How Does Charge Density Inside a Conductor Dissolve Over Time?

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th5418
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Homework Statement


We know that free charges inside a conductor will eventually move to the conductor surface. Consider a free charge initially placed inside a conductor at t=0. Show that the free charge density [tex]\rho_f[/tex] will dissolve exponentially with time. Express the characteristic time needed to dissolve the charge in terms of the conductor's dielectric constant [tex]\epsilon[/tex] and the conductivity [tex]\sigma[/tex].

Homework Equations


I think I should use charge conservation. I'm not sure...
[tex]delJ + \frac{d\rho}{dt} = 0[/tex]

The Attempt at a Solution


I know what the solution should be..
[tex]\rho (t) = \rho_0 e^{t}[/tex]
where [tex]t=\frac{\epsilon}{\sigma}[/tex]
 
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th5418 said:
I think I should use charge conservation. I'm not sure...
[tex]delJ + \frac{d\rho}{dt} = 0[/tex]

Assuming you mean [itex]\mathbf{\nabla}\cdot\textbf{J}+\frac{d\rho}{dt}=0[/tex] (i.e. div<b>J</b> not "delJ" ), that seems like a good start to me...is there some relationship between [itex]\textbf{J}[/itex] and [itex]\sigma[/itex] that might help you here?<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /><br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I know what the solution should be..<br /> [tex]\rho (t) = \rho_0 e^{t}[/tex]<br /> where [tex]t=\frac{\epsilon}{\sigma}[/tex] </div> </div> </blockquote>Surely you mean [itex]\rho(t)=\rho_0 e^{-t/\tau}[/itex], where [itex]\tau\equiv\epsilon/\sigma[/itex]...right?[/itex]
 
Just a hint: Rearranging the equation
[tex]\vec \nabla \cdot \vec J = -\frac{d \rho}{dt}[/tex]

Can you express [tex]\vec J[/tex] in terms of [itex]\rho (t)[/itex] ?
 
[tex]\vec \nabla \cdot \vec J = -\frac{\partial\rho}{\partial t}[/tex]

and you should use the relation:

[tex]\vec J = \sigma\vec E[/tex]

where [tex]\vec\nabla \cdot \vec E=\frac{\rho}{\epsilon}[/tex]
 
Ohm's law:

[tex]\vec{J}=\sigma \vec{E}[/tex]

is not valid on the relevant time scale for this problem.
 
No! Ohm's law is still valid. Only when the time is shorter than [tex]\tau[/tex] (which we will figure out when we solve the DE) Ohm's law turns out an invalid assumption. Because after time [tex]\tau[/tex], electrostatic equilibrium is reached, and finding [tex]\tau[/tex] is our concern.
 
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caduceus said:
No! Ohm's law is still valid. Only when the time is shorter than [tex]\tau[/tex] (which we will figure out when we solve the DE) Ohm's law turns out an invalid assumption. Because after time [tex]\tau[/tex], electrostatic equilibrium is reached, and finding [tex]\tau[/tex] is our concern.

Ohm's law is valid on time scales much longer than the typical collision time. The time scale [tex]\tau[/tex] in this problem will be many orders of magnitude less than that.