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My physic teacher told me that experimental deviations from the predictions of Coulomb's law occur at small separations because, being inverse square, Coulomb's law work best for larger values of r. Why is this the case?

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  • $\begingroup$ In, say, nucleus-on-nucleus scattering (Rutherford scattering), one does deviate from Coulomb's law, but that occurs when you penetrate the nucleus and nuclear physics starts happening. Electron-on-electron scattering shows no such deviations as of yet, which is how we place a (very small) upper limit on the size of the electron. $\endgroup$ – Jon Custer Nov 1 '16 at 20:08
  • $\begingroup$ @JonCuster I am investigating the behaviour of charges spheres of radii 1.9cm. In the case of a 1:1 charge ratio, the experimentally measured force is lower than the force predicted by Coulomb's law at small separations. My teacher explained this was because the law works best at large distances. How would you justify this statement? $\endgroup$ – Maddie Nov 1 '16 at 20:18
  • $\begingroup$ Ahhh... Once they get too close together, they stop looking like point sources at $1/r$ with respect to each other. $\endgroup$ – Jon Custer Nov 1 '16 at 20:30
  • $\begingroup$ @JonCuster could you elaborate on this? $\endgroup$ – Maddie Nov 1 '16 at 20:33
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Almost every source of electric field in our day life is more than an point-like, having complex structure on the distribution of charges. The naive aplication of the Coulumb's law assume that the sourcer is point-like or spherically symmetric. This is a good approximation if the dimensions of the sourcer is neglegible. However, when you get close to the sourcer, the structure (the charge distribution) start to be relevant, and you need to replace the naive Coulomb's law by the Coulumb's law applied in each charge of the distribtuion.

It is important to note that the law is still valid. Is the application of the law that need to be corrected by new inputs, new distribution of charges. This charges can form a continuum too. And applying the Coulomb's law to each piece of the continuum (using calculus) gives the right predictions, with controlled accurancy.

There is a systematic way to approach to corrections of the pointe-like and spherically symmetric approximation, the multipole expansion. The idea is to expand the potential field in terms of $1/r$. The next correction to the $1/r^2$ electric field is the dipole

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  • $\begingroup$ I have used the approach you suggest to create my own model. For charge ratios other than 1:1, the Coulomb's ceases to accurately predict the force because it does not account for induced image charges For a 1:1 charge ratio, however, I am struggling to explain why the exmerimental data shows a weaker force than Coulomb's law predicts, $\endgroup$ – Maddie Nov 1 '16 at 20:25
  • $\begingroup$ Yes, image charges gives deviation, but geometrical irregularities, non-homogeneities, etc. all this give deviations. $\endgroup$ – Nogueira Nov 1 '16 at 20:30
  • $\begingroup$ @Maddie as you get "close" to the source, all this deviations start to shows up. My suggestion, try to carry out this deviations by dipole, quadrupole,....corrections. This is an excellent exercise to learn multipole expansion. $\endgroup$ – Nogueira Nov 1 '16 at 20:34
  • $\begingroup$ In, other words, solve the following exercise: try to estimate the total charge and the dipole moment of your source. And the quadrupole moment?,..so on so on... $\endgroup$ – Nogueira Nov 1 '16 at 20:38

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