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Homework Help: Electric Field Problems

  1. Feb 26, 2008 #1
    These have been driving me crazy. The book is terrible at explaining this stuff, so I was hoping someone here could help me out.

    1. The problem statement, all variables and given/known data
    A rectangle has a length of 2d and a height of d. Each of the following three charges is located at a corner of the rectangle: +q1 (upper left corner), +q2 (lower right corner), and -q (lower left corner). The net electric field at the (empty) upper right corner is zero. Find the magnitudes of q1 and q2. Express your answers in terms of q.

    2. Relevant equations
    Coloumb's Law: k * (|q1| * |q2| / r2)
    Where k= a proportionality constant ~ 8.99 * 109 N * m2/C2

    Electric Field Definition: E = F / q0
    Where E is the net electric field at a point, and F is the force experienced by a small test charge represented by q0

    3. The attempt at a solution
    I really got nowhere trying to find this solution, but here's what I tried:
    --> I defined the upper right corner where the net electric field is zero as point T (for easy reference).
    --> I represented the forces exerted on point T as three vectors, all with unknown magnitudes. The vector created by q1 had a direction of 0º; the vector created by q2 had a direction of arctan(2)~63.43º (reasoning below), and the vector created by -q had a direction of 270º.
    --> I obtained the direction of the vector created by q2 by drawing a right triangle with leg lengths 1 and 2 and solving for the angle opposite the side with length 2. The leg lengths were obtained from the given data that the sides of the rectangle are d and 2d.
    --> At this point, I realized I was completely on the wrong track. I was planning on solving for the resultant of these three vectors, but I realized that it was already given in the problem that the resultant is, in effect, zero. Thus, in my line of thought, the resultant of the vectors produced by q1 and q2 (the positive charges) must be equal in magnitude and opposite in direction of the vector produced by -q. However, because the two vectors produced by these positive charges are at 0º and ~63.43º, they cannot produce a resultant at 90º, which would be needed in order to have the opposite direction of the vector created by then negative charge. Thus, to me, the problem appears impossible, unless q1 or q2 were allowed to be negative, which I don't believe they are.

    Thanks a ton for any help you can provide!
  2. jcsd
  3. Feb 26, 2008 #2


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    I think the way to go with this is to think in terms of vector components. Since the charge in the upper left only contributes along the x axis and the one in the lower right contributes along the y axis then the E-field from the other charge must cancel both these components for the E-field to be zero.
  4. Feb 26, 2008 #3
    Alright, well it turns out I wasn't the only one with a problem on this one.
    We went over this problem at the very end of the class period today, and the bell rang before we had a chance to finish it, and given the nature of the class and teacher I doubt we'll be coming back to it (much to my disappointment).

    I did as you suggested and took the vector components. However, given the data, I was only able to obtain a ratio of the forces exerted on T.
    I came up with something to the effect of: for every 1 unit of force exerted by -q on a test charge at point T (again, that's the upper-right hand corner, where the net electric field is zero), charge q1 exerts about .8 units of force, and charge q2 exerts about .4 units of force. (I'm not looking at the problem currently, so I may have mixed up q1 and q2, but I'm pretty sure the ratio is close to correct (sorry I couldn't remember more numbers after the decimal))

    Now, the trouble is going from this conclusion to the data that the question asks for. Given that these are the ratios of the forces exerted and knowing the relative distances between all four of the points, what are the magnitudes of q1 and q2 in terms of q (and, I assume, d)?
  5. Feb 26, 2008 #4


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    If you show your work it'll be easier to see where you're making a mistake. Although if you do follow my advice it should be pretty easy to come up with an answer.
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