fredholm alternative); a first-kind equation with a smooth kernel lacks it and is ill-posed, as in gravity and magnetic inversion, which therefore need regularisation.

Key takeaways

F = −∇φ is conservative, B = ∇ × A follows from ∇·B = 0, sources give Poisson's equation (∇²U = −4πGρ) and free space Laplace's. Harmonic functions have no interior extremes. State the sign convention: geophysics uses U = +GM/r, g = +∇U.
Maxwell's equations give the Helmholtz equation ∇²E + k²E = 0 with k² = ω²με − iωμσ. Insulator: k = ω/v = 2π/λ. Quasi-static conductor: k² = −iωμσ, diffusion, skin depth δ = √(2/(ωμσ)) ≈ 503 √(ρ/f) m.
Separation of variables: Cartesian gives sine with exponential decay, cylindrical ρⁿ cos nθ, ln ρ and Bessel functions, spherical (A r^l + B r^−(l+1)) P_l(cos θ). Finiteness at the centre and decay at infinity select the terms.
Point source: potential 1/r, field 1/r². Dipole: potential 1/r², field 1/r³, on-axis field twice the equatorial one. A uniform sphere acts as a point mass outside and has g ∝ r inside; an infinite sheet gives the constant 2πGσ.
Gauss: ∮F·n dS = ∫∇·F dV. Green's second identity gives the Green's representation and the first gives uniqueness: Dirichlet data fix φ, Neumann data fix it up to a constant and need zero net flux for Laplace's equation.
Images: grounded plane −q at the mirror point, force q²/(4πε0(2h)²); insulating plane +q; grounded sphere −qa/d at a²/d; interface kq, k = (ρ2 − ρ1)/(ρ2 + ρ1). Second-kind integral equations are well-posed, first-kind ones (gravity and magnetic inversion) are ill-posed.

Practice questions (17)

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.

At a large distance r along the axis of an ideal point electric dipole, how does the electrostatic potential vary with r?
1. As 1/r, in the same way as the potential of a single point charge
2. As 1/r³, in the same way as the electric field of the dipole itself
3. As 1/r², because the two opposite charges cancel the leading 1/r term
4. As ln r, in the same way as the potential of an infinite line of charge
Show answer
Answer: C — As 1/r², because the two opposite charges cancel the leading 1/r term
The 1/r terms of +q and −q cancel, leaving the dipole term V = p cos θ/(4πε0 r²), so the potential falls as 1/r² along the axis. The 1/r³ law is the fall-off of the dipole field, which is the tempting confusion between field and potential. The ln r law belongs to a two-dimensional line source.
A solid sphere of radius a has uniform density, and M is its total mass. How does the gravitational acceleration vary inside the sphere, for 0 < r < a?
1. It increases linearly with r, from zero at the centre to GM/a² at the surface
2. It is zero everywhere inside, exactly as it is inside a thin uniform spherical shell
3. It stays equal to the surface value GM/a², because the total mass of the sphere is unchanged
4. It rises as 1/r² towards the centre, because the distance to the centre of mass keeps shrinking
Show answer
Answer: A — It increases linearly with r, from zero at the centre to GM/a² at the surface
Only the mass inside radius r attracts (the shell outside contributes nothing), so g = G(4πr³ρ/3)/r² = (4/3)πGρ r, linear in r and equal to GM/a² at r = a. Zero field is the result for a hollow shell, not for a solid sphere. A constant g would need the whole mass to act at every depth.
For an ideal point dipole, the electric field magnitude on the dipole axis at distance r is compared with that on the equatorial plane at the same distance r. What is the ratio (axis to equator)?
1. One half, because the axis points along the dipole while the equator points across it
2. One, because both points are at the same distance r from the centre of the dipole
3. Four, because the field of a dipole falls as the cube of the distance from its centre
4. Two, because E = 2p/(4πε0 r³) on the axis and E = p/(4πε0 r³) on the equatorial plane
Show answer
Answer: D — Two, because E = 2p/(4πε0 r³) on the axis and E = p/(4πε0 r³) on the equatorial plane
On the axis (θ = 0) E = 2p/(4πε0 r³); on the equatorial plane (θ = 90°) E = p/(4πε0 r³). The ratio is 2. Equal distance does not mean equal field, because the dipole field is anisotropic, and the r⁻³ law fixes only how each value falls with r.
