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Electric Field and potential

  1. Oct 27, 2005 #1
    These question came on a test of mine recently all one after another so it surprises me how the first and second one as so similar (i think)

    Determine the electric field on the equipotential surface V=0, and verify that it is perpendicular to this surface at all points on this surface

    now since V = 0 , then [itex] \int E ds = 0 [/itex]
    So does this mean that the electric field is zero?? Something with zero potential surely has an electric field...
    Since V = 0 , then E must be perpendicular to the element of the length of this surface ds, thus E is perpendicular to all s o nthsi surrface.
    But is the electric field simply [itex] \frac{q}{\epsilon_{0} A} [/itex]
    Where q is the enclosed charge in this surface?

    Given that the electric field is conservative i.e. E(r) dr = -dV(r)

    Prove that hte electric field lines are perpendicular to the equipotential surfaces at every points on the equipotential surfaces

    not quite sure to do here... is nt the answer from the above question the same as this one??

    Verify that the surraces of conductors (at equilibrium) are equipotential surfaces

    completely clueless on ths ione... does this use the fact taht i = q/t ? But what is meant by the equilibrium state here??

    Than kyou for any help and advice you can offer!
     
    Last edited: Oct 27, 2005
  2. jcsd
  3. Oct 27, 2005 #2

    Physics Monkey

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    stunner5000pt,

    Remember that the zero of potential has no physical meaning, in particular the result
    [tex]
    \int^f_i \vec{E}\cdot d\vec{r} = V_i - V_f
    [/tex]
    is clearly insensitive to constant shifts in both the initial and final potentials. Simply knowing the potential is zero on one surface is not enough to determine the electric field. What other information where you given?

    What you can say in general, without any knowledge of the charge distributution, is that the electric field is perpendicular to any surface of constant potential (why?).

    For the conductors, what does equilibrium usually mean?
     
  4. Oct 27, 2005 #3
    no other info is given
    i typed the question exactly how it appeared in my test
    i'm not quite what you mena to say here..


    DOes equilibrium mean that there is no current flowing through the conductor?
     
  5. Oct 27, 2005 #4

    Physics Monkey

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    Well something doesn't add up on that first question. Simple being told the potential is some value on one surface is insufficient to calculate the electric field. I guess your teacher just wants you to write it in terms of charge enclosed as you've done. If this is true, then your answer to part one is fine though you might want to clean things up a bit. Just remember that the value of the potential has no meaning, only potential differences are defined. The only important thing here is that the potential is constant on the surface.

    You already essentially proved that the electric field is perpendicular to any surface of constant potential in part one of your answer.

    Yes, that is what equilibrium means. No charge is flowing, so what does that tell you about the electric field in the conductor and where the charge on a conductor lives?
     
  6. Oct 27, 2005 #5
    for this conductor then the charge is distributed evnely throughout. E is calculated by Gauss law??
     
  7. Oct 27, 2005 #6

    Physics Monkey

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    If the charge were distributed evenly throughout the conductor, what would the electric field be? If your answer is nonzero then won't the charges see an electric field and move (by definition a conductor is an object where charges are free to move). If charges are moving, the system isn't in equilibrium.

    Here's a hint: like charges repel each other, they want to be as far away from each other as possible. If they are free to move, where will they go?
     
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