University of Notre Dame - Electromagnetic Fields and Waves Exam Solutions (EE 30348), Exams of Electrical and Electronics Engineering

Download University of Notre Dame - Electromagnetic Fields and Waves Exam Solutions (EE 30348) and more Exams Electrical and Electronics Engineering in PDF only on Docsity! Fundamentals of Electromagnetic Fields and Waves: I Fall 2007, EE 30348, Electrical Engineering, University of Notre Dame Final Exam Solutions (30 Points) • Formulae are given at the end of this questionnaire. • Please show your steps clearly and sketch figures wherever necessary. • Points will be awarded for correct steps shown in the solutions. • There are four questions in this exam. Please attempt all questions. Problem 1 (10 Points) Answer the following short questions. a) (2 Points) Can the following vector fields represent a static Electric field? i) A = 3φaφ. Solution: For a vector field to represent a static electric field, Faraday’s law dictates that ∇ × E = 0. ∇ × A = 1ρ ∂∂ρ(3φρ)az = 3φ ρ az 6= 0. Therefore A cannot represent a static electric field. ii) B = xax + yay. Solution: By the same argument, since ∇× B = 0, B can represent a static electric field. b) (3 Points) The electrostatic potential distribution in the region (x ≥ 0, y ≥ 0, z ≥ 0) which is filled with a material of dielectric constant ǫ is given by V (x, y, z) = V0e −x/Lxe−y/Lye−z/Lz in Volts. Find the point (x, y, z) where the volume charge density ρv(x, y, z) reaches the maxi- mum magnitude, and the volume charge density |ρv|max at that point. Solution: By Poisson’s equation, the volume charge density at any point is related to the elec- trostatic potential by the relation ρv = −ǫ∇2V = −ǫ( 1L2x + 1 L2y + 1 L2z )V0e −x/Lxe−y/Lye−z/Lz . Clearly, it reaches a maximum at (0, 0, 0), and the charge density at that point is |ρv|max = ǫ( 1 L2x + 1 L2y + 1 L2z )V0. 1 c) (2 Points) The region between two parallel plates of a capacitor is filled with a material of di- electric constant ǫ and conductivity σ. Find the maximum permissible value of the conductivity such that any charge put on the plates at time t = 0 can be stored for at least t1 seconds before reducing to 1/100 of its initial value. As discussed in class, the charge decays exponentially, with the time constant given by the characteristic dielectric relaxation time τd = ǫ/σ. Solution: Since the charge decays as Q(t) = Q(0)e−t/τd , it is clear that to meet the requirement, we must have Q(t1) = Q(0)e −t1/τd ≥ Q(0)/100 → τd ≥ t1/ ln 100 → σ ≤ ǫt1 ln 100. d) (3 Points) Show that the skin depth δ for an electromagnetic wave is i) independent of the frequency of the wave if it propagates in a region of very low conductivity1, & ii) goes as the inverse square root of the frequency of the wave (δ ∝ 1/√ω) if it propagates in a region of very high conductivity. Solution: Since α = ω √ µǫ 2 [ √ 1 + ( σωǫ) 2−1]1/2, for (σ/ωǫ) << 1, α ≈ ω √ µǫ 2 [1+ 1 2 ( σ ωǫ) 2−1]1/2 ≈ 1 2 √ µσ2 ǫ . Hence the skin depth for EM wave propagation in a low conductivity material is given by δ = 1α ≈ √ 4ǫ µσ2 , which is clearly not dependent on the frequency ω of the wave. On the other hand, for a highly conductive medium, (σ/ωǫ) >> 1, α ≈ ω √ µǫ 2 [ √ ( σωǫ) 2]1/2 ≈ √ µσω 2 . Hence the skin depth for EM wave propagation in a low conductivity material is given by δ = 1α ≈ √ 2 µσω , which goes as 1/ √ ω. 1you might need the approximation √ 1 + x ≈ 1 + x/2 for x << 1. 2 x z 0 Volt V0 Volts gap θ0 grounded metal plane large conical conductor Figure 3: Problem (4) Problem 4 (8 Points) Refer to Figure 3. A large conducting cone (angle θ0) is placed on a grounded (V = 0 Volt) conducting plane with a tiny gap separating it from the plane. The cone is maintained at a voltage V0 Volts. a) Use Laplace’s equation to find the electric potential V (r, θ, φ) in the region θ0 < θ < π 2 . You might need the following in your calculation: ∫ dθ sin θ = ln(tan θ 2 ) + C, where C is a constant, and tan π4 = 1. Solution: Starting with Laplace’s equation in spherical coordinates in the region of interest, and making use of the symmetry [∂(...)/∂r = ∂(...)/∂φ = 0], we get ∇2V = 1 r2 sin θ ∂ ∂θ [sin θ ∂V ∂θ ] = 0. (3) Integrating this once, we get ∂V ∂θ = A sin θ , (4) where A is a constant. Integrating one more time, we get V = A ln[tan θ 2 ] +B, (5) where B is another constant. Using the two boundary conditions - V (θ = 0) = 0 and V (θ = θ0 = V0, we find the two constants, and then the voltage is given by 5 V (r, θ, φ) = V (θ) = V0 ln[tan θ2 ] ln[tan θ02 ] . (6) b) Calculate the electric field E in the region θ0 < θ < π 2 using your result from part (a). Solution: The electric field is given by E = −∇V = −1r ∂V∂θ aθ. This leads to E = − V0 r sin θ ln[tan θ02 ] aθ, (7) which indicates that the electric field points along the aθ direction from the conical conductor to the planar conductor (remember the ln factor is negative!). The electric field gets stronger as we approach the tip of the cone (r → 0), and weakens as we get farther from the top |E| → 0 as r → ∞. Clearly this implies that there is a charge pileup near the tip of the cone. x z 0 Volt V0 Volts θ0 grounded metal plane large conical conductor + + + + - - - Figure 4: Problem (4) c) Find the charge density on the two conductors from your result of part (b). Solution: The sheet charge density ρs on the metallic conductors is given by the boundary condition D · n = ρs. For the conical conductor, the charge is positive since the field lines orignate from there (assuming V0 > 0). It is given by ρcones (r, θ, φ) = ǫ0E(r, θ = θ0, φ) · aθ = − ǫ0V0 r sin θ0 ln[tan θ0 2 ] , (8) and the corresponding charge on the planar conductor is given by 6 ρplanes (r, θ, φ) = ǫ0E(r, θ = π 2 , φ) · −aθ = ǫ0V0 r ln[tan θ02 ] . (9) The charge and the electric field lines are sketched in Figure 4. End. (Formulae in the next page...) 7