Maxwell's Equations in Matter

Question 1

What does an ideal electric dipole moment consist of?

Answer 1

Two equal but opposite charges a distance d apart, where qd= p, the dipole moment.

Question 2

If we has a uniform cuboid aligned with cartesian axis which there are balancing charge densities +-ρ, what does the polarisation equal if we displace all the positive charges by a distance x in the positive x direction? Why does this happens?

Answer 2

Px = ρ*x. This polarisation happens because originally the positive and negative charges are on top of each other, but now the positive charges have been displaced so there is a charge imbalance on each face.

Question 3

What is the equation for the surface charge density on the cuboid surface?

Answer 3

σ(P) = P(hat).n(hat)

Question 4

What is the best way to look at the non-uniform polarisation?

Answer 4

Break the object into a series of small cuboids of length dx along the x axis, each with individually uniform polarisation.

Question 5

Once we have broken an object up into cuboids, what do each of the cuboids have as polarisation?

Answer 5

Cuboid centred on x has polarisation +-Px(x), next one centred on x+dx has polarisation charge density +-Px(x+dx), and there is a charge excess of A(Px(x)-Px(x+dx)) on the interface between them, where A is the cross sectional area of the cuboid perpendicular to x.

Question 6

What is the equation for polarisation charge density?

Answer 6

ρ(P) = (APx(x)-APx(x+dx))/Adx = -(Px(x+dx)-Px(x))/dx = -dPx/dx

Question 7

In the limit for dx -> 0 and adding similar terms for y and z, what is the final equation for the polarisation charge density?

Answer 7

ρ(P) = -∇.P

Question 8

What is an easy way to think of a sphere of radius a which has uniform P at all points?

Answer 8

Think of it as the superposition of 2 spheres of uniform volume charge densities +-ρ displaced from each other by x, such that P = ρx

Question 9

For 2 spheres of charge density +/ρ centred at +- x/2, what is the equation for the field due to the polarisation charges at r inside the sphere?

Answer 9

E(r) = -P/3ε0

Question 10

What is the equation for the polarisation of the sphere in terms of E0, the external field?

Answer 10

P = (3χ/(3+χ))ε0E0

Question 11

What is the equation for the electric field inside the sphere?

Answer 11

E = E0-P/3ε0 = 3/(2+εr) *E0

Question 12

What is the equation for the dipole moment outside of the spheres?

Answer 12

p = Qx = 4π/3 *a^3 *P

Question 13

What happens if the polarisation in a material changes with time?

Answer 13

Charges must move, which creates polarisation currents which generate magnetic fields.

Question 14

What is the equation for the polarisation current density J(P)?

Answer 14

J(P) = dP/dt (partial derivative)

Question 15

What can magnetised materials be envisaged as?

Answer 15

As having distributions of small current loops.

Question 16

What is the equation for magnetism if we split a material up into current loops?

Answer 16

M = m/V = IΔxΔy/ΔxΔyΔx = I/Δz, where each loop has area ΔxΔy, so magnetic moment is IΔxΔy

Question 17

Which direction does the magnetisation point for a material with current loops in x-y plane?

Answer 17

z(hat) direction, so M = (I/Δz) z(hat)

Question 18

What happens to the currents inside the block when the magnetisation is constant and uniform?

Answer 18

The current is constant so current from neighbouring loops cancel inside the material, leaving only current running along outside of block.

Question 19

What is the equation for j(M), the general expression for the equivalent surface current density?

Answer 19

j(M) = M X n(hat)

Question 20

What happens if the magnetisation varies with position inside the material?

Answer 20

The currents within the material will no longer cancel and a volume current density arises.

Question 21

What is the equation for the residual current created by a varying magnetisation for two current loops centred at x and x+Δx?

Answer 21

ΔI = I*(x+Δx)-I(x) ~ dI/dx Δx

Question 22

What is the equation for the volume magnetisation current J(M)?

Answer 22

J(M) = ∇ X M

Question 23

In Gauss’s law, what do we separate the total charge density ρ into? Why do we do this?

Answer 23

ρ = ρ(f) + ρ(P) = ρ(f) - ∇.P, where f is free and p is polarisation. We separate this because the free charges are the ones we can manipulate, while the polarisation charges just come with the territory.

Question 24

How can we rearrange the separated Gauss’s law to find the first of Maxwells equations in matter?

Answer 24

Put the divergence on the left hand side and multiply both sides by ε0, giving ∇.(ε0E + P) = ρ(f)

Question 25

What is maxwell’s first equation in matter?

Answer 25

Define new quantity D, known as the displacement, where D = ε0E + P, so the first equation is ∇.D = ρ(f)

Question 26

What is the integral form of maxwells first equation in matter?

Answer 26

closed integral of D.dS = integral of ρ(f) dV = Q(f) = free charge enclosed by the surface of integration on the left.

Question 27

What is the equation for P for a given E? What can we do with this equation?

Answer 27

P = ε0χE, where χ is the electric susceptibility. Sub this equation into the equation for D, so D = ε0εrE, where εr = 1+χ = relative permittivity

Question 28

What is χ (and therefore εr) in general dependent on?

Answer 28

The frequency, reflecting how fast a material can respond to a changing field.

Question 29

How can we split the total current density J in Amperes law?

Answer 29

Split into free current density, a term due to magnetisation currents and a term due to changing polarisation: J = J(f)+J(M)+J(P) = J(f)+(∇XM) + dP/dt

Question 30

What do we do with the split equation for J?

Answer 30

Sub it into amperes law, put the cur of the magnetisation on the left hand side and divide through by μ0.

Question 31

What do we define the new vector field H as and where do we put it?

Answer 31

H = B/μ0 -M, so sub this into the new Amperes law to get the new maxwell equation in matter.

Question 32

What is the second of maxwells equations in matter?

Answer 32

∇XH = J(f)+dD/dt

Question 33

What happens to the other two unchanged maxwell equations?

Answer 33

They remain unchanged: only the first and the fourth change in matter as the others involve no source terms.