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**Unformatted text preview: **1 ECE 3030 Electromagnetic Fields and Waves Fall 2009 Lecture 3 2009/9/2 Electrostatics Application of Gauss Law 1 Swartz and Rana 06/5/13 Electromagnetic Fields and Waves Fall 2009 Lecture 3: Instructor: Dr. Wesley E. Swartz Superposition Principle Fields of Some Charge Distributions q + q x E A Simple Charge Distributions A sampling of those with simple field solutions: A single charge A charged plate A ring of charge A line charge Oppositely charged plates Dipoles a pair of opposite charges 3 Swartz and Rana 06/5/13 Electromagnetic Fields and Waves Fall 2009 Lecture 3: Ways to solve problems in electrostatics: Gauss Law in integral or differential form Making use of symmetries Superposition of known solutions Field of a Point Charge Consider a point charge of q Coulombs sitting at Surround the charge by a Gaussian surface in the form of a spherical shell of radius r By symmetry , the E-field magnitude on the Gaussian surface must be uniform and pointing in the radial direction. = r q 4 Swartz and Rana 06/5/13 Electromagnetic Fields and Waves Fall 2009 Lecture 3: Using Gauss Law: ( ) q r E r o = 2 4 r r q E r q E o o r 4 or 4 2 2 = = r q Field of an Infinite Charged Plane (1) Suppose we have a charged plane. Let the surface charge density be Coulombs/m2 Symmetry Argument: The charge distribution is symmetric w.r.t. +z and z directions. Therefore, if at any point there is an E-field component in the + z direction, there must also be E-field component in the z direction. Since the field cannot have a z component x z 5 Swartz and Rana 06/5/13 Electromagnetic Fields and Waves Fall 2009 Lecture 3: Since the field cannot have a z-component pointing in both + z and z directions at the same time, there cannot be a z-component of the field. Similarly, there cannot be a y-component of the E-field So the field can only have an x-component y Field of an Infinite Charged Plane (2) Since the field can only have an x-component, Draw a Gaussian surface in the form of a cylinder of area A piercing the charged plane Total flux coming out of the cylinder ends = Total charge enclosed by the surface = x z y A E x o 2 A 6 Swartz and Rana 06/5/13 Electromagnetic Fields and Waves Fall 2009 Lecture 3: By Gauss Law: o x x o E A A E 2 2 = = x E A Boundary Conditions for Surface Charges Suppose we know the surface normal electric field on just one side of a charged plane with a surface charge density Question: What is the surface normal field on the other side of the charged plane?...

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