Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electrical potential and surface curvature

  1. Oct 11, 2003 #1
    Electrical potential is constant everyplace on the surface of a charged conductor.

    Also, on the surface of an irregularly-shaped conductor, the charge density is high in convex regions with small radius of curvature (especially, for example, at sharp points), and low in regions of large radius of curvature, and therefore, by Gauss's Law, the field intensity must be high just above a sharp point, and relatively low just above a gently curved region.

    So, does it follow that an equipotential surface located just an INFINITESIMAL distance outside the surface of the conductor will be MUCH closer to the surface above a gently curved region than it is in the vicinity of a sharp point, even though the potential is constant everyplace on the surface?

    Or, even more puzzling, if the field intensity is very high near a sharp point, would the potential near the point, just above the surface, be higher than the potential at the surface? But how can that be? What would the equipotential surface in that region look like?
    Last edited: Oct 11, 2003
  2. jcsd
  3. Oct 11, 2003 #2
    My E-M professor at GWU regaled us with the phenomenon of a spline - a very sharp metal point that spontaneously breaks off due to the high electic field at such a cusp. I suppose the situation you mention with the equipotential assumes perfect, not practical, conductance.
  4. Oct 11, 2003 #3
    This is not a quiz. I don't know the answer, and I don't know if this would happen in theory or in reality. These questions just came to mind while I was studying & I thought that someone here might enlighten me.
  5. Oct 11, 2003 #4


    User Avatar
    Science Advisor

    Think again. It's just the opposite. The electric field is high near a point, means dV/dz is high (V is potential, z is distance from metal), so for a given ΔV, Δz is SMALL, not large near a point.

    Don't know what you mean. If you have a piece of metal at a potential of 100 volts, infinity is ground (0 volts), then the potential falls monotonically to zero as you go away from the surface of the metal. It falls more quickly going away from a point on the surface than going away from a smooth part.
  6. Oct 11, 2003 #5
    Are you sure? I'm still not seeing it that way. Consider a charged conductor with point P on the surface at the tip of a sharp projecton, and point Q on the surface in a region with a large radius of curvature. So the charge density at P is very high, and at Q is relatively low. Now, take a point A outside the conductor but near point P, and B outside the conductor but near Q. And require that VA=VB (with V=0 at infinity). I'm saying that for any given V, since the field near P is stronger than the field near Q, the distance BQ must be less than distance AP. Isn't that correct?

    To clarify the second question:

    I read in my textbook that:
    "the surface of any charged conductor in electrostatic equilibrium is an equipotential surface."
    "...the electric field is large near convex points having small radii of curvature and reaches very high values at sharp points."
    but they don't give any numbers to quantify that statement.
    Now, these points don't affect the potential of the conductor, since the potential is the same everyplace in the conductor. But what about the potential just outside the conductor, near a "pointy" spot. Can the potential there be GREATER than the potential on the surface of the conductor?
  7. Oct 11, 2003 #6
    I think:
    2)No-We would have developed perpetual motion much sooner if that were true, but I may be missunderstanding, how can the potential at a point outside the potential source be higher than the source when the potential at this point is caused by/from this source? Unless... are you comparing the charge density on a surface with the charge density of a space? Which would be comparing V/m^2 to V/m^3, which doesn't make sense to me. Apples or Oranges anyone?
  8. Oct 11, 2003 #7
    Let me try to quantify this. Take a solid metal charged sphere of radius R and total positive charge Q. The potential throughout the interior and surface of the sphere is V0 = keQ/R relative to infinity. The potential just outside the sphere, say h above the surface, is V = keQ/(R+h). And as h -> 0, V -> V0. So, at a tiny, tiny height above the surface, the potential is essentially the same as on the surface. Or to put it another way, we can make the difference V-V0 as small as we want, just by making h small enough. I think we can all agree on that, right?

    OK. Now let's put a tiny, very sharp point someplace on the surface of the sphere. The total charge Q is the same. I don't know, but I guess that the potential on the sphere is the same. Can anybody confirm or contradict that? But if that is true, and if the charge density is "very high" near the sharp point, so that the field just outside the sphere near the point is much more intense than the mean strength of the field around the sphere, doesn't it follow that the potential just outside the point is going to be higher than the potential inside the sphere?

