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Electric field on each midpoint of square of side L

  1. Mar 9, 2013 #1
    θ1. The problem statement, all variables and given/known data
    What would be the electric field on the midpoint of each side of a square with point charges of +Q on diagonal corners and -Q on diagonal corners?


    2. Relevant equations

    E=kQ/r^2 Ex=Ecosθ Ey=Esinθ
    3. The attempt at a solution
    Considering the midpoint of the side at the bottom of the square, I found the x components of each of the four electric fields it would have (with each of the four corners of the square).
    E1x = E1 = -4kQ/L^2
    E2x= E2 = 4kQ/L^2
    E3x=E3cosθ= -4kQ/5L^2(cosθ)
    E4x=E4cosθ=4kQ/5L^2(cosθ)

    When I find the sum of the x components, it is zero. I found the y-components in a similar way, and believe they will cancel out, too. So, I think the electric field on the midpoint should be zero, but the solution in the back of the book indicates that it is 1.15x10^10 Nm^2/C^2 (Q/L^2) in the x-direction.
    I haven't done the calculation for each midpoint, but I feel like each one would cancel out in the same way, but the solution says that each one is the answer listed above in either the x or y direction.
     
  2. jcsd
  3. Mar 9, 2013 #2
    Hello Dexter09

    I'm a bit confused about how you worked out the components.
    How did you get the values for E1x and E2x?
    Edit : I mean the directions
     
    Last edited: Mar 9, 2013
  4. Mar 9, 2013 #3
    I had the midpoint with -Q on the left, and +Q on the right. I drew my E1 vector from the midpoint toward +Q, and the E2 vector from the -Q toward the midpoint.
     
  5. Mar 9, 2013 #4
    And since the vectors were completely horizontal, there is no y component. So the x-components are equal to the total E for each.
     
  6. Mar 9, 2013 #5

    rude man

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    Taking the top two charges for example, each corner charge exerts its force in the same direction on a unit test charge placed at the mid-point.

    So the E field = 2E1x or 2E2x depending on which corners you placed the + and - charges.
     
  7. Mar 9, 2013 #6
    Actually, I was a little unsure of how to treat the q for the midpoint since I didn't think it would have a net charge. I just used Coulomb's Law and made E = kQ/r^2 (and since the r is L/2, it became E=4kQ/L^2)
     
  8. Mar 9, 2013 #7
    Since you are calculating the electric field and not the electric force, then you do not need to consider a charge at the midpoint.

    Just use the equation for the electric field produced by a point charge.
    If you draw a diagram with the directions of the fields at your point, then you will see that the fields will not cancel.
     
  9. Mar 9, 2013 #8

    rude man

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    Using Coulomb's law correctly you get the right answer. Look at the direction of the field at the mid-point due to the two charges q and -q.

    It's not a question as to the value of the "net charge" at the mid-point. There is no charge there.
     
  10. Mar 9, 2013 #9
    I had +Q(1) in the upper left and lower right, and -Q(2) - in upper right and lower left. I was considering the midpoint on the bottom of the square (with -Q(3) on its left and +Q(4) on its right. I drew vectors E1 from -Q(3) to the midpoint, vector E2 from the midpoint toward +Q(4), and E3 on a diagonal from +Q(1) to the midpoint, and E4 from the midpoint to -Q(2). Does my diagram make sense?
     
  11. Mar 9, 2013 #10
    Your directions are not correct for the electric fields.

    If I have a positive charge, does the electric field created by the charge point away or towards it?
     
  12. Mar 9, 2013 #11
    Now I am completely confused. Should I even be drawing electric field vectors from the midpoint, if it is not a charge?
     
  13. Mar 9, 2013 #12
    Away from positive, toward negative.
     
  14. Mar 9, 2013 #13
    You should draw the field vectors from the midpoint because that is the point where you want to evaluate the electric field.

    Yes

    So, start by drawing the field vectors at the midpoint due to the 2 charges on either side.
    You should find that the vectors point in the same direction.
     
  15. Mar 9, 2013 #14
    So E1 and E2 both go from the midpoint toward the -Q corner that is to their left. And E3 and E4 each go from the midpoint on diagonals in the positive x direction (their y components cancel out). I got E1x=-4kQ/L^2, E2x=-4kQ/L2, E3x=E4x=4sqrt5kQ/25L^2
    But the sum of all the x-components still does not add up to 1.15 x 1-^10. I keep checking my math, but I can't find where I am going wrong.
    maybe I should seperate it into two problems? A point at the apex of a triangle formed with other vertices of +Q and -Q, and then a point on a line equidistant between +Q and -Q. Does that sound reasonable?
     
  16. Mar 9, 2013 #15

    rude man

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    Look again at your E3x and E4x. Pay attention to the direction of the E field due to the two bottom charges on the top middle location.
     
  17. Mar 9, 2013 #16

    rude man

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    No. See my previous post.
     
  18. Mar 9, 2013 #17
    Aighh. I took so long to type that I was logged out. Here is a quick recap:
    E1x=4kQ/L^2 = 3.60 x 10^10 Q/L^2
    E2x = -4kQ/L^2 = -3.60 x 10^10 Q/L^2
    E3x = E3 sin theta = 4kQsqrt5Q/25L^2 = 3.22 x 10^9 Q/L^2
    E4x = E4 sin theta = -4kQsqrt5Q/25L^2 = -3.22 x 10^9 Q/L^2
    so the x components cancel
    E1y = 0
    E2y = 0
    E3y= E3 cos theta = -8sqrt5Q/25 L^2= -6.43 x 10^9 Q/L^2
    E4y=E4 cos theta = -8sqrt5Q/25 L^2= -6.43 x 10^9 Q/L^2

    so the y component = 1.29 x 10^10Q/L^2

    This give a magnitude of 1.29 x 10^10 Q/L^2 to the electric field in the y direction. but I can tell there is something wrong with my signs. From my diagram I can see that both E1 and E2 are in the negative x direction, and the E3x and E4x are in the positive x direction, and E3y and E4y are equal but opposite and should cancel. But I feel like I am using the Coulomb's law formula correctly..... ????
     
  19. Mar 10, 2013 #18
    The values look ok, but the signs are messed up
    You had the correct signs in post 14.

    My answer doesn't agree with the answer you quoted. Could you check the book answer again.
     
  20. Mar 10, 2013 #19
    The book answer is: 1.15 x 10^10 Q/L^2 for two midpoints, and -1.15 x 10^10 Q/L^2 for the other two midpoints. I've tried my numbers in different combinations to see if I can find my sign error, but I just can't figure it out. Coulomb's Law has a +Q in E1 and a -Q in E2, but they both point in the same direction, so it seems to me that they should each have negative x components. E3 and E4 also have opposite signs based on Coulomb's law, but they both point in the positive x direction, so it seems like they should each have a positive x-component. But when I add all the x components together I get -6.55 x 10^10 Q/L^2.
    Thanks for all of your help. I wish I could figure it out - I really like working on these problems. I thought I was understanding, but I guess I must be missing something.
     
  21. Mar 10, 2013 #20
    The first thing to do is to make sure you have the directions correct for the electric field vectors.
    Consider the midpoint of the bottom line, (I've got +Q at top-left and bottom-right and -Q at top-right and bottom-left).
    What are the directions of the 4 electric field vectors at that point?
    (I've got the positive x-axis pointing right and the positive y-axis pointing up).

    Remember : E points away from a positive source charge and towards a negative one
     
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