1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential and Capacitance

  1. Feb 26, 2007 #1
    I’m a bit confused on the relationships among C,V, and Q after reading my text. I don’t know if my reasonings contradict each other for the following.

    1. The problem statement, all variables and given/known data

    A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V. If the charge Q on the plates is doubled to 2Q,

    a. the capacitance becomes (1/2)V.
    b. the capacitance becomes 2C.
    c. the potential difference changes to (1/2)V.
    d. the potential difference changes to 2V.
    e. the potential difference remains unchanged.


    2. Relevant equations

    C = Q/V

    3. The attempt at a solution

    The ratio Q/V is a constant, and Q is directly proportional to V. Therefore, if the charge is doubled, won’t the V, the potential difference, double also?





    1. The problem statement, all variables and given/known data
    Doubling the potential difference across a capacitor

    a. doubles its capacitance.
    b. halves its capacitance.
    c. quadruples the charge stored on the capacitor.
    d. halves the charge stored on the capacitor.
    e. does not change the capacitance of the capacitor.



    2. Relevant equations

    C = Q/V or C = (epsilon_0*A)/(d)

    3. The attempt at a solution

    Following the fact that Q is proportional to V and the ratio of Q to V does not change, the capacitance does not change?




    1. The problem statement, all variables and given/known data

    If the potential difference of a capacitor is reduced by one-half, the energy stored in that capacitor is

    a. reduced to one-half.
    b. reduced to one-quarter.
    c. increased by a factor of 2.
    d. increased by a factor of 4.
    e. not changed



    2. Relevant equations

    U = (0.5)*C*V^2

    3. The attempt at a solution

    U = (0.5)*C*V^2

    U2 = (0.5)*C*(0.5*V)^2 = 0.5*C*[(V^2)/4] = (1/2)*(1/4)*C*V^2 = (1/4)*U

    The energy decreases by one-quarter?

    Thanks.
     
  2. jcsd
  3. Feb 26, 2007 #2

    berkeman

    User Avatar

    Staff: Mentor

    Correct, correct, and correct. You don't seem confused to me. Changing the capacitance requires a physical change to the shape or size of the cap, not just Q or V changes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Potential and Capacitance
  1. Capacitance potential (Replies: 1)

Loading...