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EMLecture9

# EMLecture9 - Lecture 9 Notes 95.657 Electromagnetic Theory...

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Lecture 9 Notes, 95.657 Electromagnetic Theory I, Fall 2011 Dr. Christopher S. Baird, UMass Lowell 1. Electrostatic Equations with Ponderable Materials - We desire to find the electric fields of any system which includes non-conducting material. (Air is typically considered to act like vacuum.) - We can consider a material to be a collection of small regions with a charge distribution. - Let us find the potential due to one of these charge distribution, then integrate over all charge distributions to get the total potential. - We know that we can expand the potential due to a localized charge distribution into its mulitpole moment contributions and only keep the first few terms if we are far away. Because we will shrink the charge regions to a very small size when forming the integral, any point in space can be considered far away from the charge region. - Consider a small volume dV containing a charge density producing a potential expanded into multipole contributions: d = 1 4  0 dq r 1 4   0 p x r 3 ... - Because the charge region is infinitesimal and we want a macroscopic expression, we are far enough away that terms become negligible except the monopole and dipole terms. - We switch to absolute coordinates to allow us to add up effects at different locations. d = 1 4  0 dq x x ' 1 4   0 p ⋅ x x ' x x ' 3 - Multiply and divide by the volume dV : d Φ= 1 4 πϵ 0 1 x x ' ( dq dV ) dV + 1 4 πϵ 0 ( x x ' ) x x ' 3 ( p d V ) dV - By definition, the charge per unit volume is the charge density, dq dV =ρ( x ' ) , and the average dipole moment per unit volume is the polarization, p dV = P . d Φ= 1 4 πϵ 0 ρ( x ' ) x x ' dV + 1 4 πϵ 0 P ⋅( x x ' ) x x ' 3 dV - To get the total electric potential, we add up the effects of all the small charge regions. Mathematically, this means integrating both sides of the equation: Φ( x )= 1 4 πϵ 0 ρ( x ' ) x x ' d 3 x ' + 1 4 πϵ 0 P ⋅( x x ' ) x x ' 3 d 3 x ' - This is now the potential due to all charge regions in the material. Because we started very generally, it includes all materials except conductors, and to any type of charge configuration: free charges, excess charges, bound charges, surface charges, etc. - We want to combine the two terms.

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- Use the mathematical identity: x x ' x x ' 3 =∇ ' 1 x x '  x = 1 4  0  x ' x x ' d x ' 1 4  0 P ⋅∇ ' 1 x x ' d x ' - Apply integration by parts to the second integral: P ⋅∇ ' 1 x x ' d x ' = [ P n x x ' ] S ' P x x ' d x ' - We are integrating over all space, so the bounding surface is at infinity where there is no material. This makes the first term vanish, so that: P ⋅∇ ' 1 x x ' d x ' =− ' P x x ' d x ' - Use this identity in the potential equation: Φ( x )= 1 4 πϵ 0 ρ( x ' ) x x ' d x ' 1 4 πϵ 0 ' P x x ' d x ' - Collect terms:  x = 1 4  0  x ' −∇ ' P
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EMLecture9 - Lecture 9 Notes 95.657 Electromagnetic Theory...

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