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Force between charges with two dielectric media

  1. Jan 19, 2016 #1
    Does anyone have a formula and derivation of the force between two charges in the presence of two dielectric media. Specifically, have the two dielectric media separated by a plane at z = 0. I have a developed a solution using image charges to matched the boundary conditions at the dielectric boundary. The force on each charge is then determined by the product of the charge and electric field intensity at the charge location. However, the answer I get does not make physical sense.

    In the case where the charges are on opposite sides of the boundary the forces are equal and opposite as expected. The permitivity is replaced by an equivalent permitivity of E = 2*E1*E2/(E1+E2). My concern is that the equivalent permitivity is not a function of the distances of the charges from the boundary.

    In the case where the charges are on the same side things get even more weird. As the electric field is not directly toward the charges the forces are not equal and opposite. There is a net force toward or away from the boundary depending on the relative magnitude of the permitivities.

    I am having problems creating a simulation of a much more complex structure. However, I traced the problem to this fundamental question. Could anyone point me to a solution to this?

    Thanks
     
  2. jcsd
  3. Jan 21, 2016 #2

    Simon Bridge

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    I am guessing that the "plane" is a conductor (what thickness?) for the method of images to be appropriate.
    Is that correct?

    I think you need to be more explicit about the setup; i.e. are the charges in question equal? Do they both lie on the z axis?
    I'm guessing an orientation so that the "plane" of unspecified properties is in the x-y plane... but you have not said, only that it passes through z=0.

    If the charges are on the same side of the plane - then the method of images gives two extra effective charges right?
    If may be useful to understand what is happening by using another method.
     
  4. Jan 22, 2016 #3
    I will try for a more full explaination and show the analysis. As you expeceted the interface between the dielectrics is in the x-y plane. z=0 at all points on the interface. There is no conductor at the interface. For the image charges I have used a simplified version of the methods found in T. Takashima and R. Ishibashi, "Electric Fields in Dielectric Multi-Layers Calculated by Digital Computer," IEEE Transactions Electrical Insulation, Vol EI-13, No. 1, February 1978. This paper addresses multiple parallel dielectric boundaries and a ground plane. In my case I have only one dielectric.

    Let the space above the interface (##z>0##) have permitivity ##\epsilon_1## and that below the interface have permitivity ##\epsilon_2##. Let charge ##Q_1## be located at (x,y,x) = (0,0,a). The fields in the upper dielectric may be determined by the use of an image charge of magnitude ##\alpha_{12} Q_1## at location (0,0,-a) where ##\alpha_{12} = \frac{\epsilon_1 - \epsilon_2}{\epsilon_1 + \epsilon_2}##. The electric field (##E_1##) in the upper dielectric will the vector sum of the fields from ##Q_1 ## and ## \alpha_{12} Q_1## treated as if both of these charges were in a medium with permitivity ##\epsilon_1##. Note that as ##\epsilon_2 \rightarrow \infty## the boundary becomes a ground plane, ##\alpha_{12} \rightarrow -1## and the image charge becomes ##-Q_1## as expected for a ground plane.

    The electric field in the lower dielectric may be determined by the use of an image charge of magnitude ##\beta_{12} Q_1## located at (0,0,a) where ##\beta_{12} = \frac{2 \epsilon_2}{\epsilon_1 + \epsilon_2}##. In this region the electric field (##E_2##) is a result of only one charge treated as though it were in a medium of permitivity ##\epsilon_2##.

    When these resulting electric fields are evaluated at z=0 the necessary boundary conditions are met. The tangential components of the fields match as ##E_{t1} =E_{t2}##. The normal components match as ##\epsilon_1 E_{n1} = \epsilon_2 E_{n2}##.

    The force on a second charge may be determined by the static portion of the Lorentz force equation. In this case ##\mathbf{F} = Q_2 \mathbf{E}##. If the second charge is in the lower medium the electric field is directed to the location of the first charge. Working through the fields from the charges gives a force magnitude of ##F = \frac{Q_1 Q_2}{4 \pi \epsilon' r^2}## where ##r## if the distance between the charges and ##\epsilon' = \frac{2 \epsilon_1 \epsilon_2}{\epsilon_1+\epsilon_2}##. The force the second charge exerts on the first charge may be determined by the same analysis exchanging the subscripts. The forces on the charges are equal and directed toward or away from the charge location depending on the sign of the charges.

    If the second charge is located in the upper medium along with the first charge then the situation is more complex. As the electric field is the result of two charges (##Q_1## and ## \alpha_{12} Q_1##) the field cannot be directed at ##Q_1##. Therefore, the force has a component toward or away from the interface depending on the charge signs and the relative magnitude of the permitivities. This is the part that bothers me. There appears to be a force with no counter force. If this were to be true I could come up with a device to violate conservation of energy. This is clearly not real.

    I would appreciate any help in showing where the flaw is in my analysis. I hope that this is readable. Perhaps due to the length of the post the preview is not working and I hope that the LaTeX is correct.

    Thanks
     
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