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jtbell
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#4
Nov9-12, 09:26 AM
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a question on flux, and on field integration


Quote Quote by Keyser View Post
so what you are saying (in regards to question no. 1) is that when i get E=0, it's because im adding vectors for which the total sum is zero (opposite directions) while the flux does not "consider" that case, and is a "per case" thing - each face has it's own flux, even though total E is zero?
It doesn't make any sense to talk about the "total E" in this situation. You can only add E's that are located at the same point, produced by different charges or other sources. In your example, there's no meaningful way to add the E at the top of the cube to the E at the bottom of the cube.

how can i find the field exerted by infinite plane on a point along the Z axis, but without using Gauss law? do i just say that a dq part of the charge is equal to kσdxdy and then
[itex]\overline{E}[/itex]=[itex]\int\int kσdxdy\cdot\frac{z}{\sqrt{x^{2}+y^{2}}}[/itex] ?
Not quite. E is a vector, so when you add the contributions to the total E at a given point, you have to do each component separately, in general:

$$E_z = \int {\int {dE_z}}$$

and similarly for ##E_x## and ##E_y##. In this example, you can argue from symmetry that the total ##E_x## and ##E_y## are both zero. For ##E_z## you have to take into account the angle between the z-direction and the line that defines r. And remember, Coulomb's Law has an r2 in the denominator.

$$E_z = \int {\int {\frac {k \sigma dx dy} {r^2} \cos \theta}}$$