Physics 2 for Electrical Engineering

Ben Gurion University of the Negev www.bgu.ac.il/atomchip,www.bgu.ac.il/nanocenter Physics 2 for Electrical Engineering Lecturers: Daniel Rohrlich , Ron Folman Teaching Assistants: Ben Yellin, Yoav Etzioni Grader: Gad Afek Week 12. Outlook – Introduction • method of images • Poynting vector: examples • a paradox • transformers • Maxwell’s equations in matter • reflection and refraction • polarized light • spherical waves • interference Source: Halliday, Resnick and Krane, 5th Edition, Chap. 38; The Feynman Lectures on Physics, Chaps. 17, 21 and 27.

Introduction In the previous weeks in this course, we have offered you a basic understanding of classical electrodynamics, the theory electric and magnetic fields and forces. This theory – which Maxwell perfected – puts powerful tools in your hands. It does not include quantum theory but it is already consistent with Einstein’s special theory of relativity (although we have not talked about relativity).

Introduction In the previous weeks in this course, we have offered you a basic understanding of classical electrodynamics, the theory electric and magnetic fields and forces. This theory – which Maxwell perfected – puts powerful tools in your hands. It does not include quantum theory but it is already consistent with Einstein’s special theory of relativity (although we have not talked about relativity). In this lecture, we hope to entertain you with examples of how much you can already do with this theory, or can do with just one or two more tricks, and examples of everyday phenomena that you can now understand in detail.

Method of images In electrostatics, E and B are constant in time; so we can write E as a gradient: Gauss’s law then becomes Now suppose we have any function V0(r) with the property that Any linear combination of V(r) and V0(r) will be another solution of Gauss’s law. Which combination is the right one? The answer depends on boundary conditions.

Method of images We have solved electrostatic problems with a few charges, or with a lot of symmetry. Here is a different sort of electrostatic problem: A particle of charge q is at rest a distance R from an infinite conducting plane. What is the potential V(r) everywhere to the right of the conductor?

z R x q Method of images We have solved electrostatic problems with a few charges, or with a lot of symmetry. Here is a different sort of electrostatic problem: A particle of charge q is at rest a distance R from an infinite conducting plane. What is the potential V(r) everywhere to the right of the conductor? It can’t be just the potential of the particle…

z R Equipotential surface x q Method of images We have solved electrostatic problems with a few charges, or with a lot of symmetry. Here is a different sort of electrostatic problem: A particle of charge q is at rest a distance R from an infinite conducting plane. What is the potential V(r) everywhere to the right of the conductor? It can’t be just the potential of the particle, because the plane is an equipotential: V(0,y,z) = constant.

z R R x q –q Method of images Answer: Just put an “image” charge –q at on the x-axis at the point (–R,0,0) and write down the potential V(r) for x > 0:

z R R x q –q Method of images Answer: Just put an “image” charge –q at on the x-axis at the point (–R,0,0) and write down the potential V(r) for x > 0: The Laplacian of the “image” potential vanishes for x > 0 and also V(0,y,z) = 0!

z R Equipotential surface x q Method of images Answer: Just put an “image” charge –q at on the x-axis at the point (–R,0,0) and write down the potential V(r) for x > 0: The Laplacian of the “image” potential vanishes for x > 0 and also V(0,y,z) = 0!

z D R x q Equipotential surface Method of images This method also works for a hollow conducting sphere:

z D R' R x q q' Equipotential surface Method of images This method also works for a hollow conducting sphere:

E k B Poynting vector The relation between energy and momentum in radiation:

Poynting vector The relation between energy and momentum in radiation: Consider electromagnetic radiation normal to a metal sheet of size L2 and conductance σ. The electric field induces a current density J = σE. The magnetic field is B perpendicular to J, and like E lies in the metal sheet. It induces a force density J×B = σE×B in the direction of the radiation. The power delivered by the radiation to the metal sheet is I2R = (LJ)2/σ so the power per unit surface area is J2/σ = JE while the force per unit surface area is JB = JE/c. Although this result is for a special case, it must correspond to the energy and momentum content of the electromagnetic wave. Since the averaged Poynting vector Sav is the energy density flux in radiation, the momentum density flux in radiation – i.e. the radiation pressure – must be Sav/c.

Poynting vector Example 1: A great amount of debris – asteroids, meteors, dust – exists in interplanetary space. Although we might expect the smallest dust particles to be of molecular size, very little of the dust in our solar system is smaller than about 0.2 μm. Why?

