# Gauss's law and volumes with zero «net» charge

1. Oct 4, 2012

### ShayanJ

The integral form of gauss's law is used to determine the electric field of charge distributions which possess a certain amount of symmetry.
Now imagine using it in situations where the gaussian surface includes equal amounts of positive and negative charge.
For example,imagine a point positive charge surrounded with an spherical shell of negative charge of equal magnitude to the positive charge.
Gauss's law predicts a zero field for outside of the volume of this configuration but I calculated it directly and saw that it has a non-zero field although it is weak and dies off so fast.
This example leads me to think that gauss's law gives an approximation to the field in such situations and this makes me doubt about the nature of gauss's law.
Any ideas?

2. Oct 4, 2012

### Staff: Mentor

Is the charge on the shell uniformly distributed, and is the point charge at its center? If yes, the field outside must be zero. If no, the field outside need not be zero. Gauss's Law requires only that the integral of the field over a surface enclosing all of the charge must be zero. The field can be inwards at some points on the surface and outwards at other points, thereby making the integral over the entire surface zero.

Last edited: Oct 4, 2012
3. Oct 4, 2012

### ShayanJ

I calculated the field of a spherical shell with negative charge q,distributed uniformly over it:

$E=\frac{1}{4 \pi \epsilon_0} \int_0^{\pi} \int_0^{2 \pi} \frac{\sigma R^2 d \phi d \theta }{R^2+r^2-2Rr \cos{\phi}}=\frac{2 \pi \sigma R^2}{4 \epsilon_0 (r^2-R^2)}= -\frac{q}{8 \epsilon_0 (r^2-R^2)}$

R is the radius of the spherical shell,r is the distance of the observation point from the center of the sphere,$\phi$ is the angle between R and r and $\theta$ is the angle which when traversed from 0 to $\pi$,converts the circle to a sphere and $\sigma$ is the surface charge distribution.

But I'm not sure about it,because it has no $\pi$ at the denominator and so if we let R=0,it won't give the field of a point charge.
But any way,if you add it to the field of a positive point charge at the center of it,you won't get zero

Last edited: Oct 4, 2012
4. Oct 4, 2012

### Staff: Mentor

I don't understand what you did. A uniformly charged shell of radius R will present the same field as a point charge of equal total charge for r > R.

5. Oct 4, 2012

### ShayanJ

Yes...I remember you're right But can't find the problem in my calculation

6. Oct 4, 2012

### The_Duck

You've calculated the field wrong. I think you've forgotten a sin(theta) in the numerator of the integral (recall that the integration measure in spherical coordinates is r^2 sin(theta) dr dtheta dphi).

7. Oct 4, 2012

### ShayanJ

You're right,But with taking that into account,I got the following:

$E=-\frac{q}{4 \pi \epsilon_0 (r^2-R^2)}$

Which again if added to the field of a positive point charge in its center,doesn't give zero unless you're too far.
But the good thing is,it gives the field of a point charge if we set R=0.

8. Oct 4, 2012

### Staff: Mentor

You're trying to find $\vec E$ along the z-axis, right? Your corrected integrand (with $\sin \theta$ inserted in the numerator) gives you the magnitude of the bit of $\vec E$ contributed by a patch of charge $\sigma R^2 \sin \theta d\theta d\phi$ located at $\theta,\phi$ on the shell. But the bits of $\vec E$ from different patches are in different directions, so you can't simply add (integrate) their magnitudes.

You have to take advantage of the symmetry around the z-axis to note that the components of $\vec E$ perpendicular to the z-axis cancel out, and integrate only the z-component.

9. Oct 4, 2012

### gabbagabbahey

As jtbell points out, the electric field is a vector field. Each infinitesimal bit of charge on the surface produces a field with both a magnitude and a direction (directed from the source point to the field point). The direction of the field varies over the surface.

You can account for this without making any symmetry arguments (although the symmetry arguments make things quicker!), by finding the unit vector from a general point on the surface, to your field point and expressing it in terms of the Cartesian unit vectors. If you choose your coordinate axes so that your field point lies on the $z$-axis, you have

$$d\mathbf{E}(z) = \frac{1}{ 4 \pi \epsilon_0 } \frac{ \sigma R^2 \sin\theta d\theta d\phi }{ R^2+z^2-2Rz\cos\theta } \frac{ z\hat{\mathbf{z}} - R\hat{\mathbf{r}} }{ |z\hat{\mathbf{z}} - R\hat{\mathbf{r}}| } = \frac{1}{ 4 \pi \epsilon_0 } \frac{ \sigma R^2 \sin\theta d\theta d\phi }{ ( R^2+z^2-2Rz\cos\theta )^{ \frac{3}{2} }} (z\hat{\mathbf{z}} - R\hat{\mathbf{r}})$$

The Cartesian unit vectors are position independent and can be pulled out of the integral, but the spherical unit vectors are not. In particular, $\hat{\mathbf{r}} = \sin\theta \cos\phi \hat{\mathbf{x}} + \sin \theta \sin \phi \hat{\mathbf{y}} + \cos \theta \hat{\mathbf{z}}$ depends on both of the angles you are integrating over.

10. Oct 4, 2012

### Staff: Mentor

Yep, the x- and y- components of $\vec E$ both integrate to zero if you set up the integrals properly. When I used to teach intermediate E&M, I made my students do that sort of thing once, for a simpler situation. That usually convinced them to look for the symmetry first.

11. Oct 4, 2012

### ShayanJ

Yeah,Looks like I really need some practice in this one.
But there is still another confusion,about a dipole.
This one has a non-zero field but again gauss's law says that the field is zero if the gaussian surface includes both poles.Now what?
Thanks

12. Oct 4, 2012

### Staff: Mentor

No it doesn't. Only the integral of the field over the closed surface must be zero if the net charge enclosed is zero, not the field in general. (Reread jtbell's first post above.)