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Determine the magnitude and direction of the force problem

  1. Feb 26, 2006 #1
    I had a few questions which I'd appreciate some help with:

    Q3) A charge of 6 mC is placed at each corner of a square .100 m on a side. Determine the magnitude and direction of the force on each charge.

    I essentially found the force on one corner ( by using x and y components )

    Hence, for X = 2.29E7 + 0 + 3.24E7
    For Y = -2.29E7 +3.24E7

    I found the force to be 5.6E7. is this correct?


    Q4) A +4.75 micro C and a -3.55 micro C charge are placed 18.5 cm apart. Where can a third charge be placed so that it experiences no net force?

    not sure how to go about this. Do I say:

    Hence, F/E+F/E+F/E=0 ????
  2. jcsd
  3. Feb 26, 2006 #2


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    For Q3, realize that each of three charges imposes a force on one charge, and the resultant would point parallel with the diagonals of the square.

    In Q4, the force on the third charge must be were the electric potential is zero, or the electric force imposed by the postive charge is equal and opposite to the force imposed by the negative charge.
  4. Feb 26, 2006 #3
    Thats what I've tried doing..Since its a square, one of the charge will be in the x dimension only, the other charge will be only in y dimension and the third charge ( diagonally across ) will have both x and y component.

    I guess I understand what your saying but I do not know how to go about finding the force so as to enable me to get the equal and opposite of that.
    All I know are the charges and distances.. Do I randomly pick a spot and put a charge there and try and see if the sum of the forces = 0 in x and y dimensions?
  5. Feb 26, 2006 #4


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    Your answer to Q3 is correct. The direction is parallel with the diagonal of the square, and since the sign is the same on all charges, the force is repulsive, i.e. pointing out of the square.

    With regard to Q4, the force would be zero on any charge where the electric field would be zero, by definition of [itex]\vec{F}[/itex] = [itex]\vec{E}[/itex] * q.

    Now where would the electric force = 0? [itex]\vec{E}_1[/itex] + [itex]\vec{E}_2[/itex] = 0, . . .
  6. Feb 27, 2006 #5
    So for question 4, i'd set E1+E2+E3=0

    and then plug in F/Q for E?

    and further plug in Kq1q2/r^2 for F?

  7. Feb 27, 2006 #6


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    E1 = kq1/r12

    E2 = kq2/r22

    where r1 and r2 are the distances to the same point, where one is trying to find |E| = 0.

    Also r1 = x1 - x, where x is position, and similarly for charge 2.

    Also, the point has to be on the same line, i.e. colinear with the two charges since E is vector field and any non-colinear point would have a nonzero normal vector component.
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