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JenniferSmart1Best ResponseYou've already chosen the best response.1
I know it has something to do with electric flux
 one year ago

geerky42Best ResponseYou've already chosen the best response.0
http://hyperphysics.phyastr.gsu.edu/hbase/electric/gaulaw.html
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
how do I make the electic flux symbol?
 one year ago

geerky42Best ResponseYou've already chosen the best response.0
Google is your best friend.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
I'm looking at a Flux through a Piecewisecontinues closed surface
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1359307997058:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
E=+(200 N/C)k through the region z>0 and by E=200 N/C k through the region z<0
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1359308120623:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
Who does the E on top penetrate the surface and is still parallel to the surface?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
ok, I have an integral question though
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}\]
 one year ago

phiBest ResponseYou've already chosen the best response.1
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
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?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
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
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.1
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
ok it starts off with \[dE=dE_rr\] how do I make an r hat?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
and the E has an arrow on top of it
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1359310750008:dw
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.1
Sorry I'm typing it all...might take me a couple of mins
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[\vec{dE}=dE_r \hat{r}\] \[dE_x=dE_r\hat r \cdot \hat i=dE_rcos\theta\] \[dE_y=dE_r\hat r \cdot \hat j=dE_rsin\theta\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[dE_r=\frac{kdq}{r^2}\] and \[cos\theta=\frac{x_s}{r}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1359311776834:dw
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[tan\theta=\frac{y_p}{x_s}=\frac{y_p}{x_s}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
so \[x_s=\frac{y_p}{tan\theta}=y_pcot\theta\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
I'm beyond LOST...but it continues...I'll just continue typing because they have another integral coming up...
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[sin\theta=\frac{y_p}{r},\;\textrm{so}\;r=\frac{y_p}{sin\theta}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[dx_s=y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
ok here comes a big integral...
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2  sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}\frac{1}{r_1}\right)\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
yep ...tell me about it. I guess the most important thing is the equation at the end \[k\lambda\left(\frac{1}{r_2}\frac{1}{r_1}\right)\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
because they keep reusing that formula over and over in the examples that follow
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.1
Tell 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 \[E_R=\frac{k\lambda}{R}(cos\theta_2cos\theta_1)=k\lambda\left(\frac{cot\theta_2}{r_2}\frac{cat\theta_1}{r_1}\right)\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
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
 one year ago

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

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1359314142343:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1359314391724:dw
 one year ago