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How much work must be done to carry a +2.3 X 10^-3 C charge from 65 cm to 15 cm?

  1. Apr 25, 2008 #1
    1. The problem statement, all variables and given/known data

    How much work must be done to carry a +2.3 X 10^-3 C charge from 65 cm to 15 cm?

    From the question:
    q=+2.3 X 10^-3 C
    r1= 65 cm = 0.65 m (these must be converted to metres)
    r2= 15 cm = 0.15 m
    k= 9.0 X 10^9 Nm^2/C^2
    W=?
    W=∆Ee
    ∆Ee=q∆V

    (I'd just like to say that this question is from grade 12 physics)



    2. Relevant equations

    Ok here’s the formulas i've been given.

    Fe=Kq1q2/r^2
    E=kq/r^2
    E=Fe/q
    Ee=Kq1q2/r
    ∆Ee=Kq1q2(1/r2 - 1/r1)
    ∆Ee=q∆V
    V=Kq/r
    V=Ee/q
    E=∆V/r
    W=∆Ee=q∆V
    q=Ne

    Where K is coulombs constant (I think that’s what it’s called) K=9.0 X 10^9 Nm^2/C^2

    I hope the people who try to help know what those formulas are already. If not I can tell you what they are.

    3. The attempt at a solution

    From the question:
    q=+2.3 X 10^-3 C
    r1= 65 cm = 0.65 m (these must be converted to metres)
    r2= 15 cm = 0.15 m
    k= 9.0 X 10^9 Nm^2/C^2
    W=?
    W=∆Ee
    ∆Ee=q∆V

    So I thought ok, so I need to find ∆V now. Rearranging ∆Ee=q∆V for ∆V.
    ∆V=∆Ee/q

    What equations have ∆V in them?

    ∆Ee=q∆V (well we are trying to find ∆Ee already)
    E=∆V/r (oh so I need to find E now)

    What equations have E in them?

    E=kq/r^2 (can't use this equation thought because there is two radius's, r1 and r2, and this equation only has one.
    E=Fe/q (oh so now i need to find Fe)

    What equations have Fe in them?

    Fe=Kq1q2/r^2 (but I can’t use that equation because I only have one q)
    E=Fe/q (can’t use this equation either because I was trying to find Fe to find E from the other equation)

    Then I thought wait there’s also the equation ∆Ee=Kq1q2(1/r2 - 1/r1), but again the question only gives one q. I could only use this equation if q=q1 and q2=q1, but I cannot be sure of that.

    It’s like a circle that brings me to nowhere so I am stumped.

    Someone from another website told me this:
    "As long as you move the charge in a space where no electric field is detected the work is equal to zero. in order to give your problem a physical meaning you should precise the electric field in which the charge moves, which could be generated by another point-like charge or an uniform electric field."

    The problem is the question gives no other point charge, so I don't know what to do. I am stuck at this point. If you can help I would greatly appreciate it. This question is from a "take home quiz" from my grade 12 physics class. It is due on Monday April 28, 2008, 3 days from today, so if anyone could help sooner than later would be better. Thanks.
     
  2. jcsd
  3. Apr 25, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    missing information

    That's the same comment I would make. In order to find the work done, you need to know what force acts on the charge. Did you present the problem exactly as it was given to you, word for word? Does it refer to a previous problem or other information?
     
  4. Apr 25, 2008 #3
    Here’s the entire question, yes there is two parts, but it doesn’t make sense to me why you would use this from the other question.

    1.a) What is the electric potential from a charge of 3.0 X 10^-5 C at 15 cm and 65 cm?
    b) How much work must be done to carry a +2.3 X 10^-3 C charge from 65 cm to 15 cm?

    a) q= 3.0 X 10^-5 C
    r= 15 cm = 0.15 m
    k = 9.0 X 10^9 Nm^2/C^2
    V=?
    V=kq/r
    V=((9.0 X 10^9 Nm^2/C^2)(3.0 X 10^-5 C))/0.15 m
    V= 1800000 V
    V= 1.8 X 10^6 V

    Same for the second part except r= 65 cm = 0.65 m
    We find that V= 415384.62 V which is 4.2 X 10^5 V.

    NOWHERE on the sheet does it say to use the charge 3.0 X 10^-5 C from question a) in question b) but I cannot see any other way to do the question unless I do use that, unless you now know what to do from this new information.
     
  5. Apr 25, 2008 #4
    That is exactly what it says on the sheet for the questions.
     
  6. Apr 25, 2008 #5

    Doc Al

    User Avatar

    Staff: Mentor

    This is a two part question. Of course you're supposed to use the information in part a to do part b! Part b is meaningless without the background of part a. In fact you use the result of part a to do part b.

    Hint: If you know the meaning of the electric potential calculated in part a, part b should be a snap.
     
  7. Apr 25, 2008 #6
    ok so from a) the part where the radius is 0.15,I got v=1.8 X 10^6 V, when the radius is 0.65 m i got V= 4.2 X 10^5 V...

    so i guess...∆V=V2-V1...and in part b r1=0.65 m (where v1= 4.2 X 10^5 V) and r2=0.15 m (where v2= 1.8 X 10^6 V)...so i just go:

    ∆V=V2-V1
    =1.8 X 10^6 V - 4.2 X 10^5 V
    = 1380000 V


    ∆V = 1380000 V
    q=+2.3 X 10^-3 C
    Using the equation ∆Ee=q∆V

    ∆Ee=q∆V
    =(+2.3 X 10^-3 C)(1380000 V)
    =3174 J

    or do i do it this way?:

    From a) q1=3.0 X 10^-5 C, from b) q2=+2.3 X 10^-3 C

    Using the equation ∆Ee=Kq1q2(1/r2 - 1/r1)

    ∆Ee=Kq1q2(