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Flux through a piecewise-continuous closed surface

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I know it has something to do with electric flux
how do I make the electic flux symbol?

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I'm looking at a Flux through a Piecewise-continues closed surface
I'll draw it
E=+(200 N/C)k through the region z>0 and by E=-200 N/C k through the region z<0
Who does the E on top penetrate the surface and is still parallel to the surface?
ok, I have an integral question though
\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}\]
  • 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
Oh interesting
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 you have a min?
  • phi
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
Do they explain how to do this ?
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
ok it starts off with \[dE=dE_rr\] how do I make an r hat?
and the E has an arrow on top of it
  • phi
ok. so now you find E due to dq ? then integrate over all the dq's in the rod ?
Sorry I'm typing it all...might take me a couple of mins
\[\vec{dE}=dE_r \hat{r}\] \[dE_x=dE_r\hat r \cdot \hat i=dE_rcos\theta\] \[dE_y=dE_r\hat r \cdot \hat j=dE_rsin\theta\]
\[dE_r=\frac{kdq}{r^2}\] and \[cos\theta=\frac{-x_s}{r}\]
  • phi
yes, I think I follow it so far.
  • phi
now we need an equation ?
\[dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}\]
\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}\]
so \[x_s=-\frac{y_p}{tan\theta}=-y_pcot\theta\]
I'm beyond LOST...but it continues...I'll just continue typing because they have another integral coming up...
\[dx_s=-y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta\]
ok here comes a big integral...
\[\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta\]
\[E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2 - sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}-\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]
  • phi
that is some kind of ugly
yep ...tell me about it. I guess the most important thing is the equation at the end \[k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]
because they keep reusing that formula over and over in the examples that follow
  • phi
yes, but it seems they could have gone about it in a more straight-forward manner,
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 \[E_R=-\frac{k\lambda}{R}(cos\theta_2-cos\theta_1)=-k\lambda\left(\frac{cot\theta_2}{r_2}-\frac{cat\theta_1}{r_1}\right)\]
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
don't you mean Ey (not Ez) ?
sweet jesus...
Yeah this is Calculating E from Coulomb's Law
  • 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
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
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
Jennifer whats the name of your text?
physics for scientist and engineers by tipler and mosca
volume II
how would you have done it phi? I anything simpler than this please...
  • 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
Ahhhhaaa!!!! yeah that makes sense
I feel ....sad....I wish I understood this better
so is all that hassle to find that resultant vector?
  • phi
I find this stuff very confusing.
  • phi
so is all that hassle to find that resultant vector? yes.
  • 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
in this equation y is constant (we put our point at 0,y and the rod is at y=0)
oh I see, so x is the length of the rod...
how is y a constant?
  • 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),
why is it plus?
  • 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
it probably should be minus
no plus seems right...I guess
  • phi
the direction vector e pointing from the charge to our point is (-x,y) the x component of E is the x component of the e vector times all the rest of the stuff, so we do want -x dx rather than x dx (if that is what you are asking )
|dw:1359315838647:dw| can you show me?
  • phi