Homework #5 Solutions

Transcription

1 Homework #5 Solutions Problem 4.4 a) Laplace's equation in spherical coordinates is If V 1 = C 1 /R 2 V = (1/R 2 )d/dr(r 2 (dv/dr)) + 1/(R 2 sinθ)d/dθ(sinθdv/dθ) + 1/(R 2 sin 2 θ)d 2 V/dϕ 2 = 0 dv 1 /dr = -C 1 /R 2 dv 1 /dθ = 0 dv 1 /dϕ = 0 After some algebra, it can be shown that 2 V 1 = 0 Which proves that V 1 satisfies Laplace's equation. b) Laplace's equation in cartesian coordinates is 2 V = d 2 V/dx 2 + d 2 V/dy 2 + d 2 V/dz 2 = 0 If V 2 = C 2 z/(x 2 + y 2 + z 2 ) 3/2, then after some algebra it can be shown that 2 V 2 = 0 Which proves that V 2 satisfies Laplace's equation. Problem 4.6 For this example, we want to solve using Poisson's equation 2 V 2 = -ρ/ε = -A/(ε r) In cylindrical coordinate system (and because of symmettry) 2 V 2 =(1/r)(d/dr)(rdV/dr) = -A/(ε r) The general solution to this differential equation is V = -(A/ε)r + C 1 ln(r) + C 2 1

2 Our boundary conditions are that at V(r=a) = V 0 and V(r=b) = 0. Therefore V 0 = -(A/ε)a + C 1 ln(a) + C 2 0 = -(A/ε)b + C 1 ln(b) + C 2 Solving this system of linear equations yields our unknown constants in the expression for voltage C 1 = (A/ε (b-a)-v 0 ) / ln(b/a) C2 = {V 0 ln(b) + A/ε(aln(b) - bln(a))} / ln(b/a) Problem 4.7 a) To solve this problem we will use the method of images. By this method, we can replace the infinite conducting plane and point charge system with an equivalent system consisting of a dipole. From this, it follows that the electric field evaluated at y=0 can be expressed as E(y=0) = -a y Q/(4πεR 2 )(2sinθ) = -a y Qd / (2πε(d 2 +r 2 ) 3/2 From boundary conditions ρ s = a y εe(y=0) = -a y Qd / (2πε(d 2 +r 2 ) 3/2 b) The total charge can be found from integrating the surface charge over the entire surface 0 ρ s 2πr dr = -Q This matches what we expect from the method of images. Problem 4.11 Because the wires are above a conducting ground, we will use the method of images to analyze the problem. Therefore, I will replace the system of two wires and conducting plane with the following system of four wires 2

3 A D B 2h A' d B' In the above picture, the solid red circles represent the original wires (of radius a with linear charge density ρ l (for wire A) and -ρ l (for wire B) ) and the shaded red circles represent the image wire A' and B' with linear charge densities -ρ l and -ρ l, respectively. To find the capacitance per unit length of the original line (that is, the mutual capacitance between A and B, C AB, I will start off by applying superposition to find the voltages. When we had a single linear charge surrounded by two conductors (such as the following sketch illustrates) we were able to represent the voltage difference as ρ l a At Voltage V 0 b Grounded Conductor V 0 = ρ l /(2πε) ln(b/a) Where a is the distance from the source charge to the point at which the voltage is being evaluated and b is the distance from the source charge to the voltage reference point. (If you are confused on this part, then go back and review Chapter 3). Using this as a general formula, we can evaluate the voltage on the surface of the wire A relative to the surface of wire A', V AA', as the superposition of voltages due to all of the separate wire line charges V AA' = ρ l /(2πε) ln(2h/a) {due to wire A linear charge} + (-ρ l )/(2πε) ln(d/d) {due to wire B linear charge} + (-ρ l )/(2πε) ln(a/2h) {due to wire A' linear charge} + ρ l /(2πε) ln(d/d) {due to wire A linear charge} Because -ln(x/y) = ln(y/x), this can be simplified to 3

4 V AA' = ρ l /(πε){ ln(2h/a) + ln(d/d) } By symettry, the potential difference V A referenced to ground is going to be V A = (1/2) V AA' = ρ l /(2πε){ ln(2h/a) + ln(d/d) } Similarly, by repeating the argument for V B we obtain V B = (1/2) V BB' = (-ρ l )/(2πε){ ln(2h/a) + ln(d/d) } Replacing ρ l with ρ l1 and -ρ l with ρ l2, we can solve for ρ l1 and ρ l2 in terms of V A and V B ρ l1 = c 11 V A + c 12 V B ρ l2 = c 21 V A + c 22 V B From equations in Chapter 3 that describe systems of capacitors, we can find the mutual capacitance between the two wire system above a conducting plane to be C = -c 12 = (2πε ln(d/a))/((ln(2h/a) 2 -(ln(d/a)) 2 ) A circuit involving several capacitors which models this problem would be similar to the circuits involving mutual capacitance towards the end of Chapter 3. (Give yourself a pat on the back if you were able to figure this one out ) Extra Problem Shown below is my output for running the given code as well as my modifications % Original Code x = -4.05:.1:4.05; y = -4.05:.1:4.05; [xx,yy] = meshgrid(x,y); cz = xx + i*yy; cpot = asin(cz); pot = real(cpot); contour(xx,yy,pot,10,'k'); % Modified Code to plot Field Lines figure(2); x2 = linspace(-4.05,4.05,15); y2 = linspace(-4.05,4.05,15); [xx,yy] = meshgrid(x2,y2); cz = xx + i*yy; cpot = asin(cz); pot = real(cpot); contour(xx,yy,pot,10,'k'); 4

5 [Ex,Ey] = gradient(pot); hold on quiver(xx,yy,ex,ey) hold off From the following plots, it appears that a physical placement of electrodes could be one coming in from the left (positive) and another from the right (negative). 5