How Does Charge Density Inside a Conductor Dissolve Over Time?

Think of \sigma in Ohm's law as a "macroscopic" quantity, valid on scales much larger than the mean free path, but much smaller than the relevant time scales in this problem. This is a time scale in which the microscopic details of the material become important.In summary, the conversation discusses free charges within a conductor and their eventual movement to the surface. The free charge density is shown to dissolve exponentially with time, with the characteristic time needed to dissolve the charge expressed in terms of the conductor's dielectric constant and conductivity. The conversation also mentions using charge conservation and Ohm's law to solve the problem.
  • #1
th5418
27
0

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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  • #2
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. divJ 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?:wink:

I know what the solution should be..
[tex]\rho (t) = \rho_0 e^{t} [/tex]
where [tex]t=\frac{\epsilon}{\sigma}[/tex]
Surely you mean [itex]\rho(t)=\rho_0 e^{-t/\tau}[/itex], where [itex]\tau\equiv\epsilon/\sigma[/itex]...right?
 
  • #3
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] ?
 
  • #4
[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]
 
  • #5
Ohm's law:

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

is not valid on the relevant time scale for this problem.
 
  • #6
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.
 
Last edited:
  • #7
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.
 

Related to How Does Charge Density Inside a Conductor Dissolve Over Time?

1. What is charge density inside a conductor?

Charge density inside a conductor refers to the amount of electric charge per unit volume of the conductor. It is usually represented by the symbol ρ (rho) and is measured in coulombs per cubic meter (C/m³).

2. How is charge density related to electric field inside a conductor?

According to Gauss's law, the electric field inside a conductor is directly proportional to the charge density. This means that as the charge density increases, the electric field inside the conductor also increases.

3. Why does the charge density inside a conductor remain constant?

The charge density inside a conductor remains constant because of the principle of electrostatic equilibrium. In this state, the electric field inside the conductor is zero, causing the charges to distribute themselves evenly throughout the conductor, resulting in a constant charge density.

4. Does the shape or size of a conductor affect its charge density?

No, the shape or size of a conductor does not affect its charge density as long as the conductor is in electrostatic equilibrium. This is because the charges will always distribute themselves evenly throughout the conductor, regardless of its shape or size.

5. How is charge density inside a conductor affected by the presence of an external electric field?

The presence of an external electric field can disrupt the electrostatic equilibrium inside a conductor, causing the charge density to change. If the external electric field is strong enough, the charges inside the conductor may move and redistribute themselves, resulting in a non-uniform charge density.

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