# Electric Field of Line/wire Charge (E flux through Gs)

1. Mar 12, 2015

### Curiosity 1

Hi.

Consider the cylindrical Gaussian surface (Gs) shown in the following picture:

Using Gauss's law,
∫ E ⋅ dA= E ∫ dA = Q/ ε
E 2 π r L = λ L / ε
E= λ /(2 π r ε)

Is this true? I have a feeling like this is wrong. We have considered that there is only ONE electric field vector, E, fluxing through each da on Gs.
Now, consider a infinitesimally small area on Gs, call it da. There are too many electric field vectors that we can construct, since there are infinitely many charge on the line/wire (superposition of electric flux through a surface). I mean, there can also be a electric field (say E') vector that is not parallel to the vector of da:
da and E vectors

| \
| \ ← E'
| \
| \
-------- ←edge-on view of da (a part of Gs)
| \
| \
+-----------+ ←another charge on the line/wire which has electric field vector that is not parallel to the vector da.

Then the total E flux through the da on Gs is the superposition of all electric flux through it due to every charge on the line/wire.

Can you see where I'm wrong?

P.S. : Unfortunately, the diagram that I typed can't be view correctly after posting.... hope that you can get what I mean anyway. Just give some space between each " | " and " \ " so that you can see a triangle (a vector diagram with the base's edge replaced by "+" charge).

Thanks.

Last edited: Mar 12, 2015
2. Mar 12, 2015

### BruceW

it's true that there will be infinitely many contributions, since the wire is infinitely long. But, the magnitude due to charges very far down the wire will tend to zero very quickly. And once you do the mathematics, you see that (lucky for us), the total magnitude of the electric field takes some finite value.

3. Mar 12, 2015

### Curiosity 1

Actually I am wondering why we can assume there is only one E field fluxing through each da on the Gs. We assume that each E field due to a charge on the line/wire points radially outward from the line/wire, can't there be E field due to that charge not pointing radially outward from the line/wire (such that it's E field is not parallel to the normal of da when fluxing through the da)? If you really think about it, there will be infinitely many.

Let's consider another simpler case (Flat sheet of charge):

When we put a pillbox as Gs, electric flux occurs only at both end of the pillbox.
However, what I am thinking is that any + charge on the flat sheet CAN contribute electric flux through the dS, even if the dS is infinitesimally small. There is so many electric filed vector that we can construct from a single + charge on the flat sheet since the E field due to the charge acts in all direction (I don't know whether I'm right to say so). So, the E field fluxing through the dS is not necessarily to be parallel to the normal of dS. Then the total E flux through the dS on Gs is the superposition of all electric flux through it due to every charge on the flat sheet.

To summarize up my thoughts: I don't think the flux through the dS is only E dS cos0, but rather something like E dS cos0 + E dS cos 0.1 + E dS cos 0.2 + E dS cos 0.3+.... E dS cos 70 + E dS cos..........

4. Mar 13, 2015

### BruceW

yeah, so if we think of the sheet as being made up of an extremely large number of point charges, then the electric field due to a particular charge will not be perpendicular to the surface (like you said). And, if you want, you can do the calculation this way. You will find that in the limit of very large number of charges, each with a small magnitude of charge, all lying on a very large surface, you will get to the same answer as the 'Gaussian pillbox' method for the continuous charge distribution. Both methods will give you the same answer. In the Gaussian pillbox method, we consider the total electric field due to the entire continuous charge distribution, and we assume the total electric field will be parallel to the surface due to symmetry. So either: you can think about the total electric field due to everything. Or, think about the electric field due to each small patch of charge, and sum them up. Either way, you should get to the same answer.