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This is the problem given in Purcell and Morin's electricity and magnetism book. (Problem 3.2 If you have the book)

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Figure:

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I was able to figure out that the charges on A,B,C,D cant be held after connecting C and D because if you consider a loop from A to C and back to A through the wires, for a static electric field, from Kirchhoff's loop rule, $$\oint \vec{E}\cdot d\vec{l}=0$$

If the wires have no resistance, then the potential drops across A-C and B-D will add up and wont be zero. So the charges will disappear as soon as C and D are connected so as to make $\oint \vec{E}\cdot d\vec{l}=0$

But in what way will they cancel out? And I was not able to understand this (stated in the solution):

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Won't induced charges have equal magnitude?

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  • $\begingroup$ I have followed the homework guidelines. My doubt is about a specific concept. I have also showed my attempt. $\endgroup$ – Aditya Dev Mar 11 '16 at 18:25
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The picture in the book does not show all field lines. There will be some field lines from the positively charged sphere B to the negatively charged sphere A, also from the positively charged sphere C to the negatively charged sphere D. This can be seen from the superposition principle (electrostatics is all linear so superposition of fields and charges generally applies). Imagine there are only spheres A and B, then certainly there will be some electric field lines from B to A. The meaning of the puzzling comment from the book is that sphere A interacts with both spheres C and B. Interaction between A and D, and between C and B does not result in strong vertical component of electric field, can be neglected in this reasoning.

Here is an illustrative plot of distribution of electric potential, also showing some electric field lines, calculated with the standard formula, from four point charges, charged as in the problem statement. We can see the vertical electric field, upward on the left and downward on the right.

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So, we conclude that in this setup there is some electric field from B to A and from C to D. As soon as we add the connecting wires this electric field will push the charges in the wires until all charges on the spheres cancel out.

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I think the approach you should take is to show that the above situation is an unstable equilibrium. Consider what will happen if sphere A has a small excess of negative charge. This small excess will then move to sphere B, reducing its positive charge. This will lead to sphere D having a small excess of negative charge, which will then lead to it moving to sphere C, reducing its positive charge. Then sphere A will have a small excess of negative charge all over again. This will continue until the charges have been neutralized.

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  • $\begingroup$ If it is an equilibrium it must be unstable because there is more potential energy in the system than in the neutral case. But how do we know it is an equilibrium? $\endgroup$ – Maxim Umansky Feb 21 '17 at 5:57
  • $\begingroup$ Not necessarily. It could be in a metastable state (i.e., a local minimum, but not a global minimum). You know that it's in equilibrium because A and B were in equilibrium before C and D were connected. By symmetry, that means that C and D are in equilibrium after they are connected. $\endgroup$ – J. O'Brien Antognini Feb 21 '17 at 21:39
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Assuming charges on C induces same amount of charges on A until C and D are connected.

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Once C and D are connected, since charges on C are more closed to each other than "the distance between charges on C and charges on A", hence the self-repulsing force on C win;

Hence then the self-repulsing force decrease the amount of charges on C; and then that causing the induced charges on A decrease respectively.

The process goes on and on till there is no net charges everywhere.

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