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JenniferSmart1

  • one year ago

Flux through a piecewise-continuous closed surface

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  1. JenniferSmart1
    • one year ago
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    I know it has something to do with electric flux

  2. geerky42
    • one year ago
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    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

  3. JenniferSmart1
    • one year ago
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    how do I make the electic flux symbol?

  4. geerky42
    • one year ago
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    Google is your best friend.

  5. JenniferSmart1
    • one year ago
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    I'm looking at a Flux through a Piecewise-continues closed surface

  6. JenniferSmart1
    • one year ago
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    I'll draw it

  7. JenniferSmart1
    • one year ago
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    |dw:1359307997058:dw|

  8. JenniferSmart1
    • one year ago
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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

  9. JenniferSmart1
    • one year ago
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    |dw:1359308120623:dw|

  10. JenniferSmart1
    • one year ago
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    Who does the E on top penetrate the surface and is still parallel to the surface?

  11. JenniferSmart1
    • one year ago
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    @phi

  12. JenniferSmart1
    • one year ago
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    @hartnn

  13. JenniferSmart1
    • one year ago
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    ok, I have an integral question though

  14. JenniferSmart1
    • one year ago
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    \[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}\]

  15. phi
    • one year ago
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    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

  16. JenniferSmart1
    • one year ago
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    Oh interesting

  17. JenniferSmart1
    • one year ago
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    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?

  18. phi
    • one year ago
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    ok

  19. JenniferSmart1
    • one year ago
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    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

  20. phi
    • one year ago
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    Do they explain how to do this ?

  21. JenniferSmart1
    • one year ago
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    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

  22. JenniferSmart1
    • one year ago
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    ok it starts off with \[dE=dE_rr\] how do I make an r hat?

  23. JenniferSmart1
    • one year ago
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    and the E has an arrow on top of it

  24. JenniferSmart1
    • one year ago
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    |dw:1359310750008:dw|

  25. phi
    • one year ago
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    ok. so now you find E due to dq ? then integrate over all the dq's in the rod ?

  26. JenniferSmart1
    • one year ago
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    Sorry I'm typing it all...might take me a couple of mins

  27. JenniferSmart1
    • one year ago
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    \[\vec{dE}=dE_r \hat{r}\] \[dE_x=dE_r\hat r \cdot \hat i=dE_rcos\theta\] \[dE_y=dE_r\hat r \cdot \hat j=dE_rsin\theta\]

  28. JenniferSmart1
    • one year ago
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    \[dE_r=\frac{kdq}{r^2}\] and \[cos\theta=\frac{-x_s}{r}\]

  29. JenniferSmart1
    • one year ago
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    |dw:1359311776834:dw|

  30. phi
    • one year ago
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    yes, I think I follow it so far.

  31. phi
    • one year ago
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    now we need an equation ?

  32. JenniferSmart1
    • one year ago
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    \[dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}\]

  33. JenniferSmart1
    • one year ago
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    \[\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}\]

  34. JenniferSmart1
    • one year ago
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    \[tan\theta=\frac{y_p}{|x_s|}=\frac{y_p}{-x_s}\]

  35. JenniferSmart1
    • one year ago
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    so \[x_s=-\frac{y_p}{tan\theta}=-y_pcot\theta\]

  36. JenniferSmart1
    • one year ago
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    I'm beyond LOST...but it continues...I'll just continue typing because they have another integral coming up...

  37. JenniferSmart1
    • one year ago
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    \[sin\theta=\frac{y_p}{r},\;\textrm{so}\;r=\frac{y_p}{sin\theta}\]

  38. JenniferSmart1
    • one year ago
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    \[dx_s=-y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta\]

  39. JenniferSmart1
    • one year ago
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    ok here comes a big integral...

  40. JenniferSmart1
    • one year ago
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    \[\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta\]

  41. JenniferSmart1
    • one year ago
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    \[E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2 - sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}-\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]

  42. phi
    • one year ago
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    that is some kind of ugly

  43. JenniferSmart1
    • one year ago
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    yep ...tell me about it. I guess the most important thing is the equation at the end \[k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\]

  44. JenniferSmart1
    • one year ago
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    because they keep reusing that formula over and over in the examples that follow

  45. phi
    • one year ago
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    yes, but it seems they could have gone about it in a more straight-forward manner,

  46. JenniferSmart1
    • one year ago
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    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 \[E_R=-\frac{k\lambda}{R}(cos\theta_2-cos\theta_1)=-k\lambda\left(\frac{cot\theta_2}{r_2}-\frac{cat\theta_1}{r_1}\right)\]

  47. JenniferSmart1
    • one year ago
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    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

  48. phi
    • one year ago
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    don't you mean Ey (not Ez) ?

  49. JenniferSmart1
    • one year ago
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    |dw:1359314142343:dw|

  50. JenniferSmart1
    • one year ago
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    |dw:1359314391724:dw|