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I Potential inside a charged conductor

  1. May 27, 2017 #1
    As shown in figure, we know that at the sharp end E field is strong, and at the other end field is weak.
    Inside the conductor E=0. that means V=constant. on its surface and throughout the conductor.

    if we move unit positive charge from B to A or P to Q (as shown in fig.) then different amount of work will have to do to bring this charge (because field intensity is different at both the ends).
    (assumed distance between PQ = distance AB)

    According to the definition of Potential,
    Work done (against the field ) to bring unit positive charge from infinite distance to point in the electric field is called electric potential at that point.

    Here, Work done will be more to move B to A rather than P to Q because field is strong(at B to A) therefore more work will to do.

    So the potential is different at point A and at point Q?(it is not the same throughout the conductor.)
    If i wrong please correct me...
     

    Attached Files:

  2. jcsd
  3. May 27, 2017 #2
    To use a physical analogy, the potential of this whole setup looks similar to a Mesa:

    floatingmesa4.jpg

    The flat top is the inside of the conductor. As you point out correctly, the lack of electric field inside means the potential is the same in that whole area.

    Your question is regarding the slopes. Just like in the picture above, it doesn't matter, in terms of potential energy, whether a ball gets pushed up the left (more shallow) slope, or the more steep right slope. The end result is exactly the same. It's exactly the same with a unit charge that gets pushed towards the conductor. The weak field will extend further out, so in the end you expend exactly the same energy as pushing it from the pointy side where the field is strong (but doesn't extend as far).
     
  4. May 27, 2017 #3

    Dale

    Staff: Mentor

    The drawing is not quite clear. Are A and Q inside the conductor or outside?
     
  5. May 27, 2017 #4
    They are just on the outer surface of the conductor.
     
  6. May 27, 2017 #5

    Dale

    Staff: Mentor

    Then they are not equipotential, I believe.
     
  7. May 27, 2017 #6
    But when you put the charges on the conductor, the charges distributed on the surface of conductor such that it acquires the same potential ( charges moves on the surface from higher potential to lower potential until whole surface becomes equipotential.)
     
  8. May 27, 2017 #7

    Dale

    Staff: Mentor

    Just inside the surface is equipotential, but just outside the surface is not equipotential. I am unsure about actually on the surface.
     
  9. May 27, 2017 #8
    As point A and Q are just outside(A and Q are very close to surface) the surface , the potentials are not same. But as we move on surface potential becomes equal.
    I don't understand this phenomena.
     
  10. May 27, 2017 #9
    But here on both sides gravitational field is same and same height will have to covered. Means same energy used.
    But here on both sides have different electric field intensity means different amount of work will have to do.
     
  11. May 27, 2017 #10

    Dale

    Staff: Mentor

    Yes, if they are just outside then they are not necessarily equipotential.

    Think about the excellent analogy from @rumborak, but make it a little more extreme. On the left let there be a gentle slope and on the right let there be a steep cliff. If you step 10 cm off the left then you gain a little KE and maybe stumble. If you step 10 cm off the right then you gain a lot of KE and maybe die.
     
  12. May 27, 2017 #11
    I don't see your problem. Points A and Q may have the same potential (if they are on the surface) but P and B have definitely different potentials. So the potential difference between B and A is not the same as between P and Q. So the work done in the two situations is different even though the distance is the same. There is no contradiction.
     
  13. May 29, 2017 #12
    According to @rumborak , The potential at points P and B may have different. But the potential difference between B & A and P & Q should be same. because the work done in the two situations is same.
     
  14. May 29, 2017 #13

    Dale

    Staff: Mentor

    No, this is not correct. The work done would be different.

    Go back to the definition of work and check if something is different for the two paths.
     
  15. May 29, 2017 #14
    Remember, the absolute value of the potential is arbitrary. Defining infinitely far away to be V=0 ist convenient in many cases, but it's nothing but a choice that has no influence on the setup. You could just as well define the center of the conductor to be V=0.

    What matters in terms of work and energy is the potential *difference* between A and B (and P and Q), just like in the Mesa picture, where clearly it's easier to move a ball 5m from the left to the top than 5m from the right. If you bring the charges in all the way from infinity (and I thought that was initially the setup), then you'd expend the same amount of work. But these small segments have different amounts of work required.
     
  16. Jun 29, 2017 #15
    Work done :
    W(Infinity to P)+W(P to Q) = W(infinity to B) + W(B to A)
    I don't see any contradiction.
     
  17. Jun 29, 2017 #16
    Your equation must be true if the potential at Q and A are the same. But since W(P to Q) < W (B to A) as the field between B and A is stronger than the field between P and Q, it implies W(∞ to P) > W(∞ to B). Can you provide a reason for this?
     
  18. Jul 1, 2017 #17
    Infinity in case of potential is arbitrary (V=0). So i guess Workdone in bringing to P and B might be different.
    E= -(dV/dr) so potential will drop differently in both directions, making V=0 at different 'infinities'.
    (Note: Don't take infinity literally).

    Correct me if I'm wrong.
     
  19. Jul 1, 2017 #18
    Yes, it must be that W(∞ to P) > W(∞ to B) but I was just trying to see if someone could come up with a hand-waving argument for what Rumborak said: "The weak field will extend further out, so in the end you expend exactly the same energy as pushing it from the pointy side where the field is strong (but doesn't extend as far)."
     
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