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### HW3_solutions

Course: EEE 241, Spring 2007
School: ASU
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Fundamentals EEE241 of Electromagnetics Spring 2007 Solutions to Homework 3 P3.5: Two point charges, Q1 and Q2 are located at (1,2,0) and (2,0,0) respectively. Find the relation between Q1 and Q2 such that the total force on a test charge at the point P(-1,1,0) will have (a) no x-component and (b) no y-component. r r r r Q1 P = - 2ax - a y ; Q1 P = 5 r r r r Q2 P = - 3ax + a y ; Q2 P = 10 r r r r r r r 1...

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Fundamentals EEE241 of Electromagnetics Spring 2007 Solutions to Homework 3 P3.5: Two point charges, Q1 and Q2 are located at (1,2,0) and (2,0,0) respectively. Find the relation between Q1 and Q2 such that the total force on a test charge at the point P(-1,1,0) will have (a) no x-component and (b) no y-component. r r r r Q1 P = - 2ax - a y ; Q1 P = 5 r r r r Q2 P = - 3ax + a y ; Q2 P = 10 r r r r r r r 1 Q1 Q1 P Q2 Q2 P 1 Q1 Q2 -2ax - a y ) + 3/ 2 ( -3ax + a y ) EP = r 3 + r 3 = 3/ 2 ( 4 o Q P 4 o 5 10 Q2 P 1 (a) no x-component -2Q1 3Q Q 3 - 3/22 = 0 1 = - 3/ 2 5 10 Q2 4 2 (b) no y-component -Q1 Q2 Q 1 + 3/ 2 = 0 1 = 3/ 2 5 10 Q2 2 2 r r P3.10: Assuming that the electric field intensity is E = ax 100 x (V / m ) , find the total electric charge contained inside (a) a cubical volume 100 (mm) on a side centered symmetrically at the origin (b) a cylindrical volume around the z-axis having a radius 50 (mm) and a height 100 (mm) centered at the origin. r (a) Use Gauss's law and Cartesian coordinates. Cube has six sides of area (100mm)2. E is r r r r normal to the two sides at x = 50mm , where E = ax 5 and an = ax , and only these two r r sides contribute to E.ds . r r r ds = ax (100 10-3 )2 = 0.01ax E.ds = 2 5 0.01 = r r Q o Q = 0.1 o = 8.854 10-13 C (b) Convert to cylindrical coordinates. r r r ax = ar cos - a sin ; x = r cos r r r E = ar 100 r cos 2 - a 100 r sin cos r r dS = ar r d dz r r E.dS = 100 r 2 cos 2 d dz r r 0.05 2 E.dS = 0.25 cos 2 d = 0.025 -0.05 0 Q = 0.025 o = 6.44 10-13 C EEE 241 Solution to Assignment #3 P.3-11 A spherical distribution of charge = 0 1 - R2 /b2 exists in the region 0 R b. This charge distribution is concentrically surrounded by a conducting shell with inner radius Ri (> b) and outer radius Ro . Determine E everywhere. Ro Ri b Solution: From the symmetry, we expect E = aR ER (R) ^ Applying Gauss's law in integral form, we have D ds = 0 S S E ds = 4R2 0 ER (R) = Qencl Hence, E = aR ^ There are four distinct regions to consider: 0R<b b R < Ri Ri R < R o 1 Qencl 40 R2 Ro R < For 0 R < b we have: R Qencl = 4 E = aR ^ For b R < Ri we have: 0 1 - 0 R b 2 R 2 dR = 40 R3 R5 - 2 3 5b R3 0 R - 2 0 3 5b b Qencl = 4 E = aR ^ For Ri R < Ro we have: 0 1 - 0 R b 2 R 2 dR = 8 0 b3 15 20 b3 150 R2 E = 0 For R Ro we have: Qencl = 8 0 b3 15 20 b3 E = aR ^ 150 R2 P.3-12 Two infinitely long coaxial cylindrical surfaces, r = a and r = b (b > a), carry surface charge densities sa and sb , respectively. (a) Determine E everywhere. (b) What must be the relation between a and b in order that E vanishes for r > b? Solution: (a) From the symmetry, we expect E = ar Er (r) ^ Applying Gauss's law in integral form, we have D ds = 0 S S E ds = 2rl0 Er (r) = Qencl = l,encl l Hence, E = ar ^ There are three distinct regions to consider: 2 l,encl 20 r a b 0r<a ar<b br< For 0 r < a we have: l,encl = 0 E = 0 For a r < b we have: l,encl = 2asa sa a E = ar ^ 0 r For r b we have: l,encl = 2 (asa + bsb ) E = ar ^ (b) For E to vanish for r > b, we must have sa a + sb b = 0 or b = -sa a sb sa a + sb b 0 r 3 P3.13: Determine the work done in carrying a -2 C charge from ( P 2,1, -1) to P2 ( 8, 2, -1) in 1 r r r the field E = ax y + a y x (a) along the parabola x = 2 y 2 (b) along the straight line joining P and P2 . 