Download Physics 374 Homework 8: Calculating Electric Field and Potential using Multipole Expansion and more Assignments Physics in PDF only on Docsity! Physics 374 Homework 8 Due Nov. 10 1. A cylinder of radius R contains charge according to the following charge density: )()( 2 rRxcx −= ϑρ where c is a constant, 22 yxr += and q is the usual step function. The goal of this problem is to find the electric field for the region r>R. a) As a first step calculate the multipole coefficients. You should find only the m=0 and m=2 terms contribute. Hint: write the 2x term in polar coordinates. b) Find the potential Φ. As the m=0 term is nonzero you must pick an arbitrary 0r . c) Find an expression for the electric field. 2. For the problem in 1 find the potential for r<R. Hint: use the general solution. 3. Consider 4 line charges oriented along the z axis, 2 with a charge per unit length of +λ located at (x,y)=(d,d) and (x,y)=(-d,-d) and 2 with a charge per unit length of - λ located at (x,y)=(d,-d) and (x,y)=(-d,d). a) Show that the exact expression for the potential is given by ( )( ) ( )( ) −++−++ −++−++ −=Φ )cos(22)cos(22 )cos(22)cos(22 log2)( 4 72/322 4 32/322 4 52/322 4 2/322 ππ ππ φφ φφ λ drdrdrdr drdrdrdr r Hint: Use the alternative general expression and note that the integral just gets replaced by a discrete sum over the line charges. Recall yxyxyx ⋅++=− 222 b) Show that the multipole coefficients are given by 0=ob ( ) ( ))exp()exp()exp()exp(2 4743454 ππππ mmmm m m iiiim da −−+= Hint: Again note that that the integral just gets replaced by a discrete sum over the line charges. c) Write an approximate expression for the potential based on the multipole expansion and including up to the m=4 term. d) Test how well the multipole expression works by using Mathematica to plot the exact expression, and multipole truncated at m=2,3 and 4 as a function of r for φ=π/4. 4. It was argued in class that multipole coefficient with smallest nonvanishing m is independent of the choice of origin. Show that the m=2 multipole for the problem does not change if move the origin to the position of the charge at the lower left.