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Lect03ElectStatic

Lect03ElectStatic - 1 ECE 3030 Electromagnetic Fields and...

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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 π ε r r q E r q E o o r ˆ 4 or 4 2 2 πε πε = = 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 ε 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 ε ε 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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Lect03ElectStatic - 1 ECE 3030 Electromagnetic Fields and...

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