How does one find the potential due to two point charges? One charge +q is located at (x,y,d) and the other charge +q is located at (x,y,-d). I attempted to derive the potential by first finding the superposed electric fields and then integrating with the definition of electric potential, but strangely my answer depends on a logarithm instead of the correct answer in the text of Griffiths. I used trigonometric substitution too. Does anyone know what I could have done incorrectly? My answer is close except for that logarithm. Perhaps I chose the reference potential incorrectly? I set V(x,y,0)=0 since this is true from electric field symmetry. Thank you!
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asked Jun 17, 2018 at 2:38
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$\begingroup$ Change the reference to be at infinitely far away. $\endgroup$– AHusainCommented Jun 17, 2018 at 2:53
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$\begingroup$ Do you want to find potential on origin or on some other point? $\endgroup$– vrintleCommented Jun 17, 2018 at 3:01
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$\begingroup$ Any arbitrary point (x,y,z). Also, I just noticed that the choice of reference point does not matter because that reference potential goes to zero when integrating. $\endgroup$– Christina DanielCommented Jun 17, 2018 at 3:03
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$\begingroup$ @ChristinaDaniel, By the superposition principle, the potential due to two point charges on any arbitrary point (x,y,z) is the vector sum of the potential due to each charge. That is V (at x, y, z) = V (at x, y, z due to first) + V (at x, y, z due to second), where V represents vector form of V. Note: Integration isn't required when you are given point charges. $\endgroup$– vrintleCommented Jun 17, 2018 at 3:32
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$\begingroup$ Still confused about that logarithm, but at least I know how to get the right answer now. $\endgroup$– Christina DanielCommented Jun 17, 2018 at 3:48
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1 Answer
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The field is the sum of electrical fields created by each of the charges separately, so the potential is the sum of the potentials created by each of the charges separately, so you don't need to integrate anything, just use the expression for potential in the field of one point charge.
