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The setup is as below:

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The two conducting spheres, of radii $r_1$ and $r_2$ have charges $Q_1$ and $Q_2$. The distance between their centers is AB. BC is perpendicular to AB. The problem asks to determine the potential at the mid-point of the line BC, point P (not shown).

My approach was to determine the distance AP. Then use $V_1=k\frac{Q_1}{AP}$ (where $V_1$ is the potential at P due to $Q_1$) and then the fact that the potential due to $Q_2$ at P is the same as that on its surface, that is $V_2=k\frac{Q_2}{r_2}$.

However, the solution argues that since there is no net field inside the second sphere, the potential at P is the same as the potential at point B, the center of the sphere. And then proceeds to use $V_1=k\frac{Q_1}{AB}$.

My confusion is that since the net field inside the second sphere is 0, what is the reasoning behind $V_1=\frac{Q_1}{AB}$? By this argument, shouldn't the potential at P, due to $Q_1$ be equal to that on the surface of the second sphere (due to $Q_1$)?

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  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – user4552 Oct 13 '19 at 18:33
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By this argument, shouldn't the potential at $P$, due to $Q_1$ be equal to that on the surface of the second sphere (due to $Q_1$)?

Yes, it should. However, in this type of problems is is usually assumed that the spheres are very far away from each other or, in other words, that $AB \gg r_2$. In this way the two spheres don't influence each other so that you don't have to take into account the induction that one sphere exerts on the other. Within this approximation (which is usually explicitly mentioned in the text of the problem), $AB \approx AB - r_2$.

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  • $\begingroup$ I'm afraid this is not the case, as the numerical values given in the question are $AB=4\sqrt{3}$ m,$ r_1=1$ m,$r_2=2$ m. $\endgroup$ – Mrb Oct 14 '19 at 8:38
  • $\begingroup$ @Mrb then the solution you were given is wrong :) $\endgroup$ – lr1985 Oct 14 '19 at 8:39
  • $\begingroup$ So is this not solvable by elementary physics? $\endgroup$ – Mrb Oct 14 '19 at 8:42
  • $\begingroup$ @Mrb no, it is not. Here you can find an idea about how to get a solution, but you'll see that it is a very complicated problem. Usually this calculations are performed numerically. $\endgroup$ – lr1985 Oct 14 '19 at 8:51
  • $\begingroup$ Also, since in a system of charges, we need not find the net field and integrate, but simply sum the potentials due to individual charges, why isn't my initial approach justified? Is this because charged particles don't affect each other but spheres do, when the distance is not large? $\endgroup$ – Mrb Oct 14 '19 at 8:52

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