    If not, why not?
  9. Oct 11, 2003 #8
    Oh, I did misunderstand, I thought the sphere was hollow.
    In that case, I still think:
    2) No-same reason (I don't think a solid or hollow sphere matters for my 'proof')
    I do think that the charge density of the field just outside the point will be higher than the charge density of the field just outside any other place on the sphere, but I just don't see how the field could be stronger than the source given any topology change. If you do an experiment like this and find that it does (or if someone here says they're sure it does) then I think that would allow for perpetual motion, since you could use energy to make a source potential, but you can change the topology of the field with the sharp point and extract energy from the higher field caused by it. I think one would do this by getting a charge on another object through induction by being near this higher potential spot and then discharging this newly charged object through a load.
  10. Oct 12, 2003 #9
    Having read your disclaimer, I'm glad that you describe your proof as a 'proof'.

    I don't claim to have done an experiment to show this. I'm merely wondering about it. But your "proof" doesn't prove anything. First of all, electrical potential is not energy. It is a convenient tool for computing the amount of work that would be required to move a charge to a particular point. Having an extremely high potential doesn't mean that any additional CHARGE has magically appeared there. If you brought another (uncharged) conductor into contact with my sphere, you would merely redistribute the original charge between the two conductors so that the surface potentials of the two would be equal. There's no free energy or perpetual motion implied there.

    But none of this proves or disproves what I was saying.
  11. Oct 12, 2003 #10
    I think we're having a conflict of terms. First, electric potential is like water pressure in pipes in which the water is static. But if I route a pipe from the first to a place that has a lower pressure, ie 0 volts, water will flow, releasing the stored energy (I may be misunderstanding, but I always thought electric potential is a form of potential energy like any other kind. Is not gravitational potential merely an object that has been lifted and so has the ability to fall and have kinectic energy?). My point (ignore the perpetual motion stuff, that's just musings) is: How can a source of high potential(the sphere) be the source for a higher potential(the place just outside the pointy part) without energy input?
    Last edited: Oct 12, 2003
  12. Oct 12, 2003 #11


    User Avatar
    Science Advisor

    NO that is not correct. You just re-stated your question. I understood you the first time and gave you the correct explanation. I would explain it again, but I would repeat word for word the same explanation I gave before. Just because the field is large does not mean the potential is large. Field is gradient, or SLOPE.

    Here: I just thought of a 2-dimensional analog. You have a circular plateau of land, with at the edges of the plateau the land dropping sharply away at a 45 degree angle, leveling out to flat at infinity. Now you construct a pole sticking out from the plateau. A piece of cloth is draped over the pole (to provide a continuous surface, smoothly joining the rest of the hillside around the plateau). In this analogy, level is potential, and slope is field. Do you not see that the pole (analogous to the point on the sphere) is at the same level (same potential) as the plateau (sphere)? The fact that the field outside the point is stronger is analogous to the fact that the slope of the cloth away from the point is steeper than 45 degrees. It's steeper because for distances far from the plateau, the potential must tend to zero, and the equipotentials must revert to being circles.
  13. Oct 12, 2003 #12
    Well, obviously krab thinks he's right (and that means me too, though my reasons are crappy, but non-the-less...) and he also seems to know more than both of us since he's both sure and knows the math (you're not sure, I don't know the math), so unless Integral disagrees with him, go with what he says.
    EDIT: PS: krab: If gnome is the fool you seem to think he is, then calm down, because yelling at fools makes them stubborn. To make a horse drink, the horse must first want to drink. Then if you yell at the horse to go drink, it may not drink just to spite you. I think the bible says something to that effect, but it didn't use the example of horses and I don't remember what it actually said.
    Last edited: Oct 12, 2003
  14. Oct 12, 2003 #13
    Thanks krab, now I see where my mistake was. I was thinking that a high field value implies a high potential, rather than seeing the field strength as just the slope of the potential, which should have been obvious since E = dV/ds.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electrical potential and surface curvature
  1. Electric potential (Replies: 3)

  2. Electric Potential (Replies: 2)