What Is the Net Dipole Moment of This Three-Point Charge Distribution?

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Homework Help Overview

The problem involves calculating the net dipole moment of a three-point charge distribution consisting of charges q, q, and -2q located at specific coordinates. The original poster expresses confusion regarding the formation of dipoles from these charges, noting that dipoles typically consist of equal and opposite charges.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the arrangement of charges and the conditions under which dipoles can be formed. The original poster questions how the given charges can create a dipole, while others suggest considering the -2q charge as two -q charges to form dipoles.

Discussion Status

Some participants have provided guidance on how to approach the problem by treating the -2q charge as two separate charges. There is an ongoing exploration of the vector nature of dipole moments, with participants attempting to calculate the resultant dipole moment and discussing discrepancies in their results.

Contextual Notes

The discussion includes a comparison to a similar problem involving charges at the corners of a square, where participants are also questioning the correct interpretation of dipole moments and their vector addition.

vrinda mukund
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Homework Statement



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.?

Homework Equations



moment p =qd.


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

ehild
 

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