# Homework Help: Total charge given potential V(r)

1. May 24, 2010

### Saraphim

1. The problem statement, all variables and given/known data
Given
$$V(r)=\frac{Q}{4\pi \epsilon_0}\frac{1}{r+a},$$

find the total charge.

2. Relevant equations
$$\overline{E}=-\nabla V$$
$$\rho=\epsilon_0 \nabla \cdot \overline{E}$$

$$\rho=-\epsilon_0 \nabla^2 V$$

3. 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.

Last edited: May 24, 2010
2. May 24, 2010

### 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

3. May 25, 2010

### Saraphim

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?

Last edited: May 25, 2010
4. May 25, 2010

### ehild

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

ehild

5. May 26, 2010

### 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: $$\int_0^\infty \frac{1}{r(a+r)^3}dr$$

Last edited: May 26, 2010
6. May 27, 2010

### 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

7. May 27, 2010

### Saraphim

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

8. May 27, 2010

### ehild

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

ehild

9. May 27, 2010

### Saraphim

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.

10. May 27, 2010

### 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