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Good Problem

  1. Aug 12, 2004 #1
    Here is this problem if someone could help it will be highly appreciated.
    the answer be
    8sqrt( ((l-a)lmd)/(eAE^2(K-1)))
    Please help?
     

    Attached Files:

  2. jcsd
  3. Aug 12, 2004 #2
    What have u tried so far?

    Usual approach is to give a little push and show that
    F directly proportional to displacement

    -- AI
     
  4. Aug 12, 2004 #3
    Well i Tried to do like this but didnt work
    suppose capacitor has length x of dielectric filled then what is potential energy. then Force acting is = -dU/dx but it gives me constant force which acts on the dielectric but i should have got variable force dependnt on x to show it executes SHM.
    THe other way itried like this
    suppose dielectric moves inside by a distance dx and we apply an external force F so that it moves inside without acceleration the workdone by external force be -Fdx and done by battery is dW= dCV^2
    the total work done is equal to change in potential energy of capacitor but this also give constant force acting on dielectric due to the electric field.

    HELP i have to get it done by monday.
     
    Last edited: Aug 13, 2004
  5. Aug 13, 2004 #4
    I don't know how to do this one either. Dielectrics sounded far too practical to me and I slept through that entire part of the course.

    The dielectric block is released from rest, so there must be some force on it. I know that there is +Q on one side of the capacitor and -Q on the other. In the dielectric, there is -q on one side and +q on the other, but only in the space that is between the plates of the capacitor. This would mean there would be a net force to the left on the dielectric. However, this wouldn't explain a simple harmonic oscilator, since the amount of force would be inversley proportional to the distance the dielectric is away from the center.

    If this is the correct way to get started please let me know and I'll post some calculations as well.
     
  6. Aug 13, 2004 #5
    Well actually i thought about it but i could'nt get it.
    have to think some alternative.
     
  7. Aug 13, 2004 #6
    I am not prolly among the ones that are good at physics around here. I will just add my two cents tho,
    Note that when u are pushing a di-electric material, is there a change in capacitance?
    (Note that what we have is capacitances in parallel) ..

    More directly, if we say that a material is pushed to x distance inside the plates, what is the capacitance of the system? How does it affect the force?

    -- AI
    (P.S -> if anyone finds any errors to what i said , they are welcomed to point it out and correct it)
     
    Last edited: Aug 13, 2004
  8. Aug 13, 2004 #7
    I'm sure there is a change in capacitance, but don't see how that reflects on the force. Educate me.
     
  9. Aug 14, 2004 #8
    The key here is the battery...
    Calculate the work done by the battery (which will be [deltaC*V^2])
    Calculate the field energy of the capacitor
    If my physics is still in good shape,
    Work done by battery must be > field energy of the capacitor

    The rest of the energy is spent in resisting the force pushing the block in.

    See if u can carry on from here .....

    -- AI
    P.S-><same as my earlier post>
     
  10. Aug 14, 2004 #9
    Well this is what i have done and it goves constant force not force dependent on distance which is required for SHM. You can check it.
     
  11. Aug 15, 2004 #10

    Gokul43201

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    Yes ambuj, I think you're right, U seems to go linearly in x, which is not what we want. In fact, I think the force opposes you - tries to push the slab out of the capacitor.

    Anyway, I think the difficulty is because of the poor wording of the problem. I think what really happens is that the capacitor is charged to a potential of E, in the position shown, and then the battery is removed. If the slab is then let go, it will oscillate harmonically.

    The trick is that V also changes so that Q = CV remains constant. Clearly then there will be an x^2 dependence coming from the V^2 term.

    Really, if you find U(x) = (1/2)C(x)V(x)^2 = (1/2)kx^2, that gives you k and hence, T. You don't even have to find F = -dU/dx = kx...it's just redundant.
     
  12. Aug 15, 2004 #11

    ehild

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    The task in the original formulation of the problem is "show that the slab will execute periodic motion and find its time period". And it does not say that the battery is removed.
    A periodic motion is not necessarily harmonic. Consider a bouncing ball for example. I think the slab will move with a constant acceleration till it reaches the end of the capacitor. After that, it will decelerate at the same rate till it stops at a position, which is mirror image of the original, (if that little push at the beginning meant infinitesimal starting speed.) If it turns back the motion will be the same as before only at opposite direction, and the slab returns its original position.

    ehild
     
  13. Aug 15, 2004 #12

    Gokul43201

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    No, if the battery remains in place the force pushes the slab out of the plates if you displace it slightly - there is no question of periodic motion. Remember, U = (1/2)CV^2 is maximum with the slab in (as C increases) and minumum with the slab out. So clearly, the slab will get pushed out.

    I know the question does not say that the battery is removed, and that's why it is WRONG ! But clearly there could have been an error of omission in the formulation of the question.

    I would go ahead and solve it this way if only I knew what ambuj means by 8sqrt( ((l-a)lmd)/(eAE^2(K-1))).

    Ambuj, can you please confirm what these variables refer to (the answer doesn't seem to be dimensionally correct) - and in any case, it is surely suggestive of NON-HARMONIC behavior. So, it is quite perplexing !
     
  14. Aug 16, 2004 #13

    ehild

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    Well, the energy of a capacitor at voltage V is higher with a dielectric than without it. And also, the energy is higher if the capacitor is charged compared to the uncharged state. And in spite of that, the capacitor gets charged when connected to a battery instead of refusing the charges as they increase its energy... Why? :) The original problem shows a capacitor connected to a battery, and no indication of the battery disconnected.

    ehild
     
  15. Aug 16, 2004 #14

    Gokul43201

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    Okay, so YOU prove that the slab will undergo periodic motion if released, and calculate what the period will be.

    As I explained before, I claim there will be no periodic motion because the slab will slide out of the plates and fall on the floor and live there happily ever after. Please tell me how this is wrong.
     
    Last edited: Aug 16, 2004
  16. Aug 16, 2004 #15

    ehild

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    Prove that your statement is true....:) The energy stored in the capacitor is higher if it is filled with a dielectric than without it. This does not mean that the energy of the system battery-capacitor is also higher with the dielectric in.


    I will prove that there is an inward force acting on the slab if it was pushed to move inward at the beginning. And it will continue to move inward with constant acceleration till it reaches the end of the capacitor.

    Problem:
    A plane capacitor is connected to a battery that ensures constant voltage V across its plates.There are no losses, no resistance, no friction. The plates of the capacitor are of length l, width w, and they are d distance apart. The area of one plate is A = l*w. The electric field intensity inside the capacitor is E=V/d.
    A dielectric slab of dimensions l x w x d and relative dielectric constant [tex]\kappa[/tex] is placed so that its front face is inside the capacitor, at distance "a" from the edge. It is pushed a little inward. What happens?

    Consider a small displacement [tex]\delta x [/tex] of the slab inward. The intensity of the electric field, just as the voltage remains unchanged. The capacitance will increase, and so will the energy of the capacitor. At the same time, some free surface charge will disappear because of the dipole-chains built up in the dielectric. To maintain constant voltage, the battery will supply extra charges, but it has to exert work to do this. Moreover, the KE of the slab can change.

    Work of the battery= change of the energy of the capacitor + change of the KE of the slab. [tex]\delta W_B=\delta W_C + \delta KE [/tex].

    The work of the battery is:

    [tex] \delta W_B=\delta Q*V [/tex]

    The change of the surface charge on the planes of the capacitor is equal to the change of the electric displacement, D, multiplied by the increament of the surface. When the dielectric replaces the vacuum and the electric field, E, stays the same D changes by [tex](\kappa - 1)\epsilon_0*E [/tex]. So

    [tex]\delta W_B=\delta Q*V=(\kappa-1)\epsilon_0*E^2*w*d*\delta x[/tex]

    The capacitance changes by [tex]\delta C=(\kappa-1)\epsilon_0*\delta x*w/d [/tex], and the energy stored in the capacitor would change by
    [tex]\delta W_C=0.5*\delta C*V^2=0.5*(\kappa-1)\epsilon_0*\delta x*w*d*E^2[/tex].

    From the condion of work-energy balance we get:

    [tex](\kappa-1)\epsilon_0*E^2*w*d*\delta x =0.5*(\kappa-1)\epsilon_0*E^2*w*d*\delta x+\delta KE [/tex]

    Rearranging the equation:

    [tex]\delta KE=0.5*(\kappa-1)\epsilon_0*E^2*w*d*\delta x[/tex].

    The time rate of change for the kinetic energy is

    [tex]dKE/dt=m*v*\dot v =0.5*(\kappa-1)\epsilon_0*E^2*w*d*v[/tex]

    According to the formulation of the problem, the slab is given a little push inward. So v is small at the beginning, but positive. The diection of the acceleration is also positive. The speed will increase, there is an inward force acting on the slab.
    This little push is crucial, as we could not say if the slab starts to move or stays in its original position using the argument above. But there is really an inward force on it, coming from the "fringe effect" at the edges of the capacitor plates. The field near the edges is inhomogeneous and produces an inward force onto the slab. But to prove this is beyond the "college level".

    Anyway, the slab will not "slide out of the plates and fall on the floor and live there happily ever after move out of the capacitor". If it moves out, we have [tex]1-\kappa [/tex] in the formula for dKE/dt, so it would be negative. The slab would slow down, that means an inward force again.

    The slab was assumed moving inward at the beginning. We have deduced that it will continue to move inward with constant aceleration ,

    [tex]\dot v=0.5*(\kappa-1)\epsilon_0*E^2*w*d/m[/tex]

    till its front face reaches the edge of the capacitor. From there on, the situation is reversed, new free charges appear on the plates as the slab moves outward, and the capacitor feeds back charges to the battery. Its energy decreases but to feed back the charges, additional energy is needed on the account of the KE of the slab. At the end the slab will stop. Now it is the fringe effect that will start the slab to move inward again.

    The time now.

    The displacement is l-a, the time needed is

    [tex]t=\sqrt{\frac{2*(l-a)}{\dot v}}
    =2*\sqrt{\frac{(l-a)*m}{(\kappa-1)\epsilon_0*E^2*w*d}[/tex].
    The time priod is four times longer:
    [tex]T=4*t=8*\sqrt{\frac{(l-a)*m}{(\kappa-1)*\epsilon_0*E^2*w*d}[/tex]

    The solution quoted by ambuj was 8sqrt( ((l-a)lmd)/(eAE^2(K-1))) which is very similar to my result, but dimensionally not correct.

    ehild
     
  17. Aug 16, 2004 #16

    Gokul43201

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    ehild, you are absolutely right. I was wrong, and foolish. I take back what I said.
     
  18. Aug 17, 2004 #17

    ehild

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    It is all right Gokul.

    ehild
     
  19. Aug 18, 2004 #18
    wow! Good Job ehild!

    (Note : I am really sorry at misleading the peeps when i said "the battery does does a resistive act" ... should have seen that coming !! :( )

    Anyways ehild,
    i am not going to push u through the edges on this one since u specifically mentioned that "proving the *fringe_effect* was beyond college level" and i am not particularly the physics_guy. Though i would like to know one thing that bothers me .....

    Any system i believe tries to achieve a low field potential, then how come this bizarre event of capacitor actually pulling the dielectric block in ?? (assuming that my knowledge base : *that the field potential increases as the block moves in is correct*?? or is it that we relate the field potential to the actual potential difference across the plates which is constant here and hence the effect is not actually violating any rules as such)

    (Pardon me if this doubt is hidious .. i lose my mind in the gutters of Diagon Alley at times)

    -- AI
     
  20. Aug 18, 2004 #19

    ehild

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    Hi Al,

    That event -I mean the capacitor pulling in the dielectric- is not that bizarre. You see a similar effect with a coil connected to a battery and a piece of iron road. The coil will suck the piece of iron in, just like the capacitor does with the dielectric slab. And the energy density inside the coil 0.5*B*H will increase if the magnetic permeability increases and the current stays the same. Try and see yourself.

    As for your statement that "the field potential increases as the block moves in" - I do not really understand what you mean on "field potential". There is a potential in a point in an electric field. There is a potential difference between the two plates of the capacitor. But it does not tend to minimize itself. It is a positive charge which tends to occupy a place with lower potential. The field inside the capacitor has got an energy density, and this multiplied by the volume is the same as the "energy of the capacitor" 0.5*V^2*C (well, if we don't bother ourselves with the field outside the capacitor. Maybe we do it wrong.) Yes, this would increase if the slab moves in, but the capacitor is connected to the battery, and it maintains a constant potential difference across the capacitor. A systems tends to occupy the lowest energy if it is let alone, but this capacitor is not "let alone".

    I am not an expert on electric field calculations and Thermodynamics and so on, but I try to imagine that I am inside that battery and watch what happens. There is an electrolyte and two electrodes, say a zinc and a carbon one, and the zinc ions would like to go into the electrolyte, they like to be there, because of some crazy chemical desire, I never understood Chemistry, but they can not go any more, as the electrons left behind the electrode are pulling them back... then the electrodes of the battery are connected to the capacitor. The electrons happily run there to occupy the empty space, more zinc ions can dissolve and at the end the process stops again when the capacitor is charged to the voltage of the battery. Poor zinc ions, left on the electrode, should stay there longing for the cool electrolyte in vain. But then a dielectric slab is pushed between the plates of the capacitor. The electric field would polarize its molecules or atoms, and align them into dipole chains, and the ends of the chains neutralize some charges on the capacitor plates. The voltages would fall, but new charges rush over from the battery, so the result is that the voltage stays constant. This goes on till the dielectric totally fills the place between the capacitor plates. It looks as if the battery would prefer the dielectric in, as it has more place for the charges and more zinc can dissolve. A battery likes to spread out its charges, that means it loses more energy then the capacitor gaines.
    (I know this was not a physical explanation, so Gokul, if you read this, please do not force me again to sweat out a more exact derivation. :) It is not so easy for me.)

    ehild
     
  21. Aug 18, 2004 #20
    Thanks to everyone who worked on a solution for this one, I really wanted to see how it was done.
     
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