1 Solution: r r W = - q E.d l = - q ( ydx + xdy ) 2 (a) Along the parabola x = 2 y 2 , dx = 4 ydy W = - q 6 y 2 dy = - 14q = 28 J (b) Along the straight line joining P ( 2,1, -1) to P2 ( 8, 2, -1) , the work done should be same as 1 before. Along the straight line, x = 6 y - 4, dx = 6dy W = - q (12 y - 4 ) dy = - 14q = 28 J 1 2 1 P.3-15 Three charges (+q, -2q, +q) are arranged along the z-axis at z = d/2, z = 0, and z = -d/2, respectively. (a) Determine V and E at a distant point P (R, , ). (b) Find the equations for equipotential surfaces and streamlines. (c) Sketch a family of equipotential lines and streamlines. (Such an arrangement of three charges is called a linear electrostatic quadrupole.) Solution (a) V = whereby the cosine law 2 r1 = r 2 + q 4o 1 2 1 - + r1 r r2 d 2 d 2 = 2 1 r r [ + - 2] 4o r r1 r2 (1) - rd cos 2 (2) (3) 2 r2 = r 2 + + rd cos If r d, then r = r1 1 1+ d 2r 2 = [1 - - d r cos + 3 4 d cos + r d 2r 2 ]- 2 2 1 (4) 1- 1 d d2 - cos + 2 2 r 4r 1+ 2 d d2 - cos + 2 r 4r (5) (6) d2 d cos + 2 3 cos2 - 1 2r 8r Similarly d d 2 -1 r ] 2 = [1 + cos + r2 r 2r d2 d cos + 2 3 cos2 - 1 1- 2r 8r Hence for d r (7) (8) 1 V = qd2 3 cos2 - 1 16o r3 (9) The corresponding E-field of the quadrupole is E = -V V ^1 V + 1 V + ^ = - r ^ r r r sin 2 3qd ^ [^ 3 cos2 - 1 + sin 2] r = 16o r4 4 (10) (11) (12) EEE 241 Solution to Assignment #4 P.3-22 The polarization in a dielectric cube of side L centered at the origin is given by P = P0 (^x x + ay y + az z). a ^ ^ (a) Determine the surface and volume bound-charge densities. (b) Show that the total bound charge is zero. z y x Solution: The six faces and associated normals of the dielectric cube of side L centered at the origin are given by: x = L : an = ^x a 2 ^ y = L : an = ^y a 2 ^ z = L : an = ^z a 2 ^ (a) Surface bound-charge densities: ps = P an ^ at x = at at at at at P0 L L 2 : ps = 2 x = - L : ps = P0 L 2 2 P0 L L y = 2 : ps = 2 y = - L : ps = P0 L 2 2 P0 L L z = 2 : ps = 2 z = - L : ps = P0 L 2 2 1 Volume bound charge density: p = - P = - (b) Qps = total bound charge on faces P0 L = 6 L2 = 3P0 L3 2 Qpv = total bound charge in bulk = V Pz Px Py + + x y z = -3P0 p dv = -3P0 L3 Hence, Qps + Qpv = 0 P.3-33 A cylindrical capacitor of length L consists of coaxial conducting surfaces of radii ri and ro . Two dielectric media of different dielectric constants r1 and r2 fill the space between the conducting surfaces as shown in Fig. 3-42. Determine its capacitance. Solution: Assume +Q distrubuted on the surface of the inner conductor and -Q distributed on the surface of the outer conductor. Note that the charge will not be distributed uniformly over the entire surface because of the discontinuity of the dielectric material. If we neglect fringing, we can assume that the electric flux density D and the electric field intensity E are radially directed. Furthermore, we can deduce that the electric field intensity will be the same in both dielectrics since the electric field is tangential to the boundary. Thus, we have E1 = E2 = ar Er (r) ^ 2
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