A potential satisfies Laplace's equation in the strip 0 < x < a, y > 0. It is zero on x = 0, on x = a and as y → ∞. Which form is an acceptable separated solution?
1. sinh(nπx/a) sin(nπy/a), which is the product of a hyperbolic function of x and a sine function of y
2. sin(nπx/a) exp(−nπy/a), a sine in x that vanishes at both edges and a decay in y
3. cos(nπx/a) exp(−nπy/a), a cosine in x that decays as y increases
4. sin(nπx/a) exp(+nπy/a), a sine in x that grows exponentially with y
Show answer
Answer: B — sin(nπx/a) exp(−nπy/a), a sine in x that vanishes at both edges and a decay in y
Zero at x = 0 and x = a forces X = sin(nπx/a), so Y″ = +(nπ/a)²Y and Y is exponential. Vanishing as y → ∞ keeps only exp(−nπy/a). The other three satisfy Laplace's equation but not the conditions: the cosine is not zero at x = 0, the growing exponential diverges as y → ∞, and sinh(nπx/a) sin(nπy/a) is not zero on x = a and does not vanish at large y.
Which of the following functions of the plane coordinates, with r = √(x² + y²), satisfies Laplace's equation in two dimensions away from the origin?
1. x² + y², whose second derivatives with respect to x and y are both positive
2. r itself, which increases smoothly and symmetrically away from the origin
3. 1/r, which decays smoothly and symmetrically away from the origin
4. x² − y², whose x and y second derivatives are equal and opposite
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Answer: D — x² − y², whose x and y second derivatives are equal and opposite
For x² − y² the second derivatives are +2 and −2, so ∇² = 0. For x² + y² it is 4, for r it is 1/r and for 1/r it is 1/r³ in two dimensions, so none of these is harmonic. The 2-D radially symmetric harmonic function is ln r, not 1/r, which is the tempting import from three dimensions.
A uniform conducting earth is probed at low frequency, so that the displacement current is negligible against the conduction current. For the electric field E with time dependence exp(iωt), which statement is correct?
1. E obeys the wave equation ∇²E = με ∂²E/∂t² and travels without any damping at the speed 1/√(με)
2. E obeys a diffusion equation, equivalent to ∇²E + k²E = 0 with k² = −iωμσ, so it is damped
3. E obeys Laplace's equation ∇²E = 0 because the fields are effectively static at such low frequencies
4. E obeys the Helmholtz equation with a real k = ω√(με), so that only its phase changes with depth
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Answer: B — E obeys a diffusion equation, equivalent to ∇²E + k²E = 0 with k² = −iωμσ, so it is damped
Dropping ∂²E/∂t² leaves ∇²E = μσ ∂E/∂t. With exp(iωt) this is ∇²E + k²E = 0 with k² = −iωμσ, so k is complex with equal real and imaginary parts and the amplitude decays over a skin depth. The wave equation and the real-k Helmholtz equation belong to an insulating medium, and Laplace's equation would ignore the time variation entirely.
A function φ satisfies Laplace's equation inside a closed region V and φ = 0 on the whole boundary S. What does Green's first identity show about φ inside V?
1. φ can still be positive inside, because a harmonic function is allowed to have an interior maximum
2. φ can be non-zero inside, because Laplace's equation alone cannot fix a solution without initial conditions
3. φ = 0 everywhere in V, since ∫|∇φ|² dV = ∮ φ ∂φ/∂n dS = 0
4. φ is a constant fixed only up to an additive constant, as for Neumann data on the boundary
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Answer: C — φ = 0 everywhere in V, since ∫|∇φ|² dV = ∮ φ ∂φ/∂n dS = 0
Put ψ = φ in the first identity: ∫(φ∇²φ + |∇φ|²) dV = ∮ φ ∂φ/∂n dS. With ∇²φ = 0 and φ = 0 on S, the right side vanishes, so ∫|∇φ|² dV = 0 and φ is constant, hence zero because it is zero on S. Interior maxima are forbidden for harmonic functions. The additive constant belongs to the Neumann problem, where φ itself is not given on S.
