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Homework Help: Electric point charges on x-axis

  1. Jun 29, 2007 #1
    1. The problem statement, all variables and given/known data

    There are three charges on the x-axis labled q1 = 9.9 uC, q2= -5.1 uC, q3 = ?
    The distance between q1 at the origin and q2 is 0.1 m, what is the equilibrium distance between q3 and q1, and does that distance depend on the polarity of q3 ?

    2. Relevant equations

    F = (k*|q1|*|q2|) / (r^2), k=8.99E9

    E = F / q

    3. The attempt at a solution

    In this system there are 4 different force vectors, for euilibrium they must cancle out F1+F2+F3+F4= 0.

    q1--------q2--------q3; I don't think that d matters if q3 is + or - even if q3 where between
    q1 and q2 like this
    Last edited by a moderator: Jun 29, 2007
  2. jcsd
  3. Jun 29, 2007 #2


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    Why do you say there are 4 force vectors to consider? I think there are only two. What are they? And why do you put absolute values on the charges? The direction of the force depends on the sign of the charges. Put q1 at x=0 and q2 at x=0.1 and cancel the two forces on q3. There are two different distances from q1 depending on the sign of q3.
  4. Jun 29, 2007 #3


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    You know, thinking it over. You are right. The distance from q1 is independent of the sign of q3. Both forces have to be equal and opposite and reversing the sign of q3 doesn't change that. Can you ever forgive me? I still wonder about the '4 forces' though. And you can definitely say whether q3 is between q1 and q2 or not. Just think about the direction and magnitude of the forces. Can there be equilibrium between??
    Last edited: Jun 29, 2007
  5. Jun 29, 2007 #4
    Is it posible for one point charge to experience a force from another point charge that is directed at it but with another point charge in between the them ?
  6. Jun 30, 2007 #5


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    Yes. Charges don't block the fields of other charges.
  7. Jul 4, 2007 #6
    Sorry for my confusion as well, actually there exist more than 4 force vectors in this system however force vectors of equal magnitude pointing like this ---> <--- must cancle out. In equilibrium must the individual net forces on each charge, q= 0, or are the individual net forces of all the three charges in this system added so that the total F = 0. For example in order for this system to be in equilibrium must the net force on q3 = 0, net force on q2=0 and net force on q1=0 ?


    It is true that q3 is not between q1 and q2, but why can't there be an equillibrium in between q1 and q2 ?
    Last edited by a moderator: Jul 4, 2007
  8. Jul 4, 2007 #7
    Hi force
    Suppose q3 is between q1 and q2. Now if q3 is +ve, then q1 will repel q3 and q2 will attract q3, which means no matter where you put q3 in the region between q1 and q2, q3 will always accelerate towards q2.
    If q3 is -ve, then it will accelerate towards q1. Therefore, there is no point in this region where q3 remains stationary (i.e in equilibrium), due to the fact that electric field is not 0 at any point between q1 and q2.
    The equilibrium position, therefore, will lie either beyond q1 or beyond q2, depending on which charge has greater field. Here q1>q2, therefore Equil will lie at some point beyond q2, like below


    Had q1<q2, then Equil would have been at some point beyaond q1, like below


    In your question, equilibrium will lie beyond q2. Now, distance of q1 from q3 is more, so repulsion reduces. Whereas, distance of q2 from q3 is less, so attraction increases. Now there will be a point C beyond q2 where the attraction will be exactly equal to repulsion, and net field on q3 will be zero.

    That is,
    Field due to q1 = field due to q2

    I think that is enough of a hint.
    I hope my post helped.

    warm regards
    Mr V
    Last edited: Jul 4, 2007
  9. Jul 4, 2007 #8


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    I don't think they are asking that all charges be in equilibrium positions. That's not possible. Just q3. So only worry about the two forces on q3.
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