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Total Work done on a Capacitor

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  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data

    fttfUkr.png

    I am currently trying to work on the extra credit question, the capacitance of the capacitor is 2 microfarads
    2. Relevant equations
    W = 1/2CV^2

    3. The attempt at a solution
    I've done the previous sections, and have gotten 4.23 * 10^-11 J as the answer for the work done on the capacitor. Now I have no idea how to proceed. From the formula W = 1/2CV^2, how can the work ever be negative? This is where I'm stuck
     
  2. jcsd
  3. Jan 31, 2016 #2

    gneill

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    Staff: Mentor

    Much like work done by a force acting on a moving object, where the direction of the force with respect to the motion counts, the work done on the capacitor depends upon the direction of the current. If charge (current) is flowing onto the capacitor then work is being done on the capacitor. If charge (current) is flowing out of the capacitor, then it's the capacitor that's doing work on the external circuit.

    The work formula comes from the integral ##W = \int_{v_i}^{v_f} C V~dV~~~##.

    So, how might you shift your voltage vs time curve up or down in order to make the "out work" equal to the "in work"?
     
  4. Jan 31, 2016 #3

    I can see that shifting the voltage graph down will shift the power graph down, and at a certain point this would balance out the positive and negative portions giving me a work of 0, but I'm still confused as to how to figure out that specific shift value. What sort of formulas should I be looking at to calculate the exact value?
     
  5. Jan 31, 2016 #4

    gneill

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    Staff: Mentor

    I'm thinking that, assuming the capacitor at some time in the distant past began life uncharged, then increasing its potential difference does work on it and decreasing the potential gets work out. I'd say you'd want to end with the same potential you started with for the books to balance to nil.
     
  6. Jan 31, 2016 #5
    What you're saying makes sense, but I'm now sure how to make that happen. From working out the problem, I have a voltage vs time graph that is linearly increasing until 4 micro-seconds, and then linearly decreasing from 4 to 7 microseconds. The starting voltage is at 0, and ending is at .0065V. Attempting to shift the end point around would also shift the starting point, right?
     
  7. Jan 31, 2016 #6

    gneill

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    Right. The whole thing would move by the initial voltage.

    I'm beginning to think that they are intending a definition of "work done on a capacitor" that involves the signs of the voltages and currents. If i*v is positive then work is being done on the capacitor. If i*v is negative then the capacitor is doing work. If this is so, then consider the following.

    Now, i*v is the instantaneous power. Integrating it over time would then yield the net work, and area area below the zero level would be work done by the capacitor (supplying power to move the charges).

    So I think you can write expressions to represent i*v over the two time intervals where the current is +4 and -1 mA. You'll have to incorporate an offset voltage, say Vo, into the voltage expressions.

    For example, assuming units of mA, mV, and μs, then over the first 4 time units the current is a constant +4 and an expression for the power during that interval is:

    ##4(V_o + \frac{8}{4}t)##

    Do something similar for the second interval where the current is -1. Note that the voltage begins where it left off in the last interval.

    Integrate them both over their respective time intervals. You'll want to find the Vo that makes their sum zero.
     
  8. Jan 31, 2016 #7
    Very helpful, thank you. I think i see how to go about it now, and I ended up getting a value of -1.2631 Volts. Messing with the units was troublesome, but I think I finally understand. I really appreciate your help, thanks again.
     
  9. Jan 31, 2016 #8

    gneill

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    Staff: Mentor

    I found something closer to 3 or 4 mV, so one or both of use should verify their work :smile:

    Re the units, that's why I assumed the units and went for a unitless expression. In the end I knew that whatever voltage I found would be in mV. Line slopes are actually in mV/μs, hence the slope of 8/4 in my example; the voltage increases by 8 mV over a time of 4 μs.
     
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