Electric field insulating rod Question

  • Thread starter ToonBlue
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  • #1
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Homework Statement


upload_2015-9-7_0-18-58.png


upload_2015-9-7_0-53-36.png

Homework Equations




The Attempt at a Solution



This is problem I have found online and this is the part I don't understand how they come up with this.
upload_2015-9-7_0-21-54.png

How did they get this for the x component since y component cancel out due to being symmetry ?
upload_2015-9-7_0-22-57.png
 
Last edited:

Answers and Replies

  • #2
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I think they are trying to find dE not E...
 
  • #3
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upload_2015-9-7_0-36-54.png

The above is the formula given in my lecture note .

In the solution provided in the problem , it is given as
upload_2015-9-7_0-21-54-png.88285.png
which means that
upload_2015-9-7_0-22-57-png.88286.png
is the unit direction of the component.

In this case , since there is no y component (symmetry) , it has only x component but how do they get
upload_2015-9-7_0-22-57-png.88286.png
?
 
  • #4
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The y component will cancel out I think when you take the Integral over all points towards point O. I believe the formula ( at the end ) provided in the solution is trying to calculate the dE at point O.
Wait for more qualified members to reply. I am taking Electricity and Magnetism course right now :)
 
  • #5
NascentOxygen
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The diagram in your first post shows you dE; it's the field you'll see at O due to the charge contained in the short length, dl. The electric field at any point has both a magnitude (illustrated by the arrow having a length) and a direction (shown by the angle of arrow dE to the reference axes).

It's not until you sum the fields due to multiple fragments of charge will you see cancellations, this entails integrating dE for all of the line charge.
 
  • #6
rude man
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  • #7
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May I know why does the dE point away and not toward the dL?
 
  • #8
rude man
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May I know why does the dE point away and not toward the dL?[/QUOThe arc L is negatively charged. The direction of the E field is the direction of the force on a positive unit of charge. So the E field is directed towards the arc L, as you say.
If you got the impression it's directed away from L it's probably because of the picture. However, the picture is still right since the answer is negative.
 

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