Poynting vector Example 1: A great amount of debris – asteroids, meteors, dust – exists in interplanetary space. Although we might expect the smallest dust particles to be of molecular size, very little of the dust in our solar system is smaller than about 0.2 μm. Why? The sun radiates energy at the rate P = 3.90 × 1026 W. Assume each dust particle is spherical, with density ρ = 3.0 × 103 kg/m3 (density of the earth’s crust), and absorbs all energy incident on it. The mass of the sun is M = 1.99 × 1030 kg, and Newton’s gravitational constant is G = 6.67 × 10–11 m3/s2·kg. Assume the distance of the dust particle from the sun is R.

Poynting vector Example 1: A great amount of debris – asteroids, meteors, dust – exists in interplanetary space. Although we might expect the smallest dust particles to be of molecular size, very little of the dust in our solar system is smaller than about 0.2 μm. Why? The sun radiates energy at the rate P = 3.90 × 1026 W. Assume each dust particle is spherical, with density ρ = 3.0 × 103 kg/m3 (density of the earth’s crust), and absorbs all energy incident on it. The mass of the sun is M = 1.99 × 1030 kg, and Newton’s gravitational constant is G = 6.67 × 10–11 m3/s2·kg. Assume the distance of the dust particle from the sun is R. What is its radius r if the gravitational attraction of the sun just balances the radiation pressure from the sun?

Poynting vector Example 1: A great amount of debris – asteroids, meteors, dust – exists in interplanetary space. Although we might expect the smallest dust particles to be of molecular size, very little of the dust in our solar system is smaller than about 0.2 μm. Why? Answer: The attraction of the sun is The energy flux from the sun per unit area is the Poynting vector S = P/4πR2. The radiation pressure S/c = (P/4πR2)/c if the dust particle absorbs all the energy. The radiation force is the cross-section of the particle times S/c: Frad = πr2(P/4πR2)/c. If Frad = Fgrav then so r = 3P/16πρcGM. .

Poynting vector Answer: The attraction of the sun is The energy flux from the sun per unit area is the Poynting vector S = P/4πR2. The radiation pressure S/c = (P/4πR2)/c if the dust particle absorbs all the energy. The radiation force is the cross-section of the particle times S/c: Frad = πr2(P/4πR2)/c. If Frad = Fgrav then so r = 3P/16πρcGM = 3(3.90 × 1026 W) / 16π (3.0 × 103 kg/m3) (3.00 × 108 m/s) × (6.67 × 10–11 m3/s2·kg) (1.99 × 1030 kg) = 0.19 μm.

Poynting vector Answer: The attraction of the sun is The energy flux from the sun per unit area is the Poynting vector S = P/4πR2. The radiation pressure S/c = (P/4πR2)/c if the dust particle absorbs all the energy. The radiation force is the cross-section of the particle times S/c: Frad = πr2(P/4πR2)/c. If Frad = Fgrav then so r = 3P/16πρcGM = 3(3.90 × 1026 W) / 16π (3.0 × 103 kg/m3) (3.00 × 108 m/s) × (6.67 × 10–11 m3/s2·kg) (1.99 × 1030 kg) = 0.19 μm. So why is interplanetary dust mostly larger than 0.2 μm?

a B I Poynting vector Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 12): The wire shown has length L, resistance R and radius a, and carries a current I. (a) Show that the Poynting vector S on the wire surface points inward. (b) Show that the integral of S over the wire surface is –I2R.

S × E a B I B Poynting vector Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 12): The wire shown has length L, resistance R and radius a, and carries a current I. (a) Show that the Poynting vector S on the wire surface points inward. (b) Show that the integral of S over the wire surface is –I2R. Answer: (a) E is parallel to the wire, hence S = E×B/μ0 points into the wire.

S × E a B dA I B Poynting vector Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 12): The wire shown has length L, resistance R and radius a, and carries a current I. (a) Show that the Poynting vector S on the wire surface points inward. (b) Show that the integral of S over the wire surface is –I2R. Answer: (b) E = V/L = IR/L and B = μ0I/2πa, thus S = I2R/2πaL and ∫ S·dA = 2πaLS = –I2R.

S × E a B dA I B Poynting vector Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 12): The wire shown has length L, resistance R and radius a, and carries a current I. (a) Show that the Poynting vector S on the wire surface points inward. (b) Show that the integral of S over the wire surface is –I2R. This looks strange at first – electromagnetic energy enters the wire from the sides? But maybe it is not so strange – heat energy leaves the wire from the sides….

L q A paradox Assume two uniformly (but oppositely) charged cylinders, both infinitely long. One fits inside the other but they have almost the same radius r. Outside, a distance L from their common axis, is a particle of charge q at rest. The two cylinders rotate at the same angular speed but in opposite directions. But mutual friction gradually stops them from rotating, starting at t = 0.

