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Dielectric in a capacitor in equilibrium

  1. Mar 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the situation shown in the figure. The width of each plate is b. the capacitor plates are rigidly clamped in the lab and connected to a battery of emf E. All surfaces are frictionless. Calculate the value of M for which the dielectric will stay in equilibrium.

    2. Relevant equations
    Tension T = mg
    Force due to electric field = tension
    Capacitance = KAε/d

    3. The attempt at a solution
    What I don't get is how the electric field could pull the dielectric. The field is vertical, and the tension is horizontal, so shouldn't the dielectric just slide off? What force could possibly act in the horizontal direction other than tension?
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2016 #2

    gneill

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    Staff: Mentor

    Usually, when looking at the field between the plates of a parallel plate capacitor we assume that the plates are relatively large compared to the plate separation and ignore the "edge" or "fringe" effects at the perimeter of the plates. In fact the field at the plate perimeter is not strictly linear and "bows out" somewhat into the surrounding space. Compared to the essentially uniform field in the rest of the capacitor volume it is a negligible contribution in a practically designed capacitor.

    In this case you see the results of the "edge effects" of two juxtaposed capacitors, one with an air dielectric and the other with some different dielectric. Think of the dielectric slab as having induced charges on its surfaces due to dipole polarization in the electric field. These charges will be attracted to the opposite charges on the nearby plate surface, but also to a lesser extent the charges further along the plate. The result is a net force that tries to pull the dielectric further into the space between the plates.

    Take a look at the potential energy stored in the system (you have two capacitors in parallel with the same applied potential). How does the potential energy stored in the net capacitance change as the dielectric moves into the space between the plates? How is force related to potential energy?
     
  4. Mar 1, 2016 #3
    That makes a lot more sense, thank you.
    I found the net capacitance as
    bε((K-1)x +l) /d
    If I assume the dielectric is at x distance within the capacitor.
    I found U = 1/2 CV^2 and evaluated -dU/DX to get the force on the capacitor at that instant.
    Could you explain how the force was suddenly accounted for in this case by using dU/DX? It is not mentioned that the fringe effects are significant, so I guess we have to make that assumption, but is there anything else special about the situation given, that the force shows up even on differentiating?

    Secondly if I use U = -dU/DX, my mass is correct but it's negative. What is my mistake in thinking here?

    Thanks
     
  5. Mar 1, 2016 #4

    gneill

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    Staff: Mentor

    The underlying assumption is that energy conservation is going to hold for the system as a whole. The work done in moving the slab and thus lifting the mass should have the same magnitude as the change in potential energy stored in the capacitor. The power supply is providing the energy to do work. Thus the magnitude of the change in potential energy with respect to the change in position should yield the magnitude of the force acting.

    So, the bottom line is that after determining the direction of the force on the slab by inspection I looked at the magnitudes of the quantities involved and assumed that the signs would take care of themselves :smile: I relied on the conservation principle to link the magnitude of the capacitor's potential energy change to that of the gravitational PE change. Note that we're looking at a static equilibrium situation, so dealing with force magnitudes is the easy way to go. If the system were to be release from a non-equilibrium position there'd be kinetic energy to deal with too, and oscillations.
     
  6. Mar 1, 2016 #5
    Thanks for the help :)
     
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