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What is the charge of the 10 billion electrons?

  1. Jan 20, 2016 #1
    1. The problem statement, all variables and given/known data:
    I have a battery and lightbulb circuit (in series). 12V battery rated at 35Ah.

    Ten billion electrons pass a planar cross section of the wire. What is the charge of the 10 billion electrons?

    2. Relevant equations
    Q=Ne
    I=dq/dt
    1A=1C/s


    3. The attempt at a solution
    Is this as simple as using Q=Ne or not? I may be overthinking this, but this doesn't seem right.

    Q = Ne
    Q=(#electrons)(electron charge)
    Q= (1E10)(1.6E-19 C)
    Q = 1.6 E -9 C
     
    Last edited: Jan 20, 2016
  2. jcsd
  3. Jan 20, 2016 #2

    Mark44

    Staff: Mentor

    Not an advanced physics problem, so moved to Intro Physics
     
  4. Jan 20, 2016 #3

    Doc Al

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    Assuming you have the question right, that's all there is to it. (Get the sign right.)
     
  5. Jan 20, 2016 #4
    Ok. Follow up question. How many Joules of energy does the battery I mention store? I would use .5CV^2 or .5QV, but this is what I learned in E&M physics, so I think there is another way. This is EGR 110 intro to electrical and computer engineering, so there are no pre-recs. Is there a way to do this using unit conversions?
     
  6. Jan 20, 2016 #5

    Ray Vickson

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    Homework Helper

    Always state your units; chances are the question would be marked wrong without them. Otherwise, it looks OK.
     
  7. Jan 20, 2016 #6

    Doc Al

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    Hint: If the battery delivered a current I, what power would it be delivering?
     
  8. Jan 20, 2016 #7
    Yeah I was thinking about using P=IV, but I'm not sure we can. Can we still use that if we are given current in Amp hours?
    12V battery rated at 35Ah.
     
  9. Jan 20, 2016 #8

    Doc Al

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    Using P = VI is the first step. Now how do you go from power to energy? (Don't plug in numbers yet, just play with the equations symbolically.)
     
  10. Jan 20, 2016 #9
    Oh I see. P=IV will give you units of WattHours and then you just do unit conversions and get it to Joules. Got it! Thanks!
     
  11. Jan 20, 2016 #10

    Doc Al

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    Power will be in units of Watts; if you multiply by the time, then you'll have energy.
     
  12. Jan 20, 2016 #11
    Yeah, that's what I did. P= (12V)(35Ah) =420WattHours*(3600)
     
  13. Jan 20, 2016 #12

    Doc Al

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    Good.
     
  14. Jan 21, 2016 #13
    Ran into another problem. Assume the battery is supplying .5A of current to a light bulb. How long will the battery last? (Battery is 12V rated at 35Ah) I am supposed to use the 35Ah to find the time or do I have to somehow use the .5A. I don' t think there is a way to find how long it will last with just Amps. You need Amp hours to do this right?
     
  15. Jan 21, 2016 #14

    Doc Al

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    You need to use both Amp-Hours and Amps to solve for the time.
     
  16. Jan 21, 2016 #15
    Ok so, I used P=IV to find that the battery supplies 6W. I can do the same thing to find that it provides 420Wh. So can I just use unit conversions to find the hours? 6W/420Wh. Watts cancel and you are left with 1/70h. So 70 hours? Is this correct?
     
  17. Jan 21, 2016 #16

    Doc Al

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    Sure.

    All you have to do is realize that the ampere-hours are given so: 35 A-hr = I*T = .5A*T. Solve for T and get 70 hours.
     
  18. Jan 21, 2016 #17
    Can you explain that a bit more? Where did you get the "35 A-hr = I*T = .5A*T." part from
     
  19. Jan 21, 2016 #18

    Doc Al

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    The battery capacity, expressed in terms of Amp-Hours, equals Current*Time. Draw a lower current and the battery can run for a longer time.
     
  20. Jan 21, 2016 #19

    Doc Al

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    35 A-hr = (.5 A)*(70 hr) = (1 A)*(35 hr) = ... etc.
     
  21. Jan 21, 2016 #20
    Ok, that makes sense. More sense than what I did. Is the way I did it a bad way to go about this? I don't know if I would have noticed to use that equation BC=AT on a test or quiz. I was just looking at units and trying to get hours through unit conversion.
     
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