1why doesnt energy conservation work sir?
Figure shows a comfortable electrostatic configuration, consisting of four conductors with charges ±Q situated so that the plusses are near the minuses.
What happens if we join them via conducting wires as shown.
Will the system remain as it is because of electrostatic attraction. Or will the charges cancel?
Justify your answer.
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18 Answers
9thanks ive realised the mistake
but still isnt V =F(Q,r ) ??
66Who said that the charge density does not matter for potential. That will be true only if all the charges are located at the same distance from the point where we are calculating the potential. As a counter example, suppose the charge density on the surface of a sphere varies as σ = σ0 cos θ, where θ is the polar angle. According to you, the potential should be zero (since the total charge is zero) everywhere. But that's not true.
9but still wont
V =kQ/r ( charge density wont matter in potential )
66you cannot use that formula. The point is that, charge density on the four spheres won't be uniform because of induction effects, and you cannot treat them as point charges.
9sir addition of potentials bcom 0
V =kQ/r
i ve taken kQ out as common
66@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.
9oh 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
66The 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 Ï, 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.
1sir should I answer it or leave it for others to try , you had already discussed it in class?
66@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.
9i can definitely say the charges wont cancel because then PE=0
whereas PE initial is -ve
but wat abt the + ve becoming more +ve , isnt there a possibility of that sir ??
9adding wires doesnt make any difference in attraction of charges
(they just provide a strict pathway)
so in a sense if they are initially stable they remain so even after connecting with wires
66@rajat let the others try for some time...
@celestine, justification?
9ans 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 ???