Hello everyone! I need help with a problem. My class has just begun a topic on electrostatics and we have been discussing the role of Gauss' Law in determining the electric field strength due to a charge distribution over an area (or volume). We proved that the double-integral of E . da = Q/e0. Now when dealing with point charges is it true that we choose a surface that makes our calculations easy, ie. a spherical one that completely encloses the point charge so that the electric field is always perpendicular to the surface? If so, this means that E . da = EAcos(q) = EA because cos(0) = 1. (this being that q = 0 degrees because the direction of dA is outward). One can therefore expect that for spherical coordinates E = E(r)r. So it is relatively easy to work out solid/thin spheres of charge. My question remains... How would you go about calculating the electric field at points i) near a long line of charge? ii) near a plane? iii) near any arbitrary shape? Intuition tells me that for (i) instead of int(E . dA) = E(4pi r^2) it would equal E(2pi r l). The same for a plane... = 2EA. Unfortunately i need a better explanation for this! I need a great explanation for this!! I know you can do it. For (iii) what would the integral be???? What if it was just any distorted shape??? Thanks in advance for any explanation of Gauss' Theorem anyone can give me. I find that reading other people's explanations can sometimes be better than reading from any textbook! Cheers. Ben.
your intuition is correct. the argument is the same as for the spherical case: symmetry. the spherical source has spherical symmetry, therefore the field must be spherically symmetric. the long wire has rotational symmetry about the axis, and translational symmetry along the wire, and so the field must also. the plane has translational symmetry in 2 dimensions. for example, in the planar problem, translational symmetry says that the field must be independent of x and y (if the plane is the x-y plane). now let s assume the field has some nonvertical component (i.e. some component lying in the x-y plane). let it be along the x-axis. now, the source charge plane is also symmetric under reflection about the y-axis, which takes the field from some E<sub>x</sub> to -E<sub>x</sub>. since the charge is invariant under this transformation, the fields must be the same. i.e. E<sub>x</sub>=-E<sub>x</sub>. but then E<sub>x</sub>=0. so the field has only a z-component. then to use Gauss law, you just draw a parallelipiped around the plane. a similar argument works for the line and the sphere. arbitrary charge distributions? you can t calculate the field this way for arbitrary charge distributions. simply can t be done. Gauss law still holds, its just not useful for calculating electric field. here, the only approach is the multipole approximation.
there is an even quicker proof for the spherical case: assume that the field has some nonradial component. this component is a tangent vector field to the sphere. the hairy ball theorem says that any tangent vector field on the sphere must be zero somewhere. spherical symmetry requires that it be the same everywhere, so it is zero everywhere. there is no nonradial component.
also, if don t know if this will really help you on this issue, but check out my thread on stoke s theorem, which Gauss law is a special case of.
by the way, you can t calculate the field using Gauss law near any arbitrary charge distribution, but if all you are interested in is the integral of the field, then that is easy! it s just Q/ε<sub>0</sub>
Thankyou for the great explanation lethe. (I like the notion of the hairy-ball theory!). So symmetry is the key to Gaussian surfaces? If you can apply an imaginary surface around a charge that is symmetrical, most examples should work. Here is another... Gauss' law deals with flux through a closed surface. Why not open surfaces? Is it that an outside charge source could now affect the overall distribution of charge lines around the original source? Or is it that it simply does not work??? And another... (note: e0 is the constant) From what you said I worked out the electric field near a plane to be the following.. int(E . dA) = 2EA = Q/e0 = (sigma)(A)/e0 (because Q = sigma . A) Remark: This tells me that field is uniform for points close to its surface and very far away (which makes sense ;-)) rearranging... E = sigma/2e0 However I would have thought that it should be sigma/e0. Why is it a factor of 2 out??? Gauss's Law int(E . dA) = sigma/e0 This is the proper law so why should it be different? Again, any help is much appreciated! Cheers.
aha, I was typing when you were typing! I was just getting to the fact that an arbitrary surface is Q/e0. Thanks for the additional thread sounds like a good read. (this is starting to get interesting!)
symmetry is the key to Gauss law in practice. but note that the theorem is true quite generally, it s just not useful for calculating the field. knowing the integral of a function is not sufficient for knowing the value of a function unless the function is highly symmetric. the information is simply not there. field lines end at charges. how many field lines tells you how much charge. if you don t have a closed surface, then you aren t counting all the field lines when you calculate your flux, some of them might be escaping out the other side, and so you can t really use that to judge how much charge there is. ah yes.... a very common question. your calculation is correct. the formula you mention is only true for a conducting plane, where there is no factor of 2. for an insulating plane, there is a factor of 2 in the denominator. see if you can figure out why (hint: the field inside a conductor is always zero)
also, outside charges don t contribute for a closed surface, because all field lines from an outside charge that enter the closed surface must eventually exit as well. so it is as you say, if the surface is not closed, then you have no guarantee that those field lines will exit your surface. so in addition to not counting all the field lines from the "enclosed" charge (in quotes, since you can t enclose anything with a surface that is not closed), you are also counting extra field lines from outside charges.
If the field inside a conductor is zero then all the charge must lie on the outside (ie. the surface) otherwise the charges in the conductor would move around. Which would mean that the electric field lines only leave on one side. For insulator do they leave on both sides??? If they do then I see where the factor of 2 comes in. Would appreciate a better explanation of this Please Register or Log in to view the hidden image! Am I correct to assume that this comes from the omega (charge density)? It would seem so because we are talking about two different ways of defining omega.
another way to see why the surface has to be closed: gauss theorem says that the integral of the divergence of the field in some volume is equal to the flux through the boundary of the volume. it is impossible for a boundary to not be closed.
yes, that s about it. for a conductor, the charge density refers to the charge density on the surface. it has two surfaces, so there is twice as much charge in a box through the plane. alternatively, draw your box through only surface of the conductor. same result. you are correct that the definitions of charge density are slightly different. for an insulator, the charge density means the charge density throughout the material, instead of just on the surface. this is because with a conductor, the charge resides only on the surface. for an insulator, it is valid to approximate the sheet as being infinitely thin, having no depth. in fact, since the formula for a true 2 dimensional insulator is the same as that for a sheet with finite thickness, it is not even an approximation, it is an exact formula. for the conductor it is not. it is possible to imagine a conducting sheet that really is 2-dimensional, and has no depth, and therefore no interior, and then this argument would not apply, and the formulae would be identical. both would retain the factor of 2. but this is not physical, in the real world, any conductor must have some depth.
