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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.

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Physics 2 for Electrical Engineering

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  1. 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 11. Electromagnetic radiation – • Divergence theorem • Maxwell’s equations in differential form• Current conservation • Electromagnetic waves • Poynting vector Source: Halliday, Resnick and Krane, 5th Edition, Chap. 38.

  2. (Gauss’s law) (Faraday’s law) (Ampère’s law as modified by Maxwell) This set of four fundamental equations for E and B, are Maxwell’s equations in integral form.

  3. In 1929 – 65 years after Maxwell’s prediction – M. R. Van Cauwenberghe was the first to measure “displacement current”, on a round parallel-plate capacitor of capacitance C = 100 pF and effective radius R = 40.0 cm. The potential across the capacitor, which alternated with frequency f = 50.0 Hz, reached V0 = 174 kV. We calculated that the maximum field B measureable at the edge of the capacitor was 2.73×10–8 T. You ask: What was so difficult about this experiment, that it took 65 years to do? After all, the two terms I and ε0 dΦE/dt in the Ampère-Maxwell law yield the same B, and it was not difficult to verify Ampère’s law!

  4. In 1929 – 65 years after Maxwell’s prediction – M. R. Van Cauwenberghe was the first to measure “displacement current”, on a round parallel-plate capacitor of capacitance C = 100 pF and effective radius R = 40.0 cm. The potential across the capacitor, which alternated with frequency f = 50.0 Hz, reached V0 = 174 kV. We calculated that the maximum field B measureable at the edge of the capacitor was 2.73×10–8 T. Two physicists who did the measurement in 1985 (thinking they were the first) answered this question in their paper:

  5. From the integral form of Maxwell’s equations, we anticipate electromagnetic radiation: Intuitively, we see that a changing electric flux will generate a transient magnetic field around it, which will generate a transient electric field, etc. etc. while the wave spreads through space.

  6. Electromagnetic radiation is a phenomenon that spans more than 18 orders of magnitude (in wavelength and frequency)! To calculate all the features of electromagnetic waves, we need Maxwell’s equations in their differential form.

  7. With the aid of Stokes’s theorem, we converted two of Maxwell’s equations to differential form: To convert the other two of Maxwell’s equations, we need the “divergence theorem”.

  8. E The divergence of a vector field E(x,y,z), written , is the following: Note that is a scalar field. Note that the notation is reasonable because is (formally) the scalar product of the vectors and E = (Ex, Ey, Ez), just as is (formally) the vector product of the same two vectors.

  9. E Example 1: What is

  10. E Example 1: What is Answer: This sum of second derivatives is called the “Laplacian” and written as “del-squared”:

  11. E Example 2: What is

  12. E Example 2: What is Answer:

  13. E Example 3: E(r) is the electric field of a uniformly charged ball of radius R with total charge q. What is

  14. E Example 3: E(r) is the electric field of a uniformly charged ball of radius R with total charge q. What is Answer: As we know, the electric field E(r) of any electrostatic configuration is minus the gradient of the electric potential V(r), i.e. Hence where . For r ≤ R we have

  15. E Example 3: E(r) is the electric field of a uniformly charged ball of radius R with total charge q. What is Answer: As we know, the electric field E(r) of any electrostatic configuration is minus the gradient of the electric potential V(r), i.e. Hence where . For r ≥ R we have so

  16. E Example 4: What is (A is any vector field.)

  17. E Example 4: What is (A is any vector field.) Answer: We write out : so

  18. ∂V V Divergence theorem The divergence theorem is Stokes’s theorem in 3D, but without curvature. We extend our notation: If V denotes any volume, then ∂V denotes the boundary of the volume (the surface bounding V ).

  19. Heart made of cubes Double pyramid reduced to cubes Divergence theorem To prove the divergence theorem, we assume that we can reduce any volume V to infinitesimal cubes:

  20. Divergence theorem The divergence theorem tells us that over any closed surface ∂V is equal to the sum over all the cubes in the volume V that has ∂V as its boundary.

  21. Divergence theorem To see why the divergence theorem is true, all we have to do is look at two neighboring cubes and at for each one:

  22. Divergence theorem To see why the divergence theorem is true, all we have to do is look at two neighboring cubes and at for each one: The contributions of their common side cancel!

  23. Divergence theorem When we put all the cubes together, the only sides that contribute (that are not cancelled by facing sides) are the faces along ∂V ! The contributions of their common side cancel!

  24. Divergence theorem When we put all the cubes together, the only sides that contribute (that are not cancelled by facing sides) are the faces along ∂V ! The divergence theorem thus tells us , i.e. that theintegral over ∂V equals the sum of the contribution of each cube. We will now compute the contribution of an infinitesimal cube. In the limit of infinitely many cubes, the sum over cubes will become an integral over the volume V.

  25. Ey(x,y+Δy/2,z)ΔxΔz y z Ez(x,y,z+Δz/2)ΔxΔy x Ex(x+Δx/2,y,z)ΔyΔz –Ex(x–Δx/2,y,z)ΔyΔz –Ez(x,y,z–Δz/2)ΔxΔy –Ey(x,y–Δy/2,z)ΔxΔz Divergence theorem We compute the flux for an infinitesimal cube. The cube is centered at a point r = (x,y,z) with the area elements dA pointing along these axes. There are six contributions to the total flux:

  26. Divergence theorem We compute the flux for an infinitesimal cube. The cube is centered at a point r = (x,y,z) with the area elements dA pointing along these axes. There are six contributions to the total flux, and they add up to

  27. Divergence theorem We compute the flux for an infinitesimal cube. The cube is centered at a point r = (x,y,z) with the area elements dA pointing along these axes. There are six contributions to the total flux, and they add up to The sum of over all infinitesimal cubes then goes over to a volume integral, and we obtain…

  28. Divergence theorem …the divergence theorem: i.e. the flux of E through the boundary ∂V of the volume V equals the integral of over the volume.

