An easy electrostats

@celestine, how do get the potential at the conductor surface to be zero for that value of r?
And as far as the uniqueness theorem is concerned, that is a theorem. That means, it can be proved rigorously. No exceptions.

oh yes nice one

i initially took them wrongly to be point charges

but then this uniqueness thm that u r talking abt isnt a satisfying explanation

consider an exceptional case when :

1/r = 1/r₁+1/r₂ - 1/√r₁² +r₂²

where r - radius of spherical conductor
r1,r2 - dimensions of rectangle framework

in this case V at conductor surface is 0 only always !!!!

so no movement

The energy conservation is working. What will happen is that the initial energy will be lost as heat (actually some part will also go with the electromagnetic radiations which are bound to occur).

The proof that the result will be what I said takes us deeper into electrostatics. There is a particularly (in fact there are two) uniqueness theorem. The second one of these says that:

In a volume V surrounded by conductors and containing a specified charge density ρ, the electric field is uniquely determined if the total charge on each conductor is given. (The region as a whole can be bounded by another conductor, or else unbounded.)

In the present situation, after we join the balls with wire, there are now effectively two conductors, and the total charge on each is zero. One possible way to distribute zero charge over these conductors is to have no accumulation of charge anywhere, and hence zero field everywhere By the second uniqueness theorem, this must be the solution: The charge will flow down the tiny wires, canceling itself.

However, to answer the given question, we need not go into the uniqueness theorems. It suffices to look at from the point of view of the fact that conductors in electrostatic equilibrium need to be equipotentials.

sir should I answer it or leave it for others to try , you had already discussed it in class?

@celestine, you tried well. Unfortunately, your answer is not correct.
In fact, in this case, the charges will flow down the wire to cancel out and finally we shall have just the four conducting balls without any charge. The reason is that once we connect them, the two balls on the left becomes a single conductor and so do the two balls on the right. So the two balls on the left must be at the same potential while the two balls on the right should again be at the same potential. The charges will thus flow between the left two balls, because no extra effort is needed for the purpose. The same thing goes true for the right balls as well.

@rajat let the others try for some time...
@celestine, justification?

ans is as it is if the charges without wires are already at eq but i think that shud be mentioned in Q ????

does Comfortable mean they are in eq ???

as it is