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Electric Fields charge problem

  1. Mar 2, 2004 #1
    I'm having trouble with this question. Where am I going wrong?

    Three charges are at the corners of an equilateral trangle. The bottom left corner has a charge of 2.00e-6 C, bottom right -4.00e-6 C, and top with 7.00e-6 C and separated by 0.500m.

    Calculate the electric field at the position of the 2.00e-6 C charge due to the 7.00e-6 C and -4.00e-6 C charges.


    E1 = [(K_e)(q1)(q2)/(r^2)]

    =[-(8.99e9)(7.00e-6)(2.00e-6) / (.500m)^2] * [-cos(60)i - sin(60)j]

    = 0.25172 i + 0.8660 j


    E2 = [(K_e)(q1)(q2)/(r^2)]

    = [(8.99e9)(2.00e-6)(-4.00e-6) / (.500m)^2] * [cos(0)i]

    = -0.28768 i


    (0.25172 i - 0.28768 i) = -0.03596 i

    So, -0.03596 i and 0.8660 j


    I know this isn't right, where did I go wrong?

    Any help is greatly appreciated
    Last edited: Mar 2, 2004
  2. jcsd
  3. Mar 2, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You messed up the direction of the field. Why do you have a minus sign ("-(8.99e9)...")? The charge is positive.

    Always draw yourself a picture of how the field points as a sanity check for your answer.
  4. Mar 2, 2004 #3
    Oh okay yeah I see it. That was actually part B of the question, which asks to find the force on the 2e-6 C charge.

    I'm getting that right now but how do I find the electric field at the 2e-6 C charge?

    What I did was:

    E1 = [(K_e)(q)/(r^2)]

    =[(8.99e9)(7.00e-6) / (.500m)^2] * [-cos(60)i - sin(60)j]

    = (-1.26e5 i - 2.18e5 j) N/C


    E2 = [(K_e)(q1)/(r^2)]

    = [(8.99e9)(2.00e-6) / (.500m)^2] * [cos(0)i]

    = (7.19e4 i) N/C


    E3 = [(K_e)(q1)/(r^2)]

    = [(8.99e9)(-4.00e-6) / (.500m)^2]

    = -1.44e5 N/C


    I then added them all together and got something completely different from the book.

    The book has (18.0 i - 218 j) kN/C

    How did they get that?

    Should there be a negative sign on E3? Meaning, [-(8.99e9)(-4.00e-6) / (.500^2)]

    I understand how find the force but having trouble with understanding how to find the E field if I'm not supposed to get the product of the charges.

    Thanks in advance.
    Last edited: Mar 2, 2004
  5. Mar 3, 2004 #4
    I'm not quite sure why you're trying to find the E field if they're asking for the force, but they want you to ignore the electric field due to the 2e-6 charge. If you don't, then you're going to get a separation of 0 and division by 0 always causes problems.

    Electric fields are conservative and therefore obey the superposition principle. In order to find the electric field due to the two other charges at the 2e-6 charge, just imagine that the 2e-6 charge isn't there. Calculate the electric field at the position of the 2e-6 charge due to the first charge, then repeat this for the second charge, and then add the two together.

    Then, in order to find the force, just multiply the electric field you just calculated by the charge of the 2e-6 charge.

    Alternatively, you could have just found the force straight from Coloumb's Law and bypassed the electric field entirely. But since you've alread calculated it...

  6. Mar 3, 2004 #5
    Cookiemonster is correct in the method to solve the problem for the force at the position of the 2.00uC charge.
    Here is a little more of a push in order to help you solve.

    Your E1 ((-1.26e5 i - 2.18e5 j) N/C) is correct for the E-field created by the 7.00uC charge.

    Your E2 ((7.19e4 i) N/C) should not be figured into this problem due to Cookiemonsters explanation.

    Your E3 (-1.44e5 N/C) should not be negative. Remember that the 4.00uC charge is negative, so that means the field will be acting towards the charge, which is to the right in the layout of this problem.

    Now, you are correct to add the vectors.
    So, using your designations above (after you change the sign of E3):
    E=(-1.26E5 i -2.18E5 j) N/C + (1.44E5i) N/C

    Let us know how you do.
  7. Mar 3, 2004 #6

    Doc Al

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    Staff: Mentor

    Yes, you were using Coulomb's law for forces, not E field, but I forgot to point that out.
    By adding the fields due to the other two charges.
    No! This one's meaningless, since you are trying to find the field right where this charge is. Don't include it.
    Wrong sign. The charge is negative, so the field points toward the charge.
    You made two mistakes. The big mistake is adding the field due to the 2e-6 C charge.
    Not sure what you are doing with all the minus signs. Always draw a simple diagram to find the direction of the field, then apply the appropriate signs to the components.
    The field and the force are different things, but obviously related.

    The field (kq/r^2) is independent of any charge at the location; the force (kqQ/r^2) depends on the field as well as the charge at the location.
  8. Mar 3, 2004 #7
    Ooooh okay. Thanks everyone, I got the right answer and understand it clearly now!

    Thanks again!! :)
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