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Dipole of a charge distribution

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data

    three point charges q,q and -2q are located at (0,-a,a), (0,a,a) and (0,0,-a) respectively. what is the net dipole moment of these charge distribution.?

    2. Relevant equations

    moment p =qd.


    3. The attempt at a solution

    Here we have to consider only the yz plane as x is always 0. when draw we can see the arrangement as three charges situated at the corners of a triangle. buy how can these charges form a dipole. in a dipole charges should be equal and opposite, rite? but none of the pairs formed between these three charges satisfies that requirement. how to solve this problem? help me plz
     
  2. jcsd
  3. Sep 23, 2011 #2

    ehild

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    Consider the -2q charge as two -q charges at the same place. This way, you have two dipoles, one with q charge at (0,-a,a) and -q at (0,0,-a), the other one with q at (0,a,a) and -q at (0,0,-a). Do not forget that the dipole moment is a vector, it points from the negative charge to the positive one. Add them as vectors.


    ehild
     
  4. Sep 24, 2011 #3
    Thank u sir for ur guidance. now i understood how they form a dipole. but still i am not getting the answer. by calculating in the way that u said i got that length of each dipole is (a*square root of 5) and dipole = (q*a*square root of 5). then the resultant dipole should be along Z axis, i.e. along k hat direction and total magnitude should be 2(q*a*square root of 5). but the four options given for the question does not include this result.
     
  5. Sep 24, 2011 #4

    ehild

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    The dipoles are vectors, their components add instead of the lengths. The dipoles in this problem are p1=q(0,-a,2a) and p2=q(0,a,2a) What is their sum?

    ehild
     
  6. Sep 28, 2011 #5
    thank u sir,i got the answer as 4QA(k hat).
    but sir i again got confused with another problem of very similar type. in this case the charges are placed at the corners of a square. Q at (0,0,0), Q at (a,0,0), -2Q at(a,a,0) and Q at (0,a,0). in this case also i think there are only two dipole. considering the -2Q charge at (a,a,0) as two -Q charges we have two dipoles, one between -Q charge at (a,a,0) and Q at (a,0,0) and the second dipole between the other -Q charge at (a,a,0) and Q at (0,a,0).

    in this case i got the answer as -qa(i hat)-qa(j hat). but my friend was saying that answer will be qa(i hat)+qa(j hat) . which is the right answer?
     
  7. Sep 29, 2011 #6

    ehild

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    The dipole moment is a vector, that points from the negative charge to the positive one.

    ehild
     
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