L q A paradox Assume two uniformly (but oppositely) charged cylinders, both infinitely long. One fits inside the other but they have almost the same radius r. Outside, a distance L from their common axis, is a particle of charge q at rest. The two cylinders rotate at the same angular speed but in opposite directions. But mutual friction gradually stops them from rotating, starting at t = 0. According to Faraday’s law, the decay of the magnetic flux in the cylinders creates an electric field that accelerates the charged particle.

L q A paradox Assume two uniformly (but oppositely) charged cylinders, both infinitely long. One fits inside the other but they have almost the same radius r. Outside, a distance L from their common axis, is a particle of charge q at rest. The two cylinders rotate at the same angular speed but in opposite directions. But mutual friction gradually stops them from rotating, starting at t = 0. According to Faraday’s law, the decay of the magnetic flux in the cylinders creates an electric field that accelerates the charged particle. What happened to conservation of momentum?

z x L q A paradox Let the magnitude of the magnetic field inside the cylinders be B(t). The force on the charged particle is qE(t) where hence the final momentum of the charged particle is qr2B(0)/2L.

z x L q A paradox Let’s integrate the initial Poynting vector, assuming L >> r. We need Ex(0,0,z) at the initial time: so the integral of the Poynting vector is which reduces (using z = L tan θ) to qr2B(0)/2ε0μ0L = qc2r2B(0)/2L.

z x L q A paradox So the integral of the Poynting vector S is qr2B(0)/2ε0μ0L = qc2r2B(0)/2L. Now S/c is energy density; and if we assume that S/c2 is momentum density, then the electromagnetic momentum was initially qr2B(0)/2L, exactly the final mechanical momentum. Momentum is conserved, after all!

Transformers Power transmission lines lose I2R = IV of power on their way to their destination. A way to reduce power losses is to reduce I while increasing V. 350-kV power lines are common. At the destination, where it is safer to use a smaller potential, a transformer reduces V. How does a transformer work? Get yours today!

ΔV1 ΔV2 Soft iron S Secondary (output) Primary (input) R2 N1 N2 Transformers Power transmission lines lose I2R = IV of power on their way to their destination. A way to reduce power losses is to reduce I while increasing V. 350-kV power lines are common. At the destination, where it is safer to use a smaller potential, a transformer reduces V. How does a transformer work?

ΔV1 ΔV2 Soft iron S Secondary (output) Primary (input) R2 N1 N2 Transformers With the switch S open, the potential across the primary coil (with N1 windings) is ΔV1 = – N1dΦB/dt, by Faraday’s law. If the coils link with same magnetic flux ΦB then the potential on the secondary coil is ΔV2 = – N2dΦB/dt. The relation between the potentials is ΔV1 = ΔV2 N1/N2. If N2 > N1 then ΔV2 > ΔV1. This is a step-up transformer.

ΔV1 ΔV2 Soft iron S Secondary (output) Primary (input) R2 N1 N2 Transformers With the switch S open, the potential across the primary coil (with N1 windings) is ΔV1 = – N1dΦB/dt, by Faraday’s law. If the coils link with same magnetic flux ΦB then the potential on the secondary coil is ΔV2 = – N2dΦB/dt. The relation between the potentials is ΔV1 = ΔV2 N1/N2. If N2 < N1 then ΔV2 < ΔV1. This is a step-down transformer.

ΔV1 ΔV2 Soft iron S Secondary (output) Primary (input) R2 N1 N2 Transformers With the switch S closed, the current in the secondary coil is I2 = ΔV2 /R2. In an ideal transformer, there are no losses and I1ΔV1 = I2ΔV2 .

Transformers Example: A power station in Beersheba delivers 20 MW of power to a neighborhood 1 km away. The resistance in the transmission line is 2.0 Ω and the industrial price of electricity is₪ 0.30/kWh. The power station produces the electricity at 22 kV but steps it up to 230 kV before transmission. (a) What is the (approximate) daily transmission cost? (b) What would be the cost without stepping up the potential?

Transformers Example: A power station in Beersheba delivers 20 MW of power to a neighborhood 1 km away. The resistance in the transmission line is 2.0 Ω and the industrial price of electricity is₪ 0.30/kWh. The power station produces the electricity at 22 kV but steps it up to 230 kV before transmission. (a) What is the (approximate) daily transmission cost? (b) What would be the cost without stepping up the potential? Answer: (a) We approximate using dc relations. The current is power divided by potential: I = 20 MW / 230 kV = 87 A. The power lost in the line is I2R = (87 A)2(2.0 Ω) = 15 kW; in one day, the power loss is 360 kWh, worth ₪110.