A point charge q is at distance d from the centre of a grounded conducting sphere of radius a, with d > a. Which image charge reproduces the zero potential on the sphere?
1. −qa/d, placed at distance a²/d from the centre on the line joining the centre to q
2. −q, placed at the mirror point of q in the surface, at distance 2a − d from the centre
3. −qd/a, placed at distance a²/d from the centre on the line joining the centre to q
4. +qa/d, placed at distance a²/d from the centre on the line joining the centre to q
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Answer: A — −qa/d, placed at distance a²/d from the centre on the line joining the centre to q
The image is q′ = −qa/d at a²/d from the centre. Check at the near surface point on the line: the distance to q is d − a and to the image a − a²/d = a(d − a)/d, so the potentials are proportional to q/(d − a) − (qa/d) × d/[a(d − a)] = 0. The far surface point works the same way, with d + a and a(d + a)/d. The planar rule (−q at the mirror point) holds only for a plane, and +qa/d would add to the potential instead of cancelling it.
Recovering the density distribution from a measured gravity anomaly is described by a first-kind Fredholm equation. Why is this inverse problem called ill-posed?
1. Small errors in the data can produce large changes in the recovered density, and many densities fit the data
2. The kernel of the integral is not symmetric, so no solution of the equation can exist for any gravity data
3. The equation has an identity term outside the integral, which makes the solution grow without bound
4. Gravity is a vector field, so the density can be recovered only in the direction of the measured component
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Answer: A — Small errors in the data can produce large changes in the recovered density, and many densities fit the data
The gravity kernel smooths, so inversion amplifies noise (instability), and by Gauss's theorem and the equivalent-layer idea different density distributions give identical external fields (non-uniqueness). A second-kind equation, with an identity term outside the integral, is the well-posed type. Symmetry of the kernel and the vector nature of g are not the reason.
Which of the following statements about Maxwell's equations and the Helmholtz equation are correct?
1. Because ∇·B = 0, the magnetic induction can always be written as B = ∇ × A for some vector potential A
2. In an insulating uniform medium, for exp(iωt) time dependence, each Cartesian component of E obeys ∇²E + k²E = 0 with k = ω√(με)
3. In a time-varying field E remains curl-free, so it can always be written as −∇V alone, with no vector-potential term
4. In the quasi-static limit the wavenumber k = √(−iωμσ) is real, so the fields propagate without attenuation
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Answer: A — Because ∇·B = 0, the magnetic induction can always be written as B = ∇ × A for some vector potential A; B — In an insulating uniform medium, for exp(iωt) time dependence, each Cartesian component of E obeys ∇²E + k²E = 0 with k = ω√(με)
∇·B = 0 gives B = ∇ × A, and the insulating-medium wave equation reduces to the Helmholtz equation with real k. In a time-varying field ∇ × E = −∂B/∂t ≠ 0, so E needs the −∂A/∂t term as well. In the quasi-static limit k = √(ωμσ/2)(1 − i) is complex, which means attenuation over the skin depth.
Which of the following statements about Green's identities and boundary-value problems for a source-free interior region are correct?
1. Green's second identity is ∫(φ∇²ψ − ψ∇²φ) dV = ∮(φ ∂ψ/∂n − ψ ∂φ/∂n) dS
2. The interior Dirichlet problem for Laplace's equation, with φ given on the boundary, has a unique solution
3. The interior Neumann problem for Laplace's equation has a solution that is unique only up to an additive constant
4. For the interior Neumann problem of Laplace's equation, ∂φ/∂n may be prescribed arbitrarily on the boundary
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Answer: A — Green's second identity is ∫(φ∇²ψ − ψ∇²φ) dV = ∮(φ ∂ψ/∂n − ψ ∂φ/∂n) dS; B — The interior Dirichlet problem for Laplace's equation, with φ given on the boundary, has a unique solution; C — The interior Neumann problem for Laplace's equation has a solution that is unique only up to an additive constant
The second identity is stated correctly, the Dirichlet problem is unique (the difference of two solutions is zero by the first identity), and the Neumann solution is unique up to a constant. Neumann data cannot be arbitrary: with no source inside, the divergence theorem forces ∮ ∂φ/∂n dS = ∫∇²φ dV = 0.
