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Problem with electrostatics

  • #1

Homework Statement


upload_2018-9-9_19-9-35.png


I'm suppose to set a general function of the electric field, at any point in the xy plane. As the picture shows, there are 2 point charges with 1 charge, and 1 point charge with -3 charge. Each of them are equally distanced from the origin, and all 3 are evenly spaced apart.

Homework Equations


I'm used Coulumbs law
upload_2018-9-9_19-11-28.png


The Attempt at a Solution


First I defined all the point charge locations, along with an arbitrary point, to represent the unknown point of the function.
upload_2018-9-9_19-12-29.png

upload_2018-9-9_19-12-34.png


Then I merely use Coumlumbs law, to add up all the expressions:
upload_2018-9-9_19-13-8.png

upload_2018-9-9_19-13-33.png


The final expression isn't pretty at all though:
upload_2018-9-9_19-14-16.png


The field as displayed by maple (if one fixes the fieldstrength, otherwise some arrows become very very small) looks like this:
upload_2018-9-9_19-17-5.png


Which to me looks about right. However the vector function isn't divergence free, which suggests I have messed up something. What am I doing wrong?
 

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Answers and Replies

  • #2
Charles Link
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The two dimensional function will not be divergence free, but if you included the z-component of the electric field, it would then be divergence free except at the locations of the electrical charges. ## \\ ## I didn't carefully check your equations, but they seem to be correct.
 
  • #3
The two dimensional function will not be divergence free, but if you included the z-component of the electric field, it would then be divergence free except at the locations of the electrical charges. ## \\ ## I didn't carefully check your equations, but they seem to be correct.
Ahh alright then. Thanks a bunch! :)
 
  • #4
Charles Link
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Ahh alright then. Thanks a bunch! :)
What is also the case is that when you draw lines of flux for a 3-D case in two dimensions, the result is inherent inaccuracies where the density of lines doesn't fall off as inverse square, etc. The line density will be ## \frac{1}{r} ## in a 2-D diagram of a 3 dimensional case if they are drawn as continuous lines.
 
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  • #5
Charles Link
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And an addition to post 2: You can see this with a single point charge: If you let ## E=\frac{Q}{4 \pi \epsilon_o}(\frac{x}{(x^2+y^2)^{3/2}} \hat{i} +\frac{y}{(x^2+y^2)^{3/2}} \hat{j}) ##, the divergence is non-zero everywhere, but if you write it as ## E=\frac{Q}{4 \pi \epsilon_o}(\frac{x}{(x^2+y^2+z^2)^{3/2}} \hat{i}+\frac{y}{(x^2+y^2+z^2)^{3/2}} \hat{j}+\frac{z}{(x^2+y^2+z^2)^{3/2}} \hat{k} ) ## and compute ## \nabla \cdot E ##, you will find it is indeed zero everywhere except at ## (0,0,0) ##.
 

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