Charge distribution in a conductor (using maxwell's equations)

  • #1

Homework Statement

Show that any charge distribution in a conductor of conductivity σ and relative
permittivity κ vanishes in time as ρ = ρ0exp(−t/ζ) where ζ = κǫ0

Homework Equations

Maxwell's equation
∇ · D = ρfree

equation of continuity for a free charge density
∇ · Jfree = −∂(ρfree)/∂t

ohms law
J = σE

The Attempt at a Solution

I can see that ρ will get smaller and smaller as time 't' increases according to
ρ = ρ0exp(−t/ζ) and clearly some sort of substitution is required of the equations but I can't see how substitution will result in a exponential appearing. Basically I don't know where to start so any help would be appreciated or a push in the right direction.


Answers and Replies

  • #2
Try expressing J in terms of D then combine your two maxwell's equation into a single DE for [itex]\rho[/itex]
  • #3
I could sub the first eq into the second to get the divergence of current density equals the negative differential of the divergence of the electric displacement field, or

∇ · Jfree = −∂(∇ · D)/∂t

Can Jfree and total J be considered to be the same thing in this example? Also I will lose ρ if i do this.
  • #4
Is there ever any bound current in a conductor? If not, then Jfree and J are the same right?

And you don't want to combine the equations in that manner...start with expressing D in terms of E... there should be an equation for that
  • #5
Okay so Jfree=J. also D=εE so E=D/ε
and therefore J=σD/ε


∇ · σD/ε = −∂(ρfree)/∂t
(∇σ/ε)· D + (σ/ε)(ρfree) = −∂(ρfree)/∂t

Doesn't feel like I am getting anyhere
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  • #6
Not sure if your allowed to bump threads on this forum...o:)
  • #7
∇ · σD/ε = −∂(ρfree)/∂t
(∇σ/ε)· D + (σ/ε)(ρfree) = −∂(ρfree)/∂t

Doesn't feel like I am getting anyhere

σ/ε is a constant, so (∇σ/ε)=0 and therefor (σ/ε)(ρfree) = −∂(ρfree)/∂t

Also, is ρfree a function of any other variables besides time, inside a conductor?...if not, then−∂(ρfree)/∂t=−d(ρfree)/dt and you have a seperable ordinary differential equation for ρfree.
  • #8
aha. I get it, Thanks a lot that's brilliant. Don't think I would have ever got that on my own:smile:

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