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Electric field on a plane

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data

    point A is between two charges, one to the left (+2 microcoul) and one to the right (-2 microcoul). point A is midway between the two charges. the two charges are 5cm apart. what is the electric field in the plane of the page (diagram) at a point 5cm left from point A. give in terms of x and y components of the electric field

    assume positive x is directed to the right, and positive y is directed upward of point A

    2. Relevant equations

    electric field = kq/r^2 where k is 9*10^9, q is charge, r is distance

    3. The attempt at a solution

    E1 = kq_l/r_1^2
    = (9*10^9)(+2*10^-6) / (0.025^2)
    = +2.88*10^7 n/c

    E2 = kq_r/r_2^2
    = (9*10^9)(-2*10^-6) / (0.075^2)
    = -3.2*10^6 n/c

    Enet = E1 + E2 = (+2.88*10^7) + (-3.2*10^6) = +2.56*10^7 n/c

    so Enet is directed to the right, purely x direction, no y direction

    (Ex, Ey) = (+2.56*10^7, 0) n/c

    my solution is incorrect. what did i do wrong? did i interpret the question correctly (plane part), if so could you explain what the "plane" part means?
     

    Attached Files:

  2. jcsd
  3. Sep 28, 2008 #2

    Doc Al

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    Staff: Mentor

    Looks like you have the charges interchanged. (In your diagram and calculations, the positive charge is on the right.)
     
  4. Sep 28, 2008 #3
    no, the positive charge in the diagram is on the left, not the right. how would having the charge on either side affect the result?
     
  5. Sep 28, 2008 #4

    Doc Al

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    OK, I misread the diagram. But in your calculation, the charges are reversed.
    It changes the direction of the field.

    I think the problem is that you are using the formula somewhat blindly to calculate the field and its sign. Instead, use the formula to get the magnitude of the field, then figure out the direction based on the sign of the charge. The field always points away from a positive charge and toward a negative charge.
     
  6. Sep 28, 2008 #5
    i can't see how i got the charges wrong in my calculations...q_l = q_left = left charge = positive charge = +2*10^-6 coulombs, and the distance is 2.5cm from the positive charge, so 0.025m.

    as for the q_r = q_right = negative charge = -2*10^-6 coulombs, and the distance is 7.5cm from the negative charge, so 0.075m

    am i not seeing something?
     
  7. Sep 29, 2008 #6

    Doc Al

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    Which way does the field from that positive charge point? Should the sign of its field be positive or negative?

    Which way does the field from that negative charge point? Should the sign of its field be positive or negative?
     
  8. Sep 29, 2008 #7
    the positive charge electric field points away from the charge, it should be negative

    the negative charge electric field points into the charge, it should be positive

    is i change the signs in my original calculations, will my answer be correct?
     
  9. Sep 29, 2008 #8

    Doc Al

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    That should fix it.
     
  10. Sep 29, 2008 #9
    thanks so much , it worked!! it was the sign all along, that's something i need to remember
     
  11. Sep 29, 2008 #10
    just another quick question. if was to determine the electric field at point P, is this how i would do it.

    E1 = kq_l/r_1^2
    = (9*10^9)(-2*10^-6) / (0.025^2)
    = -2.88*10^7 n/c

    E2 = kq_r/r_2^2
    = (9*10^9)(+2*10^-6) / (0.025^2)
    = +2.88*10^7 n/c

    so Enet = E1 + E2 = 0

    so (Ex, Ey = 0, 0) correct? apparently it is wrong.

    i know there is no field in the y direction so it has to be zero, and the distance from either charge is equal and only along the x axis, and since they are opposite directions, shouldn't they cancel out to zero?
     
  12. Sep 29, 2008 #11

    Doc Al

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    Once again, check your signs. Which way does the field from the positive charge point? The negative charge?

    (I assume you're talking about point A, right in the middle between the two charges.)
     
  13. Sep 29, 2008 #12
    i thought my signs were correct this time? i made them such that they were in the same directions as the first problem, after we corrected them. if i reverse the signs, the magnitude will be same, and wouldn't they just cancel out?

    and yes i was talking about point A, in the middle

    when is said 'opposite directions' i meant away from the positive, towards the negative. or is that incorrect wording? the positive points away towards the negative, and the electric field of the negative flows inwards to itself

    so the positive charge, sign is negative, and the negative charge, the sign is positive. how would that not cancel out if you get same magnitude?
     
  14. Sep 29, 2008 #13

    Doc Al

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    The situation is different when the point changes. (But the principle for figuring out the field direction remains the same, of course.)

    Actually, if you stuck to that last sentence you'd be OK. :wink:

    While your previous sentence was OK, this conclusion is false. (Actually draw the field arrows on a diagram.)
     
  15. Sep 29, 2008 #14
    since the point is now between the charges, the positive points to the right so it is positive, and the negative charge's field is positive direction too
     
  16. Sep 30, 2008 #15

    Doc Al

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    Exactly. The fields don't cancel, they add up.

    (If both charges where the same sign, then the field in the middle would be zero.)
     
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