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

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• #1
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## 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
σ

## 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.

Thanks

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• #2
gabbagabbahey
Homework Helper
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Try expressing J in terms of D then combine your two maxwell's equation into a single DE for $\rho$

• #3
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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
gabbagabbahey
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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
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Okay so Jfree=J. also D=εE so E=D/ε
and therefore J=σD/ε

edit:
so

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

Doesn't feel like im getting anyhere

Last edited:
• #6
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• #7
gabbagabbahey
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∇ · σD/ε = −∂(ρfree)/∂t
or
(∇σ/ε)· D + (σ/ε)(ρfree) = −∂(ρfree)/∂t

Doesn't feel like im 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
16
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aha. I get it, Thanks alot thats brilliant. Don't think I would have ever got that on my own

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