Lect03ElectStatic - 1 ECE 3030 Electromagnetic Fields and...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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?...
View Full Document

This note was uploaded on 11/26/2009 for the course ECE 3030 at Cornell University (Engineering School).

Page1 / 4

Lect03ElectStatic - 1 ECE 3030 Electromagnetic Fields and...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online