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Determing the Potential Difference Ratio

  1. Mar 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Two parallel-plate capacitors have the same plate area.
    Capacitor 1 has a plate separation half that of capacitor 2, and the quantity of charge you place on capacitor 1 is six times the quantity you place on capacitor 2.
    How do the two Potentials differences of the capacitors compare V1/V2?
    2. Relevant equations
    C=Q/V

    3. The attempt at a solution
    The one way I was looking at this is that I know that as the as the distance increases the potential difference increases. I was figuring that since no actually numbers were given it would be conceptual passed. I also assumed that C would be the same though I am not sure how distance affects capacitance. V1 = 6q/C and V2= q/C. This is what I'm stuck
     
  2. jcsd
  3. Mar 16, 2015 #2

    Doc Al

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    Don't assume that! Look it up! (The capacitance of a parallel plate capacitor depends on area and separation distance.)
     
  4. Mar 16, 2015 #3
    which I'm given no values for. All that is given is that the distance for 1 is half that of the 2nd one. The area is never given. That's why I'm having difficulty. I just hate when given things with no values. Numbers I can deal with.
     
  5. Mar 16, 2015 #4

    SammyS

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    What do you know about (parallel plate) capacitors in general?

    How is capacitance related to area and plate separation?

    What's the relationship among Q, C, and V ?
     
  6. Mar 16, 2015 #5
    I know q is the charge that is place, and the v is the potential difference between two points. C is the availability for something to hold a charge. q is in coulombs, v is well... volts and c is Farads. I know that C*V=Q. Based of this I know that C*V_1 = 6Q and that C*V_2 = Q. I don't know how capacitance is related to area or plate seperation, as my instructor just copies formulas from the book, and the book does a horrible job going over this. But based of those I can get 6Q/C = v_1 and Q/C=v_2. Here is where I don't know what to do in terms of the separation, but the area is not given.
     
  7. Mar 16, 2015 #6

    SammyS

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    As Doc Al said, C1 ≠ C2 .

    The capacitance is inversely proportional to plate separation. (This consistent with what you stated about potential difference increasing with separation in the OP.) Therefore, C1 = 2C2 .

    BTW: Regarding plate area; Capacitance is directly proportional to plate area.
     
  8. Mar 16, 2015 #7

    Thank you for the information and the help. I'm writing this into my notes so I can study it up. First question I asked on here, and I like how it worked. Let's me actually try and work through the problem, while getting help. Going to be useful when I get into Quantum Mechanics. I got the answer.
     
  9. Mar 16, 2015 #8

    SammyS

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    I should have said it earlier,

    Welcome to PF !
     
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