aee1_2008_a03

Aee1_2008_a03 - AEE1 Assignment 3 Fall 2008 Prof S Peik AEE1 Assignment 3 Deadline next Wednesday Name Prerequisite Electrical Fields Charge

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Unformatted text preview: AEE1 Assignment 3, Fall 2008 Prof. S. Peik AEE1 Assignment 3 Deadline: next Wednesday Name: ______________________________________ Prerequisite: Electrical Fields Charge Distributions cartesian, cylindrical and spherical coordinate systems, To be studied: Lecture Notes Chapter 3 Sadiku Coordinate Systems 4.7 and 4.8 Ulaby 4.5 Griffiths 2.3 and 2.4 Objectives of this Assignment: To get familiar with the concept of potentials Understand the relation of E-fields and electric Potantial Problem 1: Two point charges with charge -q and q each are located on the z-axis at z = a, respectively, as shown. Medium is vacuum. Use cylindrical coordinates to solve the problem. z a -q y x -a q 1. Find the electric potential V (, , z) everywhere in space. 4 points 2. What is the direction of the E-field on the x-y-plane? No calculation required! 2 points 3. Find the E-field vector E(, , z = o) in the xy-plane? 4. Show, that for large values of the e-Field decays with 1 3 4 points 2 points Problem 2: Given the electric field V ^ ^ E = (z + 1) cos + (z + 1) cos + sin ^ z m (1) Problem Max. Points Points 1 12 2 4 3 - 4 - Total Points: Points: of 16 determine the work done in moving a 4 nC charge from the point (1, 0o , 0) to (4, 30o , -2) 4 points ^ ^ Hint: Break the integration into several steps with one direction ( i.e. ,,or z) ^ only. AEE1 Assignment 3, Fall 2008 Prof. S. Peik AEE1 Assignment 3 Deadline: next Wednesday Name: ______________________________________ Prerequisite: Electrical Fields Charge Distributions cartesian, cylindrical and spherical coordinate systems, To be studied: Lecture Notes Chapter 3 Sadiku Coordinate Systems 4.7 and 4.8 Ulaby 4.5 Griffiths 2.3 and 2.4 Objectives of this Assignment: To get familiar with the concept of potentials Understand the relation of E-fields and electric Potantial Solution 1: V= q 4 1 2 + (z + a)2 - 1 2 + (z - a)2 (1) The E-field is directed in positive z-direction, there is only a z component of E E = -grad V = q 4 q 4 when z = 0 : E = q 4 q 4 q 2 2 + a2 a 2 + a2 a z ^ 3 2 + a2 3 3 (2) 2 + (z + a)2 z+a 2 + (z + a)2 3 3 - + 2 + (z - a)2 (z + a) 2 + (z - a)2 ^ + z ^ 3 3 ^ + z ^ (3) (4) (5) - + 2 + a2 a 2 + a2 3 3 (6) (7) (8) = When the summand a is neclegtible hence Problem Max. Points Points 1 12 2 4 3 4 Total Points: Points: of 16 Solution 2: calculate the integration 1 r3 We move in three steps first in -direction from 1 to 4, second in -direction from 0deg to 30deg, third in z-direction from 0 to -2. AEE1 Assignment 3, Fall 2008 Prof. S. Peik Since the coordinates are orthogonal, we have to integrate only the component in the direction of integration movement. E.g. in the first step we integrate ^ (z + 1) sin only. W12 first integral = = Q( Z 4 1 E(, 0, 0)d + Z 30 0 E(4, , 0)d + Z -2 0 E(4, 30o , z)dz) (9) (10) (11) (12) (13) Z 4 1 0 0d = 0 4 cos d = 16[sin ]30 = 8Q 0 4 sin 30o dz = 4 sin 300 (-2) = -4Q second integral = Q third integral = Q W12 = Z 30 0 Z -2 4Q=16nJ ...
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This note was uploaded on 05/04/2010 for the course EECS 320256 taught by Professor Peik during the Fall '09 term at Jacobs University Bremen.

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