Variable Voltage Source

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

    As the voltage in the circuit shown below is increased (but not to the breakdown voltage), the capacitance

    [​IMG]


    a. increases.

    b. decreases.

    c. does not change.

    d. increases, decreases, or does not change, depending on the charge on the plates of the capacitor.

    e. does none of these.


    2. Relevant equations

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

    3. The attempt at a solution

    If the voltage increases at a specific rate, then the charge would also increase since Q and V are directly proportional and the ratio between the two is a constant. Therefore, the capacitance does not change????

    Thanks.
     
  2. jcsd
  3. The capacitance should remain the same as it is a function of geometry and the material between the plates.

    C = (epsilon * A) /d
     
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