When we calculate the electrostatic potential energy for discrete point charges we make sure that while adding potential energy for individual charges we don't take the same charge and square it up by declaring $i$ not equal to $j$. But when we measure potential energy for continuous charge distribution,we don't avoid self interaction. (at least in the formula). Why?**
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Qmechanic
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2Please format equations using MathJax instead of uploading a sideways picture. – BioPhysicist Feb 04 '19 at 17:12
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4Possible duplicate of Confusion about calculating electrostatic energy using the electric field – Michael Seifert Feb 04 '19 at 17:25
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Consider what an integral is. One way of describing an integral is to take some level of resolution, say $r$, chop the region over which the integral is being calculated into subregions with side lengths less than or equal to $r$, calculate the values for each subregion, and then add them up. Then take the limit as $r$ goes to zero.
If we have a resolution of $r$, then the proportion of points that are within $r$ of each other will go as $r^3$, while the potential energy due to any particular pair will go as $1/r$. Thus the total energy of pairs within $r$ of each other will go as $r^2$. So when we let $r$ go to zero, the self-interaction energy goes to zero as well.
For point particles, on the other hand, once $r$ gets to be smaller than the smallest separation between the points, the proportion of points that are within $r$ of each other is constant, leaving the $1/r$ to go to infinity as $r$ goes to zero.
Thus, we need to worry about self-interaction for point charges because their self-interaction term diverges, but we don't need to worry about it for continuous distributions because it converges to zero as our resolution goes to zero.
Acccumulation
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More precisely, we do not need to worry for volume or surface distributions of charge, but we do have a problem for linear or point distributions. – Ján Lalinský Feb 04 '19 at 21:22
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@AaronStevens the problem is with infinite Coulomb energy of such line. – Ján Lalinský Feb 06 '19 at 13:19
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Electrostatic potential energy is defined as the work that is required to either assemble or disassemble a group of charges and take them to infinite separation ensuring no change in kinetic energy. If I take an example of two spheres and am interested in finding the Electrostatic potential energy of my system. One way of doing this is, imagining the two spheres to be at infinite separation initially. We are depositing charge on the surface of the sphere and everytime we do so, we are doing extra work against the charges already present on the sphere. The same applies to the other sphere as well. Once we have charged our spheres and bring them such that they are under a mutual force of attraction or repulsion, we are also doing work against the charges on the other sphere. Thus the total Electrostatic potential energy becomes the sum of self energy and mutual energy as you have said. In case of point objects, the only work required is the work that needs to be put in to bring the charges to some separation and thus does not include self energy terms as a point charge can be considered to be like a single charge. Hope this answers your doubt
Aditya Sriram
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