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Net Electric Field between two line charges (different densities)

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    We are given that one infinite line of charge is placed horizontally on the x axis. The other line charge is placed parallel to the first at y = 0.4m. The first line has charge density 4.8 * 10^(-6) C/m, and the second -2.4 * 10^(-6) C/m. Find the net electric field at the following points. A) At y = 0.2m and B) At y = 0.6m.


    2. Relevant equations
    The relevant equation would be lambda / (2 * PI * epsilon - nought * radius).

    Where epsilon - nought = 8.85 * 10^(-12) C^2 / (N*m^2)


    3. The attempt at a solution

    Well.. since we are said to find the net electric field. I assume that we have to sum the two fields at the given point. So we find the field at the given point with respect to the first line, and then the second, then add them. Also do the same for the second point. However, the answer they give in the back of the book is 8.05 * 10^5 N/C, which is a single answer to a two part question. (Young and Freedman University Physics Ed 12) I am at a total loss as how they can get a single answer to a two part question....

    Anyhow, taking the zero point as the origin (0.2m) I did (4.8 * 10^(-6)) / (2 * 3.14... * 8.85 * 10^(-12) * 0.2m) + (-2.4 * 10^(-6)) / (2 * 3.14 * 8.85 * 10^(-12) * -0.2m)). I tried all variations of this, including negating the 0.2m in both equations and adding them. I did the same type of calculation for the y = 0.6m charge as well, except for line one 0.2 is changed to 0.6 and for line two it is kept at 0.2m.

    Any help would be appreciated. Thank you!
     
  2. jcsd
  3. Sep 14, 2008 #2

    Doc Al

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    Staff: Mentor

    Those two parts will certainly have different answers. (I might have that book. What's the problem number?)
     
  4. Sep 14, 2008 #3
    The chapter is 22 and the problem is 21, for edition 12.


    Thanks
     
  5. Sep 14, 2008 #4

    Doc Al

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    Must be a typo. Note that it says "toward negatively charged sphere", which has nothing to do with this problem.
     
  6. Sep 14, 2008 #5
    They seem to have a ton of typos in this book. The book is great.. it's just the answer key that pretty much sucks 10% of the time. However, is the method I am using in the problem, correct?
     
  7. Sep 14, 2008 #6

    Doc Al

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    I'm not quite sure I follow what you're doing (especially regarding signs). Describe the magnitude and direction of each field contribution.
     
  8. Sep 14, 2008 #7
    Ok well.. I am finding the sum of the two contributions of both electric fields due to the charged lines (wires). I am taking the point y=0.0m to be the origin so that the origin lies on the origin of the Cartesian plane.

    The contribution from the first line charge (plugging into the equation) would be :

    4.316 * 10^5 N/C. The first line charge is positive. Since the point is above the first line charge, the radius is 0.2m - 0.0m. The radius from the first line charge is 0.2m. This makes sense.

    For the second line charge :

    -2.158 * 10^5 N/C. The second line charge at (y = 0.4m) is negative. The radius would be 0.4m - 0.2m = 0.2m. The electric field contribution is negative.

    If I sum the two contributions I should get the net electric field:

    = 2.158 * 10^5 N/C. This makes sense to me. The electric field would be a positive value at y = 0.2m which makes sense. The bottom line charge is stronger than the top line charge.
     
  9. Sep 14, 2008 #8

    Doc Al

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    What direction does it point?

    What does "negative" mean? What direction does it point?

    Hint: With respect to the line charge, which way does the field point for a positive line charge? For a negative line charge? Figure out the actual direction of the fields first, then assign the proper sign wth respect to your y-axis.
     
  10. Sep 14, 2008 #9
    I think I get what you are saying. Basically the positive line charge pushes the probe at that point away because its a positive charge. The negative line charge pulls the probe toward it because it is negative. Therefore the contribution of the positive line charge is positive (the probe is being pushed directly upward) and with respect to the same probe, the contribution from the negative line charge should be positive as well because the line charge is negative (pulling toward). So to get the positive out of the negative contribution should I change the radius to be 0.2m - 0.4m = -0.2m? Then would I sum the fields again to get something on the order of 6.5 * 10 ^5N/C?
     
  11. Sep 14, 2008 #10

    Doc Al

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    Exactly.
    No. Radius is always positive. Don't try to plug into equations blindly. (The negative sign in the general equation just means toward the line charge; you have to give the field the proper sign according to its actual direction and your coordinate system.)

    The negative line chage exerts a positive field contribution (in part a). Period.
    Yes. (In what direction?)
     
  12. Sep 14, 2008 #11
    It would be in the direction of the negative line charge (positively upward), with respect to where it is , directly between the two line charges.
     
  13. Sep 14, 2008 #12

    Doc Al

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    Right. But express it as acting in the +y direction.
     
  14. Sep 14, 2008 #13
    Right. The final answer would be 6.5 * 10 ^ 5 N/C (estimate, Ill do the actual calculation out) in the +y direction. I think I can figure out part b of the problem, given the help.

    Thank you!
     
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