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Electostatistics- non/conducting sphere/sheet in E field

  1. Aug 4, 2009 #1

    I was reading Griffiths- Intro to electrodynamics, and I came cross some difficulties in understanding some cases which are not covered in the book well. I really appreciate your insight.

    Here are my questions;

    1) If we bring a charge near a grounded infinite metal sheet, it will induce charge (we can find the potential using images method) . Now, can we say that the E field on the other side of the sheet is zero? basically does this sheet shield the E field on one side produced by the charge on the other side? It is obvious the div of V is non zero for the region that charge does not exist, but this solution is not for that region.

    2) Under a uniform ext E field, if we place a non conducting sphere with total charge Q, then I would expect it to move along the direction of the field, is that right?

    3) What happens if this is a metal sphere with total charge Q? will it also move? I know that it will redistribute its charge such that it will cancel the E field inside, but does that affect the over all motion?

    4) How can be the E field inside a metal sphere be still zero if the ext E field is very strong that there is no charge to balance the field?

    5) If the sphere is grounded, will it still make E field inside zero? and can it hold against any amount of ext E field? what happens if the field is very strong?
  2. jcsd
  3. Aug 4, 2009 #2
    :) Well, this was neither a homework, nor a course work question, but just curiosity one.
  4. Aug 4, 2009 #3


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    1. If there are no charges on the other side of the conducting plane and the plane is infinite, the electrostatic potential on the other side will be zero. Proof: Zero is a solution to Laplace's Eqn. and this solution matches all the boundary conditions, therefore by the power vested in us by the Uniqueness Theorem it is the solution. What the conducting plate essentially does is to bring electrostatic "infinity" (where the potential is zero) near the point charge.
    2. Correct.
    3. It will not move in a uniform E field. There will be an induced dipole moment on the sphere and dipoles experience no net force in uniform E-fields.
    4. There are gadzillions of free charges that will always move around to cancel the E field inside a conductor. Think of it this way: If you create an electric field inside a conductor, the free charge carriers will experience an electric force and start moving ... and keep on moving until they have no more reason to move, i.e. until the electric field inside the conductor is zero. If the electric field is extremely strong you will start ripping electrons off the conductor's surface. This is known as field emission.
    5. Grounding means that you provide a conducting path to an infinite reservoir of electrons (the Earth) that sits at zero potential. So grounding is a way to specify that the potential of a grounded conductor is zero no matter what else is happening around it.
  5. Aug 5, 2009 #4
    Thank you for explanations.

    For #1, I thought as the potential difference, at inifinity potential is zero, and on the surface of the sheet it is also zero. So [tex]\int_{\infty}^{Sheet} \vec E \cdot d\vec l =V(sheet)-V(\infty)=0[/tex]. Since the solution has to be unique, regardless of the path chosen, this integral holds and therefore E field must be zero. I am not sure if it is reasonable to approach like this,though.

    Your way- approaching from Laplace is a much solid way. Since, in Laplace eq, extremes only occur at the boundaries, which are zero in this case, the potential has to be zero too, and so does the E field. It makes more sense now.

    Regarding #3, I am little bit confused. The metal sphere is connected to nowhere, so the total charge has to be conserved, and it is +Q. I would understand if the sphere was neutral, then E field creates a dipole like -q <--> q, which remains still in E field (except the dipole rotates). But you have extra +Q charge on the sphere. Regardless of its distribution, I would expect it to be positively charged, which should cause a motion along the E field, while cancels the E field inside.

    Is there an easy way to prove the charge distribution of a metal sphere with total charge +Q under a uniform E field?
  6. Aug 5, 2009 #5


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    You are correct for 3. If there is net charge on the conducting sphere, there will be a net electric force. I must have had a mental lapse. Sorry about that.

    I am not sure I understand what you mean by "prove the charge distribution." If you mean "find what it is", yes there is a way. First find the potential outside the sphere as if the sphere has no charge. This is something that is done in many (if not all) textbooks as an example. Note that this potential is zero on the sphere (it can't be zero at infinity because the uniform electric field extends all the way out there - an artifact of the problem.) Add a new term to your potential (what could it conceivably be?) to take into account that there is charge Q on the sphere. Derive the surface charge density from this potential. Prove that the new potential is a solution to Laplace's equation and satisfies all the boundary conditions. Note: part of this step involves integrating the surface charge density over the sphere, that is why you need to find the charge density first. Also note that putting charge on the sphere is equivalent to asking "Suppose I connect the sphere to a battery that raises the potential of the sphere to V0. What is the charge Q that is added to the sphere?"
  7. Aug 7, 2009 #6
    Thank you for your reply. I'll work on the potential one.
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