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Point charge electric field help find direction?

  • Thread starter awertag
  • Start date
  • #1
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Homework Statement


Ok i got the hard parts, a and b. For some reason I'm stuck on part c, the direction of the electric field.

Consider point a which is 70 cm north of a -4.2 µC point charge, and point b which is 76 cm west of the charge.


http://www.webassign.net/giancoli5/17_23alt.gif <------FIGURE

(a) Determine Vba = Vb - Va.
1Your answer is correct. V
(b) Determine Eb - Ea (note this is a vector minus a vector).
Magnitude
2Your answer is correct. N/C
Direction
3Your answer is incorrect.° (counterclockwise from east is positive)

Homework Equations





The Attempt at a Solution



I tried all the combinations of inverse tangent to try to get theta. Came up with 42.647 47.353 132.647 and 137.353...all wrong. What am I missing? Any good help is appreciated.
--aweg
 

Answers and Replies

  • #2
Andrew Mason
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Do you just want to find the direction of the field? If so, draw the vectors for the field at a and for the field at b and add them tail to head. The direction of the field at a is the direction in which a positive charge placed at a will move in response to the field produced by Q.

AM
 
Last edited:
  • #3
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that's what i thought at first but I want the direction of Eb-Ea. I don't know if that's different than the direction of the E field?
 
  • #4
Andrew Mason
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that's what i thought at first but I want the direction of Eb-Ea. I don't know if that's different than the direction of the E field?
To determine [itex]\vec{E_b} - \vec{E_a}[/itex] draw the vectors for the field at a and for the field at b; then reverse the direction of the vector for the field at a (ie. multiply by -1); and then add them tail to head.

I think I see what your problem is. You seem to be using displacement vectors rather than vectors representing the magnitude of the field. You have to determine the magnitude of the field at a and at b (not the potential) and then draw vectors representing those field values. Give us the magnitude and directions of the vectors you get for Eb and Ea.
 
  • #5
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well for part b I had found that Ea=-77142.57 (south) and Eb=-62326.87 (west)
 
  • #6
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so then i added their i and j components, did pythag, and got 99174.892 N/C
 
  • #7
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does that help atall? I'm sorry, I just don't understand your directions :(
 
  • #8
Andrew Mason
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well for part b I had found that Ea=-77142.57 (south) and Eb=-62326.87 (west)
Show us how you get Eb = -62326.87 N/C. I get a different answer. I also get a different direction for Ea. Both fields are toward Q.

AM
 
  • #9
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really? well i know 62326.87 is right because i used it to get part b which was correct
 
  • #10
Andrew Mason
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really? well i know 62326.87 is right because i used it to get part b which was correct
So why not show us how you got it? I get:

Eb = kQ/r^2 = 9e9 x 4.2e-6/(.76)^2 = 37.8e3/.5776 = 6.544e4 N/C

AM
 
  • #11
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yeah i got 6.544e4 now as well. I guess it didn't affect my previous (magnitude) answer because the percent error was tiny (due to the enormity of the numbers). But now, with the direction, i guess that error was enough to mess it up. Thanks for bearing with me...in case you're curious, I got a final angle of 49.69 degrees . (with inverse tangent)
 
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