anonymous
  • anonymous
Flux through a piecewise-continuous closed surface
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I know it has something to do with electric flux
geerky42
  • geerky42
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html
anonymous
  • anonymous
how do I make the electic flux symbol?

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geerky42
  • geerky42
Google is your best friend.
anonymous
  • anonymous
I'm looking at a Flux through a Piecewise-continues closed surface
anonymous
  • anonymous
I'll draw it
anonymous
  • anonymous
|dw:1359307997058:dw|
anonymous
  • anonymous
E=+(200 N/C)k through the region z>0 and by E=-200 N/C k through the region z<0
anonymous
  • anonymous
|dw:1359308120623:dw|
anonymous
  • anonymous
Who does the E on top penetrate the surface and is still parallel to the surface?
anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
ok, I have an integral question though
anonymous
  • anonymous
\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}\]
phi
  • phi
The flux is the part of E normal to the surface. So to answer your question how does the E on top penetrate the surface and is still parallel to the surface? It doesn't the flux through the sides is 0
anonymous
  • anonymous
Oh interesting
anonymous
  • anonymous
can you help me understand an example in my book. It's a different problem. It's about electric field due to a line charge of finite length. It doesn't involve gauss...do you have a min?
phi
  • phi
ok
anonymous
  • anonymous
A thin rod of length L and charge Q is uniformly charged, so it has a linear density of \[\lambda=\frac Ql\] Find the electric field at point P, where P is an arbitrarily positioned point
phi
  • phi
Do they explain how to do this ?
anonymous
  • anonymous
Yes, they draw it out and everything. I can write their explanation and we can talk about it. I kinda get it, but I need to be able to reproduce this on my own
anonymous
  • anonymous
ok it starts off with \[dE=dE_rr\] how do I make an r hat?
anonymous
  • anonymous
and the E has an arrow on top of it
anonymous
  • anonymous
|dw:1359310750008:dw|
phi
  • phi
ok. so now you find E due to dq ? then integrate over all the dq's in the rod ?
anonymous
  • anonymous
Sorry I'm typing it all...might take me a couple of mins
anonymous
  • anonymous
\[\vec{dE}=dE_r \hat{r}\] \[dE_x=dE_r\hat r \cdot \hat i=dE_rcos\theta\] \[dE_y=dE_r\hat r \cdot \hat j=dE_rsin\theta\]
anonymous
  • anonymous
\[dE_r=\frac{kdq}{r^2}\] and \[cos\theta=\frac{-x_s}{r}\]
anonymous
  • anonymous
|dw:1359311776834:dw|
phi
  • phi
yes, I think I follow it so far.
phi
  • phi
now we need an equation ?
anonymous
  • anonymous
\[dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}\]
anonymous
  • anonymous
\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}\]
anonymous
  • anonymous
\[tan\theta=\frac{y_p}{|x_s|}=\frac{y_p}{-x_s}\]
anonymous
  • anonymous
so \[x_s=-\frac{y_p}{tan\theta}=-y_pcot\theta\]
anonymous
  • anonymous
I'm beyond LOST...but it continues...I'll just continue typing because they have another integral coming up...
anonymous
  • anonymous
\[sin\theta=\frac{y_p}{r},\;\textrm{so}\;r=\frac{y_p}{sin\theta}\]
anonymous
  • anonymous
\[dx_s=-y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta\]
anonymous
  • anonymous
ok here comes a big integral...
anonymous
  • anonymous
\[\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta\]
anonymous
  • anonymous
\[E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2 - sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}-\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]
phi
  • phi
that is some kind of ugly
anonymous
  • anonymous
yep ...tell me about it. I guess the most important thing is the equation at the end \[k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]
anonymous
  • anonymous
because they keep reusing that formula over and over in the examples that follow
phi
  • phi
yes, but it seems they could have gone about it in a more straight-forward manner,
anonymous
  • anonymous
Tell me about it!!!!! this is what they wrote \[E_z=\frac{k\lambda}{R}(sin\theta_2-sin\theta_1)=k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\] and \[E_R=-\frac{k\lambda}{R}(cos\theta_2-cos\theta_1)=-k\lambda\left(\frac{cot\theta_2}{r_2}-\frac{cat\theta_1}{r_1}\right)\]
anonymous
  • anonymous
they have a drawing to go with the above statement, but I don't feel like drawing it. It's similar to the drawing I drew earlier, but it has z and R axis I guess
phi
  • phi
don't you mean Ey (not Ez) ?
anonymous
  • anonymous
|dw:1359314142343:dw|
anonymous
  • anonymous
|dw:1359314391724:dw|
anonymous
  • anonymous
sweet jesus...
anonymous
  • anonymous
Yeah this is Calculating E from Coulomb's Law
phi
  • phi
For a single charge, the E field at some point is q (the charge)/ r^2 (the distance squared between the charge and the point, all times a unit vector that points from the charge to the point (scaled by a factor) \[ E= k \frac{q}{r^2} \mathbf{e} \] For more than one charge you have to integrate the contributions from all the charges.
phi
  • phi
The hard part is setting up the integral. Instead of keeping it a function of x (which is what I would try), they made everything a function of theta
anonymous
  • anonymous
What I have a hard time picturing is the fact that it causes a single arrow electric field. I would expect it go look like this|dw:1359314691161:dw| I don't know how to explain it...I would expect several arrows
anonymous
  • anonymous
Jennifer whats the name of your text?
anonymous
  • anonymous
physics for scientist and engineers by tipler and mosca
anonymous
  • anonymous
volume II
anonymous
  • anonymous
how would you have done it phi? I anything simpler than this please...
phi
  • phi
What I have a hard time picturing is the fact that it causes a single arrow electric field. I would expect it go look like this all the individual contributions to E could be thought of like that. But they sum up to produce one vector. Just add up all the vectors to get a resultant vector
anonymous
  • anonymous
Ahhhhaaa!!!! yeah that makes sense
anonymous
  • anonymous
I feel ....sad....I wish I understood this better
anonymous
  • anonymous
so is all that hassle to find that resultant vector?
phi
  • phi
I find this stuff very confusing.
phi
  • phi
so is all that hassle to find that resultant vector? yes.
phi
  • phi
If we change the single charge equation \[E= k \frac{q}{r^2} \mathbf{e}\] to an integral over all charges \[E_x=\int\limits_{x_1}^{x_2}k\lambda \frac{x dx}{(x^2+y^2)^{\frac{3}{2}}}\]
phi
  • phi
in this equation y is constant (we put our point at 0,y and the rod is at y=0)
anonymous
  • anonymous
oh I see, so x is the length of the rod...
anonymous
  • anonymous
how is y a constant?
phi
  • phi
the rod is at y=0 and the point we are interested in does not move. it is always at (0,y) as we integrate along the rod, r^2 changes (that is in x^2+y^2),
anonymous
  • anonymous
why is it plus?
phi
  • phi
I skipped a step. the vector from the charge at (x,0) pointing at our point is (-x,y) to make it unit length divide by sqrt(x^2 +y^2) that makes the denominator (originally r^2 or x^2+y^2 become (x^2+y^2)^(3/2)
phi
  • phi