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Capacitors: charge, potential difference and stored energy

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data

    The 50Volts is going through 3 capacitors.
    C1 = 4.0 µF, C2 = 8.0 µF, and C3 = 9.0 µF,

    The URL below is the image of how the capacitors are connected:
    http://www.webassign.net/hrw/26_27.gif

    For each capacitor i need to find the charge, the potential difference and the stored energy.

    2. Relevant equations

    CV = Q

    3. The attempt at a solution
    I really don't understand what potential difference is, and I'm not sure what the problem asks for with the stored energy.
    I tried CV = Q for the charge of C1 and got 4E-6 * 50 = 2E-4, which is wrong.
     
  2. jcsd
  3. Mar 24, 2010 #2
    Remember that a capacitor is a break in the circuit. Capacitors in parallel can store a lot more charge than two in series, because the parallel combination allows each plate to fill up with its "rated" charge. In the series combination, the break in the circuit prevents this simple process. Instead, charge is stored on the first plate the voltage source comes in contact with and then charge departs the bottom plate toward the second capacitor's first plate. The second capacitor's second plate then departs charge toward to voltage source allowing for an apparent current (though no charge actually flows THROUGH a capacitor). Specific to this problem, all of the charge depleted from the voltage source due to the equivalent capacitance, Ceq, must be stored on the first two capacitors that are in parallel. The third capacitor holds the same charge as the two in parallel do (because it only receives charge out of the departure of charge from the first two capacitors which is equal to the stored charge on the first capacitor's plates)

    Voltage is the energy per unit charge needed to move a charge from point p(1) to point p(2). A person must expend energy per unit charge in moving some charge q only if there is an electric field between the points. (Remember, f = Eq. If E = 0, f = 0 and w = 0).

    The mechanical analog to voltage usually helps students visualize the concept:
    gravitational potential energy due to a gravitational field is mgh where m is the measurement of the material's sensitivity to the field's tendency, g is the field, and h is the distance over which the constant field acts. If both sides were divided by mass, we would arrive to the never used but now dubbed gravitational potential (work per mass) -- gh.

    Similarly, electric potential energy is qEd where q is the measurement of the material's sensitivity to the field's tendency, E is the field, and d is the distance over which the constant field acts. If both sides are divided by charge, we would arrive to the always used electric potential (i.e. voltage, work per charge) -- Ed.
     
    Last edited: Mar 24, 2010
  4. Mar 24, 2010 #3
    Thank you so much! I really appreciate your in depth response; it cleared up a lot.
     
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