1
$\begingroup$

Suppose we have two neutral charges connected by a conductor line. Now, we bring together two spheres with oppositely charges, and approximate it to our original system.

enter image description here

Now, imagine that we connect C and D by another line/wire, what exactly will happens?

I tried to answer this question as follow: Since A and B have the same potential, we can call it $\phi$. Now, The potential of C near the sphere A is $\phi c>\phi$, and the D potential near the B sphere is $\phi d < \phi$, so that we have $\phi c > \phi d$. But, since we are dealing with conductors, $\phi c$ and $\phi d$ are the potentials throughout (constant through it) all the surface C and D, respectively.

When we connect C and D using a wire, since C has greater potential, the electric field in the wire will be directed down, from C to D. In Such way that, since electrons flow in the opposite direction of the electric field, they will rise up through the wire. So negative charge will flow from D to C.

enter image description here

But apparently, the answer is the opposite of the mine, the charge will goes down the wire. What is wrong with my attempt?

$\endgroup$

1 Answer 1

0
$\begingroup$

You missed a very important detail while calculating the potential of the identical spheres C and D.

Initially they are neutral meaning they have equal number of oppo charged particles.

Lets consider of C for now (argument with D is exactly the same), You bring the Sphere C close to A. What happens now? Charges in Sphere C will rearrange, but how? Such that some positive charges are closer to A. But since the C is neutral, the an equal amount of negative charges will move as well! Hence in the electrostatic condition, what you have is some positive charges on the surface towards A, while equal amount of opposite charges diametrically opposite. So then take the center of C and find the net potential there, because you have equal and opposite charges at diametrically opposite points, the net potential inside Sphere C is ZERO! (same for D)

Does it sound contradictory? Absolutely no. Initially you had Sphere C at 0 Potential, now when you bring it closer to A the charges start rearraging, why does that happen? Since now Sphere A influences the Sphere C and so there is now a potential inside Sphere C, and the charges in C move so as to nullify this potential! Hence as we expect the end result would still have zero potential!

Now then if I connect C and D with a wire, what is the direction of current? This situation is no longer ELECTROSTATICS because charges are moving.

The fact that Conductor has the same potential everywhere ONLY holds in electrostatics

Here you not supposed to look at the net Potential, instead look at the local Potential i.e. Potential just about the two points where you connect the wire to the two spheres. You will now find that the local potential near wire at C is NEGATIVE (because of accumulation of electrons) and using similar argument local potential at the wire for D is POSITIVE.

And since electrons flow from negative to positive i.e. from C to D.

$\endgroup$
3
  • $\begingroup$ But as far as I understand the question, C and D are not neutral, they are the ones that have the initial charge on them. A and B are neutral as they are connected by the wire. So according to your answer, charges are induced on C and D by 'neutral' conductors A and B, which as you know is incorrect. $\endgroup$
    – Cluse
    Commented Mar 9, 2021 at 9:47
  • $\begingroup$ Exact, C and D are the charged bodies. $\endgroup$
    – LSS
    Commented Mar 9, 2021 at 10:56
  • $\begingroup$ Ahh sorry... You are right... I got confused between the the charged sources. Well, it that case your argument @LSS is absolutely right. Otherwise there would be a violation of energy conservation. So my guess is you might probably has misinterpreted the question, in which I could suggest you to post the question as it is (with the same wordings) or add the source of the question if its openly available online. $\endgroup$ Commented Mar 9, 2021 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.