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JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1I know it has something to do with electric flux

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1how do I make the electic flux symbol?

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1I'm looking at a Flux through a Piecewisecontinues closed surface

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359307997058:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1E=+(200 N/C)k through the region z>0 and by E=200 N/C k through the region z<0

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359308120623:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Who does the E on top penetrate the surface and is still parallel to the surface?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1ok, I have an integral question though

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}\]

phi
 one year 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1can 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1A 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
 one year ago
Best ResponseYou've already chosen the best response.1Do they explain how to do this ?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1ok it starts off with \[dE=dE_rr\] how do I make an r hat?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1and the E has an arrow on top of it

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359310750008:dw

phi
 one year 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 ?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Sorry I'm typing it all...might take me a couple of mins

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\vec{dE}=dE_r \hat{r}\] \[dE_x=dE_r\hat r \cdot \hat i=dE_rcos\theta\] \[dE_y=dE_r\hat r \cdot \hat j=dE_rsin\theta\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[dE_r=\frac{kdq}{r^2}\] and \[cos\theta=\frac{x_s}{r}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359311776834:dw

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[tan\theta=\frac{y_p}{x_s}=\frac{y_p}{x_s}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1so \[x_s=\frac{y_p}{tan\theta}=y_pcot\theta\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1I'm beyond LOST...but it continues...I'll just continue typing because they have another integral coming up...

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[sin\theta=\frac{y_p}{r},\;\textrm{so}\;r=\frac{y_p}{sin\theta}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[dx_s=y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1ok here comes a big integral...

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2  sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}\frac{1}{r_1}\right)\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1yep ...tell me about it. I guess the most important thing is the equation at the end \[k\lambda\left(\frac{1}{r_2}\frac{1}{r_1}\right)\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1because they keep reusing that formula over and over in the examples that follow

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Tell 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 \[E_R=\frac{k\lambda}{R}(cos\theta_2cos\theta_1)=k\lambda\left(\frac{cot\theta_2}{r_2}\frac{cat\theta_1}{r_1}\right)\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1they 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
 one year ago
Best ResponseYou've already chosen the best response.1don't you mean Ey (not Ez) ?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359314142343:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359314391724:dw