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Find electrostatic energy

  1. Jun 16, 2012 #1
    A solid sphere carries a uniform, surface-charge density. The sphere is surrounded by a linear dielectric with electric susceptibility, κ. For which system is the electrostatic energy the largest.

    a) For the sphere alone.
    b) For the sphere and the dielectric.
    c) Both systems have same electrostatic energy.

    Now I know, because my teacher told us, that the correct answer is a). I don't really understand why though. Can anyone explain? I can see that the dielectric would get polarized in such away that the total polarization is proportional to the total field (since the dielectric is linear). But that's pretty much all. Also, I am confused why it doesn't matter whether the sphere carries a positive or negative charge density.

    Also the answer seems quite weird to me. Suppose you have the sphere first and then surround it by the dielectric. According to the answer the electrostatic energy has now decreased! But where has it gone? Is the energy stored in the polarization and don't we count that or what?
     
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  3. Jun 16, 2012 #2

    TSny

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    Maybe this will help.

    Think of starting with the charged sphere alone.

    Next imagine bringing in from infinity a small cubical chunk of dielectric material and letting it approach the sphere. As you know, the dielectric will become polarized by the electric field of the sphere such that the face of the cube that is nearest the sphere will have an effective surface layer of charge. Is this layer of charge of the same sign or the opposite sign of the charge on the sphere? What about the face of the little cube that is farthest from the sphere?

    These surface charges will create an electric field inside the little cube. What is the direction of this induced field compared to the field of the sphere?

    The net field inside the cube will be the vector sum of the field from the sphere and the field due to the surface charges on the cube. Is the magnitude of the net field inside the cube less than the field due to the sphere alone or greater?

    Therefore, does the presence of the cube increase or decrease the total electrostatic field energy around the sphere compared to without the cube?

    To account for the energy change: Is there an electric force on the cube from the sphere as the cube is brought in from infinity? Is the force attractive or repulsive?

    If you were the agent bringing in the cube at a slow constant speed, would you have to do positive or negative work on the cube?

    What happens to the total energy of a system if an outside agent does negative work on the system?

    You can now imagine bringing in more and more chunks of dielectric until you fill up the space around the sphere with dielectric material.
     
  4. Jun 16, 2012 #3
    1) The field will be like that of a physical dipole. Between the ends of the cube it will point in the opposite direction of the field from the sphere whilst outside it, the field will point in the same direction.
    2) Since the field is in the opposite direction inside the cube the field will be less.
    3) Here I am not quite sure. You seem to hint me at the field inside the cube but I don't know why that would say anything about the energy of the system. I guess you can say that since negative charges assemble on the side facing the sphere, i.e. closer to the sphere than the positive charges, you have effectively lost some energy to thermal energy or something like that since it requires a negative work to bring in charges of sign + to a charge configuration of sign - (or the other way around). But I am not sure.
    I also thought of this whole idea of polarization. Suppose we have an atom in a field. When it gets polarized the electron and proton gets a potential energy relative to each other (the atom gets stretched). But where did this energy come from in the first place?
     
  5. Jun 17, 2012 #4

    TSny

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    It is possible to express the electric potential energy of a system of charges in terms of the integral of the square of the electric field over all regions of space where the field exists. (This is often covered in junior level EM courses.) I was trying to relate this to your question. But I wasn’t very clear. (I think that there are some subtleties here that have to do with distinguishing between the “microscopic” electric fields in the dielectric and the averaged out “macroscopic” field in the dielectric.)

    So, instead, maybe we can try another approach that might also relate to your question about the polarization of a single molecule. Imagine placing a molecule in an external electric field. As the charges in the molecule separate due to the forces of the external field, there is an increase in the potential energy that is associated with the attraction between the opposite charges in the molecule. However, we shouldn't overlook the fact that there is also a decrease in the potential energy of the charges associated with their movement in the external field. Overall, the total change in potential energy of the charges in the molecule turns out to be negative, as can be seen as follows.

    The reason the charges of the molecule separate from one another when placed in the external field is because the force on each charge due to the external field is stronger than the attractive force between the charges (until the charges reach their new equilibrium polarized configuration where the net force becomes zero on each charge). While becoming polarized, each charge in the molecule moves in the direction of the net electric force acting on it. But when a charge moves in the direction of the net electric force acting on it, it’s electric potential energy decreases. So, as the charges move to their polarized equilibrium position, the total electric potential energy of the charges decreases.

    This is in agreement with the general principle that particles acted on by conservative forces will tend to move toward positions of lower potential energy.
     
  6. Jun 17, 2012 #5
    Okay, so we have that the total potential energy of the cube decreases or simply gets stored in another location.
    And since we are effectively moving a bunch of charges of the sign + closer to a charge distribution of sign - than are we the bunch of charges with the same sign then I can see that we have lost energy.
    There's one thing I dont understand though. Say we place a neutral atom in empty space - effectively nothing. We then place an electron next to it whose field will polarize the atom. Now the the electron and proton have a potential energy associated with the field. Isn't that like energy has been created? Well maybe the question it silly, because my mistake is perhaps that you can't just see a + and a - as effectively nothing. Or what?
    And one last thing (you've been quite helpful) - Suppose we put the dielectric around a neutral sphere first and then add the charge to the inner sphere after that. Some energy must be gained since + charges have moved away and negative charges have moved forward - what does that energy go to?
     
  7. Jun 17, 2012 #6

    TSny

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    Let your “atom” consisting of a proton and electron be sitting alone in space and then we slowly bring in an electron from infinity until it’s near the atom. We can assume that the atom is held in place by some external force, but the proton and electron in the atom are free to move so that the atom can become polarized.

    When talking about potential energy of the “system”, we need to define the system. You could take the atom alone as the system. Then the electron that we bring nearby is to be considered as an external agent that exerts external forces on the particles in the system (i.e., the atom). These external forces cause the particles in the atom to separate. The potential energy of the atom alone then increases due to the work done by the external forces while separating the particles of the atom.

    Or, you could take the “system” to be the atom and the electron that is brought in (i.e., the entire system of charges). Then the potential energy of the system is the total potential energy of all three particles (the proton and electron of the atom as well as the electron brought in). It will now turn out that the potential energy of this 3-particle system decreases as the electron is brought in. This decrease in potential energy of the system is accounted for by the negative work that we did on part of the system (namely the electron that was brought in). As we move the electron toward the atom and the atom starts to become polarized, the atom will exert an attractive force on the electron. Thus, we have to “pull back” on the electron to prevent it from accelerating. Our force is opposite to the displacement. So, we do negative work on the 3-particle system. Or, you could say that the 3-particle system does positive work on us as we bring in the electron (because the system exerts a force on us in the direction of our displacement). So, the 3-particle system “uses up” some potential energy in order to do positive work on us.

    I didn’t quite understand your question about charging the sphere with the dielectric already in place. If you bring in positive charge and place it on the sphere, you will have to do some work. The work equals the increase in electrical potential energy of the system. You will find that you would do less work with the dielectric in place compared to adding the same charge to the sphere when there is no dielectric. So, the potential energy of the system with the dielectric is less than the potential energy without the dielectric. It is similar to charging up a parallel plate capacitor with a specified amount of charge. Less work is required to charge it with a dielectric between the plates compared to charging it without a dielectric between the plates.
     
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