Which of the following statements about the method of images are correct?
1. A point charge q above an infinite grounded conducting plane has an image charge −q at the mirror point of q in the plane
2. The total charge induced on the grounded conducting plane equals −q, the negative of the point charge
3. For a surface current electrode on a half-space under insulating air, the image has the same sign and doubles the full-space potential
4. The force on the charge above a grounded plane is q²/(4πε0 h²), where h is the height of the charge above the plane
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Answer: A — A point charge q above an infinite grounded conducting plane has an image charge −q at the mirror point of q in the plane; B — The total charge induced on the grounded conducting plane equals −q, the negative of the point charge; C — For a surface current electrode on a half-space under insulating air, the image has the same sign and doubles the full-space potential
The image is −q at the mirror point, the induced charge totals −q, and an insulating boundary needs a same-sign image, which doubles the potential (Iρ/2πr against Iρ/4πr). The force uses the distance to the image, 2h, so it is q²/(4πε0 (2h)²), a quarter of the quoted value.
A sphere of radius a has the potential φ(a, θ) = V0 cos θ on its surface, with V0 = 80 V. The potential is finite at the centre and tends to zero at infinity, and it satisfies Laplace's equation outside the sphere. The potential at r = 2a, θ = 60°, in volts, is ____.
Numerical answer — type the value.
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Answer: 10
Only the l = 1 term matches cos θ. Outside, the solution that vanishes at infinity is φ = V0 (a/r)² cos θ, which equals V0 cos θ at r = a. At r = 2a and θ = 60°: φ = 80 × (1/2)² × cos 60° = 80 × 0.25 × 0.5 = 10 V. Using the interior form (r/a) would give 80 × 2 × 0.5 = 80 V, which wrongly grows with r.
A monochromatic wave of frequency 50 Hz obeys the scalar Helmholtz equation ∇²p + k²p = 0 in a medium where the wave speed is 2500 m/s. The magnitude of the wavenumber k, in rad/km to the nearest whole number, is ____.
Numerical answer — type the value.
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Answer: 126
k = ω/c = 2πf/c = 2π × 50/2500 = 0.1257 rad/m. In rad/km this is 125.66, which is 126. Quoting 0.126 forgets the conversion to kilometres, and using f/c = 0.02 omits the 2π, which would be cycles per metre.
Let φ = x² + 2y² + 3z² (lengths in metres). The surface integral ∮ ∂φ/∂n dS of the outward normal derivative over the surface of the cube −1 ≤ x, y, z ≤ 1 equals ____.
Numerical answer — type the value.
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Answer: 96
By the divergence theorem ∮ ∇φ·n dS = ∫ ∇²φ dV. Here ∇²φ = 2 + 4 + 6 = 12, and the cube has volume 2³ = 8, so the integral is 96. Check by faces: the two x-faces give 2 × (2 × 4) = 16, the two y-faces 2 × (4 × 4) = 32 and the two z-faces 2 × (6 × 4) = 48, total 96.
A point charge q = 2.0 nC is held 0.50 m above a large grounded conducting plane. Take 1/(4πε0) = 9.0 × 10⁹ N·m²/C². The magnitude of the force on the charge, in nN, is ____.
Numerical answer — type the value.
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Answer: 36
The image is −2.0 nC at depth 0.50 m, so the separation is 2h = 1.00 m. F = 9.0 × 10⁹ × (2.0 × 10⁻⁹)²/1.00² = 3.6 × 10⁻⁸ N = 36 nN, directed toward the plane. Using h = 0.50 m instead of 2h would give 144 nN.
The function φ satisfies the integral equation φ(x) = 1 + 2 ∫₀¹ x s φ(s) ds for 0 ≤ x ≤ 1. The value φ(1) is ____.
Numerical answer — type the value.
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Answer: 4
The integral is a number C = ∫₀¹ s φ(s) ds, so φ(x) = 1 + 2Cx. Then C = ∫₀¹ s(1 + 2Cs) ds = 1/2 + 2C/3, which gives C/3 = 1/2 and C = 3/2. So φ(x) = 1 + 3x and φ(1) = 4. Check: ∫₀¹ s(1 + 3s) ds = 1/2 + 1 = 3/2.