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Electric field strength problem

  1. Sep 12, 2005 #1
    Here is my solution. I completed this problem to the best of my ability and I am asking for someone to check and correct my solution if possible. I am not sure if my answers are correct, although and my thinking might also be flawed :confused:

    q1= 6.0e-6C
    q2= 1.5e-6C

    The charges are arranged as such:


    The problem is to find the electric field at a point 0.01 meter to the left of q2.
    Here is my solution: ( I am ommiting units in calculation for clarity, units are given in answers)
    E1y = 0
    E1x = Ke q1/r^2 = 8.99e9 * 6.0e-6/.02^2 = 1.35e8 N/C
    E2x= Ke q2/r^2 = 8.99e9 * 1.5e-6/.01^2 = 1.35e8 N/C
    E3y = 0
    E3x = Ke q3/r^2 = 8.99e9 * 2e-6/.03^2 = 2.00e7 N/C
    Summing the x components I see that E2x and E1x cancel eachother out with respect to the arbitrary point 0.01m to the left of q2. So, I am left with a total electric field of E3 at a point 0.01 cm to the left of q2= 2.00e7 N/C.
    Is this correct?
    There is a second part of this problem that asks about the force on a point charge placed at the point given its charge of -2.00e-6C. Oddly enough I find finding forces a bit more straightforward. Conceptually I would set it up as a standard charge triangle problem. Seeing as there are no Y components I can just take the sum of the X components in the same way as I did the electric force? Not looking for calculations on the second part of this problem just a conceptual hint on how to go about it, but I would appreciate if someone checks my physics and calculations for the first part of the problem where I have given my answers.

    Thanks in Advance,
  2. jcsd
  3. Sep 12, 2005 #2
    Your calculations for this are correct. As for finding the force on a point charge, go back to the definition of an electric field.
    And rearrange for F. (Once you have the field, you only need to plug in a value for q to find the force).
  4. Sep 15, 2005 #3


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    what if 1 of the charges is negative?
    would the answer still be the sum of field strengths or would it require the subtraction of the negative charge's field strength?
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