Charge Density (Relaxation Time)

1. Jul 21, 2009

jeff1evesque

Statement:
If a volume charge distribution existed inside a conductor at t = 0, the charges would quickly migrate to the outside surfaces due to repulsion. The rate at which the charge density would decrease is given by:
$$\rho_{v}(t) = \rho_{v}(t = 0)e^{-\frac{t}{t_{r}}}$$ where the relaxation time, $$t_{r}$$, is $$t_{r} = \frac{\epsilon}{\sigma}$$.

Question:
But to me, it seems $$\rho_{v}(t) = \rho_{v}(t = 0)e^{-\frac{t}{t_{r}}}$$ is a function that is not defined. Where is the equation that makes the charge density decrease? Both sides of the equality (above) is a function that is not defined. I feel like this equation is like saying take the equation $$F(y) = F(y = a)$$ and solve for the function F(y = a, where a could be any of the Real's), which hasn't been defined. Does anyone know how to find the (decreasing) rate of charge density from the equation given? Or even the time it takes for a particular charge density to decrease to some fixed amount say 44coulombs/m^2?

Thanks,

JL

Last edited: Jul 21, 2009
2. Jul 21, 2009

cepheid

Staff Emeritus
The function is defined. It is an exponential function. rho at t = 0 is a number. More specifically, it is the starting density. This is what normalizes the exponential function. If you plug in t = 0 to the equation, the exponential factor becomes one and you're left with the starting value.

To the find the time it takes to decrease to "x" C/m^2, you set rho equal to x and solve for t.

3. Jul 21, 2009

jeff1evesque

Why isn't the equation written as, $$\rho_{v}(t) = e^{-\frac{t}{t_{r}}}?$$

And to solve for the time it takes to decrease to "x" C/m^2,
$$x = e^{-\frac{t}{t_{r}}}?$$

Thanks,

JL

4. Jul 21, 2009

cepheid

Staff Emeritus

5. Jul 21, 2009

jeff1evesque

And to solve for the time it takes to decrease to "x" C/m^2,the equation would be $$x = e^{-\frac{t}{t_{r}}}?$$
This decrease of charge density is the decrease of the material inserted into another medium? Or the medium (enviroment), that decreases in charge density (I'm guessing the former). What if the material (whatever one chooses) is inserted into some other medium (water, pond, big bowl of cool aid, or whatever) at time t = 0. Would 0 be substituted to the equation above for the variable t?

Thanks again.

Last edited: Jul 21, 2009
6. Jul 21, 2009

cepheid

Staff Emeritus
No. I mean, to find out when rho is equal to x, you'd...set rho equal to x, where rho is defined the way it is in your book and in my last post.

Sorry, not quite getting your question. The charge in the conductor spreads out due to electrical repulsion, and hence it becomes less dense. I don't think this would change if you immersed the conductor in anything else.

7. Jul 21, 2009

jeff1evesque

So it would be written like,
$$x(t) = Ae^{-\frac{t}{t_{r}}}$$

And I was wondering what is meant by "The rate at which the charge density would decrease"? Since the equation invovles putting some substance say hot water into a larger medium (cold water). The charge density would mean the hot water decreasing in charge density (not the cold water), correct?

And back to the equation above, if the hot water is inserted into cold water at time t = 0, then 0 would be substituted into the the variable t of the equation:
$$x(t) = Ae^{-\frac{t}{t_{r}}} = A.$$

Last edited: Jul 21, 2009
8. Jul 21, 2009

cepheid

Staff Emeritus
Interesting point. This may be bad wording on their part. Rho of t is the charge density as a function of time. I would interpret the rate of decrease of the charge density to be the derivative of that function. Oh well. Just words, really.

It does???

I dunno. I don't really have any context here.

I think we might have a communication problem. I feel like you are asking me, "at t = 0, should I set t equal to zero?" If that's not what you meant, can you clarify?

9. Jul 21, 2009

cepheid

Staff Emeritus
It sounds like the function is meant to tell you how the density evolves with time starting from the "dumping time", which is taken to be t = 0. So, I guess the answer to your question is, yes.