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Help with dielectrics, capacitors, batteries, etc

  1. Feb 9, 2006 #1

    dnt

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    1. can someone please explain the idea of a dielectric to me (or at least give me a link that explains it in laymans terms) - all i understand is that its placed between two capacitors and it increases their capacitance (by the factor of K). what does it do to the voltage? and the charge?



    2. a question states that a fully charged capacitor of capacitance 1.0 F is able to light a .50 watt bulb for 5.0 seconds before it quits - what was the terminal voltage of the battery that charged the capacitor?

    I first tried wolving for work (power = work/time) and I got W = 2.5 J but after that im stuck. i know that V = W/q but i dont know what q is. please help.



    3. the last question says to store energy to trigger an electronic flash unit for photography, a capacitor is connected to a 400 V voltage source. the flash requires 20 J. the capacitor is initially airfilled and has a capacitance of 100 uF (microfarads). you then insert a dielectric material, leaving everything else the same. what must be the dielectric constant to store 20 J.

    my solution: I started by saying that U = 1/2 CV^2 (is that correct?)

    by solving for C required I got 2.5*10^-4 F

    however, it says my capacitance is only 100 uF = 1*10^-4 F

    therefore the dielectric constant needs to be (2.5*10^-4)/(1*10^-4) = 2.5

    is that the right answer? did i do the work correctly?

    thanks for the help.
     
  2. jcsd
  3. Feb 9, 2006 #2

    Andrew Mason

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    A dielectric contains polar molecules that align themselves in an electric field so as to oppose it. Thus the insertion of a dielectric reduces the potential difference (voltage) between the plates for a given plate charge. Since : C = Q/V, the dielectric has the effect of increasing capacitance.


    The energy stored in a capacitor is power x time:
    [tex]W = \int_0^{t_f} Pdt = \int_0^{t_f} VIdt = \int_Q^0 Vdq = \int_V^0 VCdv = \frac{1}{2}CV^2[/tex]

    You don't need to work out Q (charge on plates at full voltage, V) to find V. But if you want to do so, Q=VC where V is what you just worked out and C = 1.0 F.

    Looks fine.

    AM
     
    Last edited: Feb 9, 2006
  4. Feb 9, 2006 #3

    dnt

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    thanks for the help even though i dont understand all those symbols and stuff. however, how do you find Q when you dont have V either? you said Q=VC but we only know C.

    either way, the best way to solve it would be W=(1/2)CV^2

    so 2.5 J = (1/2)(1.0 F)V^2

    and solve for V that way...is that right?

    thanks again.
     
  5. Feb 9, 2006 #4

    Andrew Mason

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    In order to get 2.5 J stored in the capacitor, you will need to have a voltage of [itex]V = \sqrt{5} = 2.24 Volts[/itex]. Q = CV, so Q = 2.24 Coulombs. So a charge of 2.24 C. will produce a voltage of 2.24 volts and a stored energy of 2.5 J.

    AM
     
  6. Feb 9, 2006 #5

    dnt

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    ok so basically you cannot solve for Q unless you solve for V first?
     
  7. Feb 10, 2006 #6

    lightgrav

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    The Electrical PE of a collection of charges (as on a capacitor) is PE = Q*V .
    But here, the Electric Potential (V) is caused by the charges (Q) themselves.
    The Electric Potential when the first charge was put on was V_start = 0 ;
    the Electric Potential when the final charge accumulated was V_end = V .
    So, for a capacitor, PE = Q_total * V_average = 1/2 Q*V .

    Capacitance is defined: C = Q/V, so it's also ok to write PE = 1/2 Q*Q/C .
    With that expression, you don't need to compute Voltage first.

    Remember that Voltage is dV = E dx ; as the dielectric is inserted,
    the E-field reduces to E/K , if the same Q are on the plates.
    So the device has the "capacity" to store K times as much charge
    for the same "electrical analogy to gas pressure" (Voltage).
     
    Last edited: Feb 10, 2006
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