Contradiction in Gauss's Law: Analytical vs Intuitive Evaluation

[r16]

I was thinking about gauss's law and ran into this contradiction.

Consider this situation in electrostatics. You have an infinite line of charge, uniform charge density [text]\lambda[/tex]. In cgs units, E at a distance r from the line of charge along the +y axis (assuming a left handed coordinate system) is $$\frac{2 \lambda}{r}$$.

Now consider a right circular cylinder in empty space as a gaussian shape. Intuitively, I will tell you that the net flux is 0 through the entire shape: all field lines that flow into the shape flow out.

But thinking about this analytically gives you a different answer:

Flux through top bottom (closest to the charge) = $$-\frac{2 \lambda}{r} (\pi r^2)$$
Flux through sides = 0
Flux through top (furthest away from the charge) = $$\frac{2 \lambda}{r + l} (\pi r^2)$$

so the net flux is tex]-\frac{2 \lambda}{r} (\pi r^2) + \frac{2 \lambda}{r + l} (\pi r^2)[/tex] which must be less than 0 because $$r > r+l$$, which would mean that there is a net negative flux, or some sort of negative charge enclosed in the gaussian surface which is contradictory to the intuitive evaluation of the shape.

Any resolution to this contradiction?

 

[Doc Al]

It sounds like you have an infinite line of charge, say along the x-axis. Then you want to consider a cylindrical gaussian surface in empty space above the line charge, say along the y-axis.

I was thinking about gauss's law and ran into this contradiction.

Consider this situation in electrostatics. You have an infinite line of charge, uniform charge density $$\lambda$$. In cgs units, E at a distance r from the line of charge along the +y axis (assuming a left handed coordinate system) is $$\frac{2 \lambda}{r}$$.

OK.

Now consider a right circular cylinder in empty space as a gaussian shape. Intuitively, I will tell you that the net flux is 0 through the entire shape: all field lines that flow into the shape flow out.

That's true.

But thinking about this analytically gives you a different answer:

Flux through top bottom (closest to the charge) = $$-\frac{2 \lambda}{r} (\pi r^2)$$
Flux through sides = 0
Flux through top (furthest away from the charge) = $$\frac{2 \lambda}{r + l} (\pi r^2)$$

Realize that the field from the line charge is radially outward. Thus the field is not perpendicular to the top and bottom--you must consider the angle to calculate the flux through the ends. Similarly, the field is not parallel to the sides--again you must consider the angle to calculate the flux through the sides.

Done correctly, the net flux will be zero.

 

[r16]

rgr that,

opening up my book and looking at it again, I see what you are saying.

I guess I am having difficulty imaging an electric field and calculating values for an electric field especially when it comes to asymmetric shapes, such as an electric field in an unsymmetrical hollowed out sphere or the electric field inside a hollow cubic pipe for example. I could do it using the definition of an electric field
$$dE = \int \frac{\rho(\vec{x_n}) d^3 x}{r^2} \hat{r}$$
but this leads to some unnecessarily complicated integrals. I'm sure if I picked a few clever Gaussian surfaces I could solve the problems a whole lot easier, but I have a hard time picking out those surfaces because I have a hard time intuitively picturing what E will look like so I make poor decisions for my surface I'm using

 

[ZapperZ]

Gauss's Law technique (i.e. constructing a gaussian surface) is useful in highly symmetric situation. This is where you can construct a surface where the electric field flux is either a constant, or zero, or a combination of the two. A uniform infinite line charge is one such example, where a cylindrical gaussian surface will give a zero net flux at the flat surfaces, and a constant flux at the curved surface. This is why we use Gauss's law method here.

If you cannot construct such a surface with the geometry of the charge given, then the method is no longer useful as far as getting an analytic solution to get the field. You have to use other techniques.

Zz.