## JenniferSmart1 Group Title Flux through a piecewise-continuous closed surface one year ago one year ago

1. JenniferSmart1 Group Title

I know it has something to do with electric flux

2. geerky42 Group Title
3. JenniferSmart1 Group Title

how do I make the electic flux symbol?

4. geerky42 Group Title

5. JenniferSmart1 Group Title

I'm looking at a Flux through a Piecewise-continues closed surface

6. JenniferSmart1 Group Title

I'll draw it

7. JenniferSmart1 Group Title

|dw:1359307997058:dw|

8. JenniferSmart1 Group Title

E=+(200 N/C)k through the region z>0 and by E=-200 N/C k through the region z<0

9. JenniferSmart1 Group Title

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10. JenniferSmart1 Group Title

Who does the E on top penetrate the surface and is still parallel to the surface?

11. JenniferSmart1 Group Title

@phi

12. JenniferSmart1 Group Title

@hartnn

13. JenniferSmart1 Group Title

ok, I have an integral question though

14. JenniferSmart1 Group Title

$\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}$

15. phi Group Title

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 Group Title

Oh interesting

17. JenniferSmart1 Group Title

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 Group Title

ok

19. JenniferSmart1 Group Title

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 Group Title

Do they explain how to do this ?

21. JenniferSmart1 Group Title

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 Group Title

ok it starts off with $dE=dE_rr$ how do I make an r hat?

23. JenniferSmart1 Group Title

and the E has an arrow on top of it

24. JenniferSmart1 Group Title

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25. phi Group Title

ok. so now you find E due to dq ? then integrate over all the dq's in the rod ?

26. JenniferSmart1 Group Title

Sorry I'm typing it all...might take me a couple of mins

27. JenniferSmart1 Group Title

$\vec{dE}=dE_r \hat{r}$ $dE_x=dE_r\hat r \cdot \hat i=dE_rcos\theta$ $dE_y=dE_r\hat r \cdot \hat j=dE_rsin\theta$

28. JenniferSmart1 Group Title

$dE_r=\frac{kdq}{r^2}$ and $cos\theta=\frac{-x_s}{r}$

29. JenniferSmart1 Group Title

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30. phi Group Title

yes, I think I follow it so far.

31. phi Group Title

now we need an equation ?

32. JenniferSmart1 Group Title

$dE_x=\frac{k\;dq}{r^2}cos\theta=\frac{k\;cos\theta\;\lambda dx_s}{r^2}$

33. JenniferSmart1 Group Title

$\int_{x_1}^{X_2} \frac{kcos\theta\lambda dx_2}{r^2}=k\lambda\int_{x_1}^{x_2}\frac{cos\theta \;dx_s}{r^2}$

34. JenniferSmart1 Group Title

$tan\theta=\frac{y_p}{|x_s|}=\frac{y_p}{-x_s}$

35. JenniferSmart1 Group Title

so $x_s=-\frac{y_p}{tan\theta}=-y_pcot\theta$

36. JenniferSmart1 Group Title

I'm beyond LOST...but it continues...I'll just continue typing because they have another integral coming up...

37. JenniferSmart1 Group Title

$sin\theta=\frac{y_p}{r},\;\textrm{so}\;r=\frac{y_p}{sin\theta}$

38. JenniferSmart1 Group Title

$dx_s=-y_p\frac{dcot\theta}{d\theta}=y_pcsc^2\theta d\theta$

39. JenniferSmart1 Group Title

ok here comes a big integral...

40. JenniferSmart1 Group Title

$\int_{x_1}^{x_2}\frac{cos\theta dx_s}{r^2}=\int_{\theta_1}^{\theta_2}\frac{cos\theta y_p csc^2\theta d\theta}{y^2p/sin^2\theta}=\frac1{y_p}\int_{\theta_1}^{\theta_2}cos\theta d\theta$

41. JenniferSmart1 Group Title

$E_x=k\lambda\frac{1}{y_p}\int_{\theta_1}^{\theta_2} cos\theta d\theta=\frac{k\lambda}{y_p}(sin\theta_2 - sin\theta_1)=\frac{k\lambda}{y_p}\left(\frac{y_p}{r_2}-\frac{y_p}{r_1}\right)=k\lambda\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$

42. phi Group Title

that is some kind of ugly

43. JenniferSmart1 Group Title

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 Group Title

because they keep reusing that formula over and over in the examples that follow

45. phi Group Title

yes, but it seems they could have gone about it in a more straight-forward manner,

46. JenniferSmart1 Group Title

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 Group Title

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 Group Title

don't you mean Ey (not Ez) ?

49. JenniferSmart1 Group Title

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50. JenniferSmart1 Group Title

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