Electric Field due to two circular line charges

  • Thread starter emmett92k
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  • #51
blue_leaf77
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Look for something like ##\hat{r} = \ldots \hat{x} + \ldots \hat{y} + \ldots \hat{z}##.
 
  • #52
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Yeah so:

##\hat{r}=cos{\theta}sin\phi\hat{x}+sin{\theta}cos\phi\hat{y}+cos\phi\hat{z}##
 
  • #53
blue_leaf77
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No, check again whether you copied it right.
 
  • #54
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Sorry dunno how I did that:

##\hat{r}=sin{\theta}cos\phi\hat{x}+sin{\theta}sin\phi\hat{y}+cos\phi\hat{z}##
 
  • #55
blue_leaf77
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Now put that into the integral in comment #44 and do the integration, remember the integration variable is ##\phi## while ##\theta## is a constant (I leave to you to express the sine or cosine of ##\theta## in terms of known quantities). You should find that there are only two components surviving the integration. Upon considering (no calculation is needed) the contribution from the other half of the ring, one more component should vanish, and the only remaining components should be enhanched by a factor of two. Good luck.
 
  • #56
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Quick question, those the upper limit stay as ##\phi##? Thanks for all the help with this question, hopefully it all pays off.
 
  • #57
blue_leaf77
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That's a mistake, see again I have corrected it.
 
  • #58
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What I end up with is:

##E_+=\frac{Q}{2\pi\epsilon_0} \frac{1}{\sqrt{a^2+z^2}}##

Is this the answer I should've gotten?
 
  • #59
blue_leaf77
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Obviously that can't be right answer as the unit doesn't match that of an electric field.
 

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