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Need help very badly please.

  1. Mar 9, 2005 #1
    Two equipotential surfaces that surround a +3.0 x 10^-7 C point charge has a radius of 0.15m, what is the potential surface

    Two capacitors are identical, except that one is empty and the other is filled with a dielectric (k= 4.50). The empty capacitor is connected to a 12.0 V battery. What must be the potential difference across the plates of the capacitor filled with a dielectric such that it stores the same amount of electrical energy as the empty capacitor?

    A resistor is connected across the terminals of a 9.0-V battery which delivers 1.1x10^5 J of energy to the resistor in six hours what is the resistance to the resister? (i solved this one but got a ridiculously small resistance of like 3.4 to the negative 8)
     
  2. jcsd
  3. Mar 10, 2005 #2
    2nd one:

    E = 1/2CV^2

    E1 = 1/2C x 144

    E2 = 4.5 x 1/2 x C x V = 1/2C x 144

    Ummm... don't know if that's right, but it looks like it. I mean, kappa basically mutliples the capacitance. C's cancel.

    3rd one: My way of thinking is: P = V^2/R

    J = V^2/R x T (In your case 3600) You know everything but R, so plug and chug.

    Sorry, forgot how to do the first one. We went over this for like a day, then went on to circuits =S

    PL
     
  4. Mar 10, 2005 #3

    nd

    User Avatar

    For the third one, http://www.sengpielaudio.com/calculator-ohm.htm.
    Your first question is unclear.
    For the second one, can you tell me the energy in the capacitor without the dielectric? Qualitatively, do you expect the required voltage to be more or less than the case with k=1?
    Do you have a textbook?
     
  5. Mar 10, 2005 #4
    Would it matter what the capacitance is? Doens't it just cancel out anyway?

    And don't give him (her?) those links. :p He'll (she?) never learn that way. :)

    PL
     
  6. Mar 10, 2005 #5
    the second question is exactly as it is out of the book, and i dont have the book, i left it at school. thnx for helping guys, i appreciate it... =P. i have a chem 102 exam tmmrw morning at 8 am too... i am screwed.
     
  7. Mar 10, 2005 #6
    crap is Watts=J/s?
     
  8. Mar 10, 2005 #7
    Completed.
    A resistor is connected across the terminals of a 9.0-V battery which delivers 1.1x10^5 J of energy to the resistor in six hours what is the resistance to the resister? (and i figured it out without the link =) i found out that P is J/s hahaha rather than just joules. haha stupid watts. thnx for helping out.)

    Thnx alot guys only 2 to go.

    Hrmm... i dont actually know if capacity matters but i think it probably does...
     
  9. Mar 10, 2005 #8

    nd

    User Avatar

    The capacitance will of course cancel since we are given no information about the capacitor itself.
    The energy store in a capacitor is [tex] U = \frac{1}{2} CV^2[/tex].
    The problem at hand states [tex] \frac{1}{2} C_1V_1^2= \frac{1}{2} C_2V_2^2[/tex]. You are looking for V_2.
     
  10. Mar 10, 2005 #9
    i believe V works out to be 5.66 =) thnx guys.
     
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