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Science Forum Index » Physics - Electromagnetic Forum » Liénard-Wiechert Electric Potential
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Message |
| blackhead |
 Posted: Mon Sep 03, 2007 4:33 pm |
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Guest
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The retarded electric potential is:
Phi (_r, t) = volume_integral (volume, [ p(_r') ] / R, dv') (1)
Where:
_r is the radius vector to the observation point
_r' is the radius vector to a volume element of charge density at t- R/
c
R = | _r - _r' |
[ p(_r') ] = p(_r', t - R/c) is the charge density at _r' evaluated at
t - R/c
Standard text books on classical EM use this result to get the Liénard-
Wiechert electric potential for a moving charge as:
Phi (r, t) = q / [ R - B._R ]
where:
_u = d/dt _r'
_R = r - r', _B = _u/c
[ ] is evaluated at t - R/c
Quote: From (1), [ p(_r')] = q at r' = location of the charge, so I don't
understand why (1) shouldn't immediately simplify to:
Phi (_r, t) = q / [ R ]. Can anyone enlighten me?
Thanks. |
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| blackhead |
| Posted: Mon Sep 03, 2007 6:40 pm |
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On 3 Sep, 22:33, blackhead <larryhar...@softhome.net> wrote:
Quote: The retarded electric potential is:
Phi (_r, t) = volume_integral (volume, [ p(_r') ] / R, dv') (1)
Where:
_r is the radius vector to the observation point
_r' is the radius vector to a volume element of charge density at t- R/
c
R = | _r - _r' |
[ p(_r') ] = p(_r', t - R/c) is the charge density at _r' evaluated at
t - R/c
Standard text books on classical EM use this result to get the Liénard-
Wiechert electric potential for a moving charge as:
Phi (r, t) = q / [ R - B._R ]
where:
_u = d/dt _r'
_R = r - r', _B = _u/c
[ ] is evaluated at t - R/c
From (1), [ p(_r')] = q at r' = location of the charge, so I don't
understand why (1) shouldn't immediately simplify to:
Quote: Phi (_r, t) = q / [ R ]. Can anyone enlighten me?
Sorry for the stupididty, this is only true when _u = 0.
> Thanks. |
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