What Are the Monopole and Dipole Moments of a Two Charge System?

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SUMMARY

The discussion focuses on calculating the monopole and dipole moments of a two-charge system consisting of charges 3q and q, separated by a distance a. The monopole moment is correctly derived as q(a + 2) in the k-direction, while the dipole moment is calculated as 2qa in the k-direction, assuming the origin is positioned midway between the charges. Participants clarify the importance of coordinate transformation and the convention of selecting the origin to achieve a zero dipole moment, emphasizing that the dipole moment is indeed a vector quantity.

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Two point charges 3q and iq are spearated by distance a as in the diagram. Find the monopole, dipole moments and the approximate potential at large (in spherical coords including both dipole and monopole contributions)

monopole moment is sum of charges 3q \hat{k} + qa \hat{k} = q(a+2) \hat{k}

dipole moment is (assume that the origin is located half way between the two cahrges

then
p= 3q \frac{a}{2} \hat{k} + (-q}\frac{a}{2} \hat{k} = 2qa \hat{k}

also
\overline{p} = p + qa\hat{k}
\overline{p} = qa \hat{k}
is this correct?

where di i got wrong? with the transofmration of coordinates? With the way i calculated p?? When i calculate the potential the dipole moment will point radially in the z direction hence \hat{k} = \hat{r} \cos\theta [/tex]<br /> <br /> thank you for your help and responses!
 
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stunner5000pt said:
Two point charges 3q and iq are spearated by distance a as in the diagram. Find the monopole, dipole moments and the approximate potential at large (in spherical coords including both dipole and monopole contributions)

monopole moment is sum of charges 2q \hat{k} + qa \hat{k} = q(a+2) \hat{k}

dipole moment is (assume that the origin is located half way between the two cahrges

then
p= 3q \frac{a}{2} \hat{k} + (-q}\frac{a}{2} \hat{k} = 2qa \hat{k}

also
\overline{p} = p + qa\hat{k}
\overline{p} = qa \hat{k}
is this correct?

where di i got wrong? with the transofmration of coordinates? With the way i calculated p?? When i calculate the potential the dipole moment will point radially in the z direction hence \hat{k} = \hat{r} \cos\theta [/tex]<br /> <br /> thank you for your help and responses!
<br /> <br /> 1. You seem to mean q and 2q in the first line. There should be no \hat k and no a in the monopole moment.<br /> 2. The usual convention for a charged object is to pick the origin so that the dipole moment is zero. I have not seen the diagram, so maybe you are told where to pick the origin.
 
oops

here is the diagram

pick a zer dipole moment... how in this case
 

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Meir Achuz said:
1. You seem to mean q and 2q in the first line. There should be no \hat k and no a in the monopole moment.
2. The usual convention for a charged object is to pick the origin so that the dipole moment is zero. I have not seen the diagram, so maybe you are told where to pick the origin.

but isn't hte dipole moment a vector?/
 
stunner5000pt said:
here is the diagram

pick a zer dipole moment... how in this case
For your diagram, the charge is 3q-q=2q.
With the origin given at the postionf of the -q charge, the dipole moment is
{\bf p}=3q{\bf\hat z}.
 

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