Transformers Example: A power station in Beersheba delivers 20 MW of power to a neighborhood 1 km away. The resistance in the transmission line is 2.0 Ω and the industrial price of electricity is₪ 0.30/kWh. The power station produces the electricity at 22 kV but steps it up to 230 kV before transmission. (a) What is the (approximate) daily transmission cost? (b) What would be the cost without stepping up the potential? Answer: (b) We approximate using dc relations. The current is power divided by potential: I = 20 MW / 22 kV = 910 A. The power lost in the line is I2R = (910 A)2(2.0 Ω) = 1.65 MW; in one day, the power loss is 40 MWh, worth ₪12,000.

Maxwell’s equations in matter Here are Maxwell’s equations, including sources (charges, currents) but not dielectric, diamagnetic, paramagnetic or ferromagnetic materials:

Maxwell’s equations in matter Here are Maxwell’s equations, including sources (charges, currents) and many dielectric, diamagnetic, paramagnetic or ferromagnetic materials:

Maxwell’s equations in matter We deduce the speed of electromagnetic waves in these materials just as we deduced the speed in the vacuum. First, we eliminate sources (charges, currents):

Maxwell’s equations in matter We find that the speed of electromagnetic waves in a medium with electric permittivity ε and magnetic permeability μ is . Faraday’s law Vector identity Ampère’s law, as modified by Maxwell Gauss’s law in a vacuum: •E = 0

Maxwell’s equations in matter The plane wave E(r,t) = E cos (k·r – ωt + δ) is often written as a complex function: where E is a constant vector that fixes the amplitude and the polarization of the wave, k = 2π/λ is the wave number of the wave, and ω is its angular frequency. (You can just ignore the imaginary part of the function.) Substituting this plane wave into we obtain a solution if and only if – k2 + μεω2 = 0, hence = ω/k = fλ is the speed of the wave in the material.

Maxwell’s equations in matter Writing a plane wave as E(r,t) = Eeik·r – iωtis very convenient! It allows us to write , so we learn that E is transverse in this case as well. The factor of i just indicates a phase difference of π/2. It also allows us to write ik × E(r,t). Now from Faraday’s law we have hence B(r,t) = k × E(r,t)/ω.

ki θi x θt θr kr kt μiεi μtεt Reflection and refraction Now let’s consider what happens when an electromagnetic wave crosses between two media, i.e. between two materials with different permittivities and permeabilities. Let the interface of the two materials be the plane x = 0, with a wave Ei(r,t) incident on the left. There will be also a reflected wave Er(r,t) on the left and a transmitted wave Et(r,t) on the right:

ki θi x θt θr kr kt μiεi μtεt Reflection and refraction On the x = 0 plane, the electric field must be continuous, which means that Ei(0,y,z,t) + Er(0,y,z,t) = Et(0,y,z,t); or in complex wave notation (after cancelling the common factor e– iωt):

ki θi x θt θr kr kt μiεi μtεt Reflection and refraction The equation must hold for all y and all z. The only way it can hold for all y and all z is if ki · (0,y,z) = kr · (0,y,z) = kt · (0,y,z), i.e. ki , kr and kt have the same y- and z-components: ki sin θi = kr sin θr = kt sin θt .

Since ki = kr we have angle of incidence = angle of reflection! ki θi x θt θr kr kt μiεi μtεt Reflection and refraction The equation must hold for all y and all z. The only way it can hold for all y and all z is if ki · (0,y,z) = kr · (0,y,z) = kt · (0,y,z), i.e. ki , kr and kt have the same y- and z-components: ki sin θi = kr sin θr = kt sin θt .

ki Snell’s law of refraction! θi x θt θr kr kt μiεi μtεt Reflection and refraction The equation must hold for all y and all z. The only way it can hold for all y and all z is if ki · (0,y,z) = kr · (0,y,z) = kt · (0,y,z), i.e. ki , kr and kt have the same y- and z-components: ki sin θi = kr sin θr = kt sin θt .

Polarized light Our plane wave solution E(r,t) = Eeik·r – iωt implies that the electric field of a wave has a certain direction, perpendicular to propagation vector k. Is this a prediction we can test? Yes, with a polarizer! A polarizer has a polarization axis, and passes only the component of light along the polarization axis. We will not usually see an effect with one polarizer, because most light is unpolarized – it is mixture of waves with E pointing in any direction perpendicular to k. But with two polarizers we see an effect: the first polarizer polarizes the light emitted by a source, and as we turn the second polarizer relative to the first, we may see all or part or none of the polarized light.

Physics 2 for Electrical Engineering