# Homework Help: Charge Density inside Conductor

1. Dec 3, 2009

### th5418

1. The problem statement, all variables and given/known data
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 $$\rho_f$$ will dissolve exponentially with time. Express the characteristic time needed to dissolve the charge in terms of the conductor's dielectric constant $$\epsilon$$ and the conductivity $$\sigma$$.

2. Relevant equations
I think I should use charge conservation. I'm not sure...
$$delJ + \frac{d\rho}{dt} = 0$$

3. The attempt at a solution
I know what the solution should be..
$$\rho (t) = \rho_0 e^{t}$$
where $$t=\frac{\epsilon}{\sigma}$$

2. Dec 3, 2009

### gabbagabbahey

Assuming you mean $\mathbf{\nabla}\cdot\textbf{J}+\frac{d\rho}{dt}=0[/tex] (i.e. divJ not "delJ" ), that seems like a good start to me....is there some relationship between [itex]\textbf{J}$ and $\sigma$ that might help you here?

Surely you mean $\rho(t)=\rho_0 e^{-t/\tau}$, where $\tau\equiv\epsilon/\sigma$...right?

3. Dec 3, 2009

### Reshma

Just a hint: Rearranging the equation
$$\vec \nabla \cdot \vec J = -\frac{d \rho}{dt}$$

Can you express $$\vec J$$ in terms of $\rho (t)$ ?

4. Dec 30, 2009

$$\vec \nabla \cdot \vec J = -\frac{\partial\rho}{\partial t}$$

and you should use the relation:

$$\vec J = \sigma\vec E$$

where $$\vec\nabla \cdot \vec E=\frac{\rho}{\epsilon}$$

5. Dec 30, 2009

### Count Iblis

Ohm's law:

$$\vec{J}=\sigma \vec{E}$$

is not valid on the relevant time scale for this problem.

6. Dec 31, 2009

No! Ohm's law is still valid. Only when the time is shorter than $$\tau$$ (which we will figure out when we solve the DE) Ohm's law turns out an invalid assumption. Because after time $$\tau$$, electrostatic equilibrium is reached, and finding $$\tau$$ is our concern.
Ohm's law is valid on time scales much longer than the typical collision time. The time scale $$\tau$$ in this problem will be many orders of magnitude less than that.