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JenniferSmart1
Flux through a piecewise-continuous closed surface
I know it has something to do with electric flux
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html
how do I make the electic flux symbol?
Google is your best friend.
I'm looking at a Flux through a Piecewise-continues closed surface
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E=+(200 N/C)k through the region z>0 and by E=-200 N/C k through the region z<0
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Who does the E on top penetrate the surface and is still parallel to the surface?
ok, I have an integral question though
\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}\]
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
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?
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
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
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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 \[cos\theta=\frac{-x_s}{r}\]
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yes, I think I follow it so far.
\[dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}\]
\[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}\]
\[tan\theta=\frac{y_p}{|x_s|}=\frac{y_p}{-x_s}\]
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...
\[sin\theta=\frac{y_p}{r},\;\textrm{so}\;r=\frac{y_p}{sin\theta}\]
\[dx_s=-y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta\]
ok here comes a big integral...
\[\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta\]
\[E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2 - sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}-\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]
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
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
don't you mean Ey (not Ez) ?
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