  29. Maxwell’s equations in differential form Gauss’s law says where ρ(r) is charge density. But the divergence theorem says so This is true for anyV, hence

  30. Maxwell’s equations in differential form Gauss’s law for B says But the divergence theorem says therefore This is true for anyV, hence

  31. Maxwell’s equations in differential form Using Stokes’s theorem and the divergence theorem, we have converted all four Maxwell equations to differential form:

  32. Current conservation What happens when we apply the div operator to Ampère’s law, with and without Maxwell’s modification? Applying to But we proved for any A, and Maxwell’s equations say E = ρ/ε0, so we conclude What does this equation mean?

  33. Current conservation The equation is called the “continuity equation”. What does it mean? Let’s integrate both sides of this equation over the volume V of an infinitesimal box. We obtain where q is the amount of charge in the box. Next, using the divergence theorem, we can rewrite this equation as

  34. Jy(x,y+Δy/2,z)ΔxΔz y z x Jz(x,y,z+Δz/2)ΔxΔy Jx(x+Δx/2,y,z)ΔyΔz –Jx(x–Δx/2,y,z)ΔyΔz –Jz(x,y,z–Δz/2)ΔxΔy –Jy(x,y–Δy/2,z)ΔxΔz Current conservation The left side of the equation is the net flux of current out of the box. The right side is the rate of change of the total charge in the box.

  35. Current conservation The left side of the equation is the net flux of current out of the box. The right side is the rate of change of the total charge in the box. Ampere’s law, without Maxwell’s modification, implies that the net flux of current out of any box vanishes: No charge can build up anywhere. That is certainly not true for a circuit with a capacitor inside.

  36. Current conservation The left side of the equation is the net flux of current out of the box. The right side is the rate of change of the total charge in the box. Ampere’s law with Maxwell’s modification simply states that charge is conserved: Anywhere that the net flux of current out of a box is not zero, charge must build up in that box, or leave that box.

  37. Electromagnetic waves Let’s learn more tricks with the curl and div operators, and with Maxwell’s equations. Example 1: What is

  38. Electromagnetic waves Let’s learn more tricks with the curl and div operators, and with Maxwell’s equations. Example 1: What is Answer: We have from which we can compute the x-component of It is

  39. Electromagnetic waves We have where we added and subtracted ∂2Ex/∂x2 on the right side.

  40. Electromagnetic waves Since it is easy to guess how the other components will turn out; we obtain Both terms on the right side are vectors, as they must be since the left side is a vector.

  41. Electromagnetic waves Now let’s consider Maxwell’s equations in the vacuum (no charge ρ or current density J):

  42. Electromagnetic waves Let’s apply to the Maxwell equation and insert the vacuum Maxwell equation We get and since in the vacuum, we arrive at a wave equation:

  43. Electromagnetic waves Let’s try a plane wave solution to this wave equation: E(r,t) = E cos (k·r – ωt + δ) , 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. Substituting this plane wave into we obtain a solution if and only if – k2 + μ0ε0 ω2 = 0, hence ω/k =

  44. Electromagnetic waves We found that E(r,t) = E cos (k·r – ωt + δ) is a solution if and only if ω/k = We can determine the speed of this wave by checking at what speed a crest of the wave moves. At a crest of the wave, the phase k·r – ωt + δ is unchanging over time. For the component of r in the k-direction, we have Around 1864, Maxwell calculated the speed of these waves. Imagine how he felt when he obtained the speed of light!

  45. Electromagnetic waves The speed of electromagnetic waves is v = ω/k = Today’s data: μ0 = 4π × 10–7 T·m/A, ε0 = 8.854187817 × 10–12 F/m, c = 2.99792458 × 108 m/s. “We can scarcely avoid the conclusion that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena.” (J. C. Maxwell)

  46. Electromagnetic waves Heinrich Hertz was the first (in 1886) to verify Maxwell’s prediction of electromagnetic waves travelling at the speed of light. Receiver Spark Gap Transmitter

  47. Receiver Spark Gap Transmitter

  48. Faraday’s law Vector identity Ampère’s law, as modified by Maxwell Gauss’s law in a vacuum: •E = 0 Electromagnetic waves Summary: Maxwell’s equations predict that electromagnetic waves (including light waves) travel at speed c in vacuum. The accepted value of c today is 299,792,458m/s.

  49. Electromagnetic waves What else can we find out about electromagnetic waves? Applying the Gauss law to our wave solution E(r,t) = E cos (k·r – ωt + δ), we obtain Since sin (k·r – ωt + δ) does not vanish, it follows that that k·E = 0: the electric field of an electromagnetic wave in a vacuum is transverse, i.e. perpendicular to the wave vector k.

  50. Electromagnetic waves What else can we find out about electromagnetic waves? Applying Faraday’s law to our wave solution E(r,t) = E cos (k·r – ωt + δ), we obtain if we try B(r,t) = B cos (k·r – ωt + δ), we obtain a solution provided k × E = ωB! Thus B is perpendicular to E and to k (i.e. B is transverse) and propagates in phase with E. Since we have seen that ω = ck, we have E = cB.

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