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Lect03ElectStatic

Lect03ElectStatic - ECE 3030 Electromagnetic Fields and...

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1 1 Swartz 06/5/13 ECE 303 – Electromagnetic Fields and Waves – Fall 2008 Lecture 3: ECE 3030 Electromagnetic Fields and Waves Instructor: Dr. Wesley E. Swartz Fall 2008 Lecture 3 2008/9/3 Electrostatics Application of Gauss’ Law Superposition Principle Fields of Some Charge Distributions q + q x E A 2 Swartz 06/5/13 ECE 303 – Electromagnetic Fields and Waves – Fall 2008 Lecture 3: Announcements We will not be able to “move” any of the Demo or Workshop sections to accommodate conflicts. If you have only a conflict for the Demo sections there is something blowing in the wind that might help. However, if your conflict results from trying to take ENGRD 2300, ECE 3100, and ECE 3030, then You should realize you are taking on a huge burden… All three of these courses are major undertakings! I recommend you have a talk with your advisor! If you have only a conflict for the Workshop sections, then You 3 Swartz 06/5/13 ECE 303 – Electromagnetic Fields and Waves – Fall 2008 Lecture 3: Simple Charge Distributions A single charge A Charged plate A ring of charge A line charge Oppositely charged plates Dipoles – a pair of opposite charges 4 Swartz 06/5/13 ECE 303 – Electromagnetic Fields and Waves – Fall 2008 Lecture 3: 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 surface must be uniform and pointing in the radial direction 0 = r q Using Gauss’ Law: ( ) q r E r o = 2 4 π ε r r q E r q E o o r ˆ 4 or 4 2 2 πε πε = = r q 5 Swartz 06/5/13 ECE 303 – Electromagnetic Fields and Waves – Fall 2008 Lecture 3: Field of an Infinite Charged Plane 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 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 x z y 6 Swartz 06/5/13 ECE 303 – Electromagnetic Fields and Waves – Fall 2008 Lecture 3: Field of an Infinite Charged Plane 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 = By Gauss’ Law: x z y A E x o ε 2 A o x x o E A A E ε ε 2 2 = = x E A
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