A dipole is a collection of two oppositely charged particles held at some distance, but if two charges are unequally charged (but oppositely charged) how do we take the dipole between them? Do we only take the parts of charge that are equal?

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    $\begingroup$ You could consider it as a combination of a dipole and a monopole (single charge). $\endgroup$ – user253751 Jun 8 '18 at 11:22
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    $\begingroup$ This is related to the en.wikipedia.org/wiki/Multipole_expansion $\endgroup$ – Luke Jun 8 '18 at 11:49
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    $\begingroup$ I guess we could take the part of the bigger charge that we need to create a dipole and treat the rest of it as another charge. $\endgroup$ – Anurag Jun 8 '18 at 12:40

Such a distribution of charges will also have a dipole moment and therefore a non-zero dipole term in the potential of the electric field produced (along with higher order pole terms). Just use the definition of the dipole moment to compute it: $$\vec{p}(\vec{r}) = \sum_i q_i (\vec{r}_i-\vec{r})$$

As people have pointed out in the comments, thanks to the principle of superposition you could treat any problem as having a perfect dipole (equal but opposite charges) plus a monopole. So that the potential is obtained by adding both contributions.

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  • $\begingroup$ Is it necessary to calculate potential by including dipole moment, or will we get the same result if we add the potential just due to individual charges. Also can you explain what are higher order pole terms ? $\endgroup$ – karun mathews Jun 10 '18 at 17:46
  • $\begingroup$ You will get a more exact result if you consider the individual charges even, since dipole refers to just one term of the multipole expansion. $\endgroup$ – ohneVal Jun 10 '18 at 18:42

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