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anonymous
 3 years ago
Flux through a piecewisecontinuous closed surface
anonymous
 3 years ago
Flux through a piecewisecontinuous closed surface

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know it has something to do with electric flux

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.0http://hyperphysics.phyastr.gsu.edu/hbase/electric/gaulaw.html

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do I make the electic flux symbol?

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.0Google is your best friend.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm looking at a Flux through a Piecewisecontinues closed surface

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359307997058:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0E=+(200 N/C)k through the region z>0 and by E=200 N/C k through the region z<0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359308120623:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Who does the E on top penetrate the surface and is still parallel to the surface?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, I have an integral question though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}\]

phi
 3 years ago
Best ResponseYou've already chosen the best response.1The 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
 3 years ago
Best ResponseYou've already chosen the best response.0can 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A 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
 3 years ago
Best ResponseYou've already chosen the best response.1Do they explain how to do this ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok it starts off with \[dE=dE_rr\] how do I make an r hat?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and the E has an arrow on top of it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359310750008:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.1ok. so now you find E due to dq ? then integrate over all the dq's in the rod ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry I'm typing it all...might take me a couple of mins

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\vec{dE}=dE_r \hat{r}\] \[dE_x=dE_r\hat r \cdot \hat i=dE_rcos\theta\] \[dE_y=dE_r\hat r \cdot \hat j=dE_rsin\theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[dE_r=\frac{kdq}{r^2}\] and \[cos\theta=\frac{x_s}{r}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359311776834:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.1yes, I think I follow it so far.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[tan\theta=\frac{y_p}{x_s}=\frac{y_p}{x_s}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so \[x_s=\frac{y_p}{tan\theta}=y_pcot\theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm beyond LOST...but it continues...I'll just continue typing because they have another integral coming up...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[sin\theta=\frac{y_p}{r},\;\textrm{so}\;r=\frac{y_p}{sin\theta}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[dx_s=y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok here comes a big integral...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2  sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}\frac{1}{r_1}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep ...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
 3 years ago
Best ResponseYou've already chosen the best response.0because they keep reusing that formula over and over in the examples that follow

phi
 3 years ago
Best ResponseYou've already chosen the best response.1yes, but it seems they could have gone about it in a more straightforward manner,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Tell me about it!!!!! this is what they wrote \[E_z=\frac{k\lambda}{R}(sin\theta_2sin\theta_1)=k\lambda\left(\frac{1}{r_2}\frac{1}{r_1}\right)\] and \[E_R=\frac{k\lambda}{R}(cos\theta_2cos\theta_1)=k\lambda\left(\frac{cot\theta_2}{r_2}\frac{cat\theta_1}{r_1}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0they 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
 3 years ago
Best ResponseYou've already chosen the best response.1don't you mean Ey (not Ez) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359314142343:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359314391724:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah this is Calculating E from Coulomb's Law