Total charge given potential V(r)

[Saraphim]

Homework Statement

Given
V(r)=\frac{Q}{4\pi \epsilon_0}\frac{1}{r+a},

find the total charge.

Homework Equations

\overline{E}=-\nabla V
\rho=\epsilon_0 \nabla \cdot \overline{E}

\rho=-\epsilon_0 \nabla^2 V

The Attempt at a Solution

I started out by finding the charge density \rho from \overline{E}, and I got the result:

\rho(r)=\frac{aQ}{2\pi r(a+r)^3}

I might as well have gone with Poisson's equation to start with, but I didn't think of that.

I have, however, no idea how to proceed from here to find the TOTAL charge. I'm assuming I somehow have to integrate over some region (all of space?), but I don't know how to do it. Any pointers at setting up the integral would be appreciated. I should be able to handle it from there.

 

[ehild]

Your ρ(r) is wrong. Do not forget that the Laplace operator is not simply the second derivative with respect to the radius.

If you have the proper formula for the charge density, you get the total charge by integrating to the whole space.

The other way is to find E and then applying Gauss' law.

ehild

 

[Saraphim]

Your ρ(r) is wrong. Do not forget that the Laplace operator is not simply the second derivative with respect to the radius.

I actually found \rho by first getting the \overline{E}-field from the potential and then using Gauss' law in differential form. I actually thought I had \rho right, so I better figure that out before I try and find the total charge.

1. Finding \overline{E}:

\overline{E} = -\nabla V

The potential is constant to \theta and \Phi, so their partial derivatives drop out and I get:

\overline{E}= - \frac{\partial V}{\partial r}\hat{r} = \frac{Q}{4\pi \epsilon_0}\frac{1}{(r+a)^2}\hat{r}​

Is this correct? Assuming that...

2. Finding \rho:

I utilize Gauss' law in differential form.

\rho=\epsilon_0 \nabla \cdot \overline{E}=\epsilon_0 \frac{Q}{4\pi \epsilon_0}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{1}{(r+a)^2}\right)

=\frac{1}{r^2}\frac{Q}{4\pi}\frac{2r(a+r)-2r^2}{(a+r)^3}=\frac{aQ}{2\pi r(a+r)^3}
​

I went over it all again and I couldn't find the error. Could you give me a pointer?

 

[ehild]

Sorry, my derivation was wrong, yours is correct. Now integrate for a sphere of infinite radius.

ehild

 

[Saraphim]

Can you give me a pointer for setting up the relevant integral? The only thing I can think of is a triple integral where the relevant r-integral doesn't converge: \int_0^\infty \frac{1}{r(a+r)^3}dr

 

[ehild]

The total charge is the volume integral of the charge density:

Q_{total}=\int{\rho dV}

Use the volume element for spherical coordinates.

ehild

 

[Saraphim]

Thanks a lot for your help! The total charge is Q. :-)

 

[ehild]

Well done! Just for fun, try it with the Gaussian surface integral of E.


ehild

 

[Saraphim]

Well done! Just for fun, try it with the Gaussian surface integral of E.

I get stuck doing that. I use Gauss' law in integral form and arrive at this:

Q_{total} = \epsilon_0 \oint_S \overline{E} \cdot d\overline{a}

The field and area elements are everywhere parallel, so I can drop the directions and thus the dot product:

Q_{total} = \epsilon_0 \oint_S E da = \epsilon_0 \oint_S \frac{Q}{4 \pi \epsilon_0}\frac{1}{(r+a)^2}da = \frac{Q}{4 \pi} \oint_S \frac{1}{(r+a)^2} da

That's about as nice as it gets, I think, but I don't know how to evaluate that integral. I guess I have to expand it into a double integral somehow. But since the integral is over my gaussian surface, which is a sphere of constant radius, the double integrals should be with respect to \phi and \theta, but how do I get rid of the r's then?

Edit: I'm sure this has something to do with my gaussian surface being at infinity, but I can't see how to procede. If I just let the fraction tend to zero, I get the result that total charge is zero, which is obviously wrong.

 

[ehild]

You need to integrate for a sphere of radius R and then take the limit to R-->infinity. Note that a in da in your integral is not the same as the parameter a. Denote it dA instead. What is the surface element in spherical coordinates?
But it is simpler than that in this case.
As the magnitude of the electric field is constant on the surface of a central sphere, you get the integral just by multiplying E(R) with the surface area of the sphere of radius R, don't you?

ehild