
#1
Nov912, 06:28 AM

P: 6

hey all,
i've recently started studying electrostatics, and i have couple of question about things that i did not fully understand, and would very much appreciate if someone could set me straight. 1) how can a cube, with a single charge in the middle of it have a flux? don't the field lines cancel each other, thus achieving 0 electric field for each plane? i mean, you have a field vector going up on the Y axis, and down on the same axis (and so on for the others), so how come the flux [itex]\Phi=A\overline{E}[/itex] isn't zero? 2) if i have a flat disk on a plane, with uniform charge density [itex]\sigma[/itex], why when i integrate, i do so for small rings, and not for very small circles? why is it: [itex]E=\int\frac{kσ2∏rdr}{(...)}[/itex] instead of [itex]E=\int\frac{kσ2∏dr}{(...)}[/itex]? thank you very much for your help 



#2
Nov912, 07:13 AM

P: 987

For second one,you will have to cover the whole area of disk(for integration) and a circle does not have any thickness so you must have to select a ring.Also note that it should have the units of area which is only in the first case,not the second. 



#3
Nov912, 08:28 AM

P: 6

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?
and another one: 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] ? thank you! 



#4
Nov912, 09:26 AM

Mentor
P: 11,254

a question on flux, and on field integration$$E_z = \int {\int {dE_z}}$$ 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 zdirection and the line that defines r. And remember, Coulomb's Law has an r^{2} in the denominator. $$E_z = \int {\int {\frac {k \sigma dx dy} {r^2} \cos \theta}}$$ 



#5
Nov912, 12:35 PM

P: 6

but doesn't the E that go in the +Y direction "cancels" the one going in the Y direction?




#7
Nov912, 03:28 PM

P: 6

so, if i want to treat E as a vector, i must relate it to a certain point? i thought the field was a vector anyhow, so i can add it however i wanted.
i'm adding a picture that might make things clear. the green arrows are E exerted by a single charge at (0,0,0). so, from symmetry, you argue that E in that axis cancels each other, but can't you say as well, that if you were to add them up (0,y,0)+(0,y,0) you'd get the same answer? note that i refer to E outside the box, as i understand that inside (from gauss law) the field is zero in any case http://img842.imageshack.us/img842/3222/cubeq.jpg thank you! edit: i think i'm getting it: i can treat E in that manner if, like you say, i have a point on which E is exerted, but if i'm talking about a sphere / cube / other as a shape, there is no point for which i can add the vectors? thank for your patience, when i get things slow, i understand them fast :) 



#8
Nov912, 04:05 PM

Sci Advisor
HW Helper
P: 1,937





#9
Nov1112, 06:46 AM

P: 987





#10
Nov1112, 08:47 AM

P: 147

For a closed surface around a charge, all the Efield lines are "poking" in or out, so there IS a net flux.
If you replace the charge with a bar magnet, some of the field lines poke out and some poke in, resulting in no net flux. Gauss' Laws. 



#11
Nov1212, 11:52 AM

P: 6

@greswd: if i put a bar inside, wouldn't i have 0 E at the top and bottom? 



#12
Nov1512, 02:20 PM

P: 6

and another question please: I understand Gauss's law in regards to plains (thanks to you). but how does it work when dealing with a sphere? i know how to integrate the whole thing, but i'm trying to get it in a more intuitive manner. when i calculate the flux , I'm looking at infinitesimal area dA, which is the perpendicular vector to the surface. so far, so good. but when I'm dealing with spheres, I integrate over infinitesimal spheres  or more accurate, over their surface area which is 3D.
what I think is, that because the flux goes through the entire sphere  say we have a 1C charge inside  and is different between these little spheres (the flux is proportional to electric field lines) so you need to sum the flux through all these little spheres in order to get the total flux, and that's why dA is the entire 3D surface area of an infinitesimal sphere. Am I correct? If not, how do these dA look like? if they are 2D (r^2), then how can i get just the flux for one such dA? Don't laugh if my question seems stupid, but I always thought of area as a 2D thing 


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