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Physics true/false questions v. isolated parallel plates

  1. Mar 4, 2005 #1
    Which of the statements below are true for two oppositely charged, isolated parallel plates? (C is the capacitance, U is the stored energy, +Q and -Q are the charges on the plates.) Note: isolated plates can not lose their charge. (Enter ALL correct statements, e.g., BCD)

    A) When the distance is doubled, U increases.
    True. Im thinking that if the distance is doubled there is more to store energy in.

    B) Inserting a dielectric increases C.
    I have no idea on this one.

    C) When the distance is halved, Q stays the same.
    True. I dont think distance will affect the charge.

    D) Increasing the distance increases the electric field.
    True.

    E) When the distance is doubled, C increases.
    False. It doesnt have any affect on the capacitance

    F) Inserting a dielectric decreases U.
    Again no idea.

    G) Inserting a dielectric increases Q.
    I just have no clue with dielectrics.

    Thanks for any help in advance.
     
  2. jcsd
  3. Mar 4, 2005 #2
    Some forumlas you need to know is

    [tex]C = \frac {Q}{V} = \epsilon_0 \frac {A}{d}, \ \ U = \frac {Q^2}{2C}[/tex]

    So when d increases, C decreases, then U increase.

    When ever you are dealing with a dielectric, the capacitance is given by

    [tex]C_d = \kappa \epsilon_0 \frac {A}{d}[/tex]

    Where [itex]\kappa[/itex] is a constant called the dielectric constant. This is always greater than 1. This says that [itex]C_d > C[/itex]. Which is exactly why dielectrics are inserted, to increase the capacitance.

    That should get you started.
     
  4. Mar 4, 2005 #3
    cool. I got it thanks.
     
  5. Mar 5, 2005 #4

    Andrew Mason

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    True. [itex]C \propto \epsilon > \epsilon_0[/itex]

    False. The field is uniform between the plates and is independent of the separation where the plate size is large compared to separation. The potential depends upon distance (V = Ed)

    True. C is proportional to d.

    True. The dielectric reduces the field between the plates. So it reduces the potential (V=Ed),

    False. Charge is constant.

    So: ABCEF

    AM
     
  6. Mar 5, 2005 #5
    I think [itex]C \propto \frac {1}{d} [/itex].
     
  7. Mar 5, 2005 #6

    Andrew Mason

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    Right you are. So ABCF.

    AM
     
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