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Homework Help: Electric Field Calculation given three point charges?

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Three point charges are located in an equilateral triangle with side 0.330 cm, where q1 = +8.30 µC, q2 = −8.20 µC, and q3 = +1.90 nC.

    A) What is the electric field at the location of q3? Enter the x component, then the y component.

    B) What is the force on q3? Enter the x component, then the y component.


    2. Relevant equations
    |Fe|= (k*|q1|*|q2|)/r^2

    k = 8.99*10^9 (N*m^2)/C^2
    Fe = Electrostatic force between a pair of charges
    r = charge separation
    q1, q2 = charge magnitudes
    E = electric field
    Q = charged object

    3. The attempt at a solution

    I've attempted this many ways and have been incorrect each time. I know the angles at the three points of the triangle will be 60deg. Knowing this and that the sides of the triangle are 0.330cm in length I calculated "r" -> the distance in respect to X and Y from q1 to q3 and from q2 to q3.

    I got r(1-3x) = 0.165cm and r(1-3y) = 0.286cm and r(2-3x) = 0.165cm and r(2-3y) = 0.286cm.

    Any help would be greatly appreciated. Even to just the first portion would help a lot.
    Last edited: Feb 18, 2014
  2. jcsd
  3. Feb 18, 2014 #2
    Careful -- to calculate the components of the force, you can't use the components of the separation. This is a pretty common mistake, I think, and not very intuitive, but you have to calculate the field vector using the full separation between the charges, and then find the components from that vector.

    The difference comes from the formula for electric field. Say for finding the y component of the field, using just the y component of the separation looks like this:
    $$E_{y} = k\frac{q_{1}q_{2}}{(rsin\theta)^{2}}$$
    Whereas finding the component of the whole field vector looks like this:
    $$E_{y} = k\frac{q_{1}q_{2}}{r^{2}}sin\theta$$
  4. Feb 19, 2014 #3
    Hi jackarms! Thanks for the help but I'm still very confused. I attempted another solution, again it's wrong.. I'll post my attempt and maybe someone can give me some guidance. I'm not sure if I'm on the right track or completely off in the wrong direction.


    Thank you!
    Last edited: Feb 19, 2014
  5. Feb 19, 2014 #4
    So not only did I use some wrong values in my previous calculation, after reading through my textbook further I approached it differently and got the correct answer for Part A :)


    I'm not sure what to do for Part B, but am going to attempt it and post my work.
  6. Feb 19, 2014 #5
    Attempted B, to find the force on q3 and it is incorrect.

    Any suggestions:

  7. Feb 19, 2014 #6
    All the calculations look right for both the field and the force, so I think one must be wrong. For the field, I'm not sure about the 7.15 x 107 value - are you sure that's right?

    Also, an easy way to go from field to force is to just use the definition of field: [itex]E = \frac{F}{q}[/itex]
    It's just nice to save some work.
  8. Feb 19, 2014 #7
    Thanks for the suggestion! I used your equation and got the same result, well, almost. The force for x was the same, the force for y was 0.134 N, so I must have had an issue with my sig figs. I entered both answers and they are correct!

    Thank you for your help!!
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