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I`ve been looking for this for some time now: I have two thin shells (S1 and S2) of radii R1 and R2, charged negatively C1 and C2 respectively, separated in space by a distance D. What is the force exerted by S1 on S2? (For my purposes, the charge can remain constantly distributed on the shell. Non-conductor, I'm guessing.)

What do I have to read to learn this? I know Coulomb's equation (F=kQ1Q2/d^2). (If relevant, I need this to emulate repulsion in a very crude ion simulation, where the attraction is emulated by considering the ions point charges. Considering the electron shells as points would be great, but obviously doesn't work... that's when the sphere approximation should do the trick.) Thank you in advance.

sammy gerbil
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Jorge Al Najjar
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    For spherically symmetric charge or mass distributions, the force on an external object is the same as though the charge were concentrated at the centre. See Shell Theorem. This can be applied mutually between the 2 shells. – sammy gerbil Mar 11 '17 at 18:55
  • Ah, I understood what you meant when I read the paper :) I mean the electric force, not gravitational :) Maybe the same principles work for both kinds, but considering the outer shell as a point mean they would always repel. If, sya, Na+ and Cl-, I calculate the attraction by making the whole ions points, and get attraction. The shells in this case would be -10 and -18, and they would never attract, but they do, showing my logic is faulty :) – Jorge Al Najjar Mar 11 '17 at 19:01
  • The electrostatic force is $1/r^2$ same as gravity, so the same principle applies, although in your case it is repulsion instead of attraction. – sammy gerbil Mar 11 '17 at 19:22
  • ok... so im doing something very wrong when people tell me to consider them points. See, in my NaCl case, the attraction would be fAttraction=(k) (+e) (-e)/d^2, being +e and -e the charges of the Na and Cl ions. If I use the same approach to the shells, I get fRepulsion=(k) * (-10e) * (-16e)/d^2. Too many electrons, too much repulsion and I have no clue on how this behaves :) – Jorge Al Najjar Mar 11 '17 at 19:36
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    No, considering them as point charges is the right thing to do if the charge distribution is spherically symmetric. However, your question says both shells are charged -ve so they repel. Now you are asking about +ve and -ve spherical ions, which attract. If you are trying to model ions with moving electrons they lose spherical symmetry when brought close. Then considering them as point charges at the centres is only approximate and gets worse as the ions get closer. – sammy gerbil Mar 11 '17 at 19:43
  • aha, as im finding the hard way :D Actually i`m not going down to details as orbital polymorphism or exclusion principle, etc. Solid shells should explain it fine. It is the repulsion before distances are small that I'm interested, and now it is ocurring to me I probably can't really separate them as 2 systems (one attracting and another repelling). Must be something else :) – Jorge Al Najjar Mar 11 '17 at 19:50

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As the comments tried to explain, if you can indeed treat your system as two conducting shells whose charge distribution is not affected by the presence of the other charge, then you can simply consider the charge localized at the center of each sphere, and calculate the Coulomb force in the normal way.

So if you have radii $R_1$ and $R_2$ and a separation $D$, then the distance between the centers is $R_c=R_1+R_2+D$ and the force is

$$F = \frac{C_1 C_2}{4\pi\epsilon_0 R_c^2}$$

You already said

a very crude ion simulation, where the attraction is emulated by considering the ions point charges

It turns out that the Shell theorem (which is valid for any situation where the inverse square law holds) means that, under the assumption that the spherical charge distribution remains uniform, the "point charge at the center" assumption is exactly valid.

Floris
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