Class_22new - 1 PHYS809 Class 22 Notes Energy and force...

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Unformatted text preview: 1 PHYS809 Class 22 Notes Energy and force considerations in dielectric media For a charge distribution in free space, the electrostatic energy is 3 1 . 2 U d x ρ = Φ ∫ (22.1) Here ( ) ρ x and ( ) Φ x are the charge density and electrostatic potential, respectively. This expression holds in the presence of dielectric media only under special circumstances. To determine how the electrostatic energy can be found in the general case, we consider the effects of adding charge ( ) δρ x to an existing localized charge distribution ( ) ρ x with associated potential ( ) . Φ x The work done to add this charge is 3 . W d x δ δρ = Φ ∫ (22.2) The added charge is related to the change in electric displacement by . δ δρ ∇⋅ = D (22.3) Hence ( ) ( ) 3 3 3 . W d x d x d x δ δ δ δ = ∇⋅ Φ = ∇⋅ Φ- ⋅∇Φ ∫ ∫ ∫ D D D (22.4) The first integral on the right can be converted to a surface integral by use of the divergence theorem. Since the charge distribution is localized, integration over all space makes the surface integral go to zero, so that 3 3 . W d x d x δ δ δ = - ⋅∇Φ = ⋅ ∫ ∫ D E D (22.5) The total electrostatic energy is equal to the work required to assemble the charge distribution starting from a state in which D = 0. Hence, we can write 3 1 . 4 D allspace U d x δ π = ⋅ ∫ ∫ E D (22.6) If the medium is linear , we have ( ) 1 , 2 δ δ ⋅ = ⋅ E D E D (22.7) so that 3 1 . 2 allspace U d x = ⋅ ∫ E D (22.8) 2 This can be transformed into (22.1) by relating E to the potential and D to the charge density and performing integration by parts....
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This note was uploaded on 12/02/2011 for the course PHYS 809 taught by Professor Macdonald during the Fall '11 term at University of Delaware.

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Class_22new - 1 PHYS809 Class 22 Notes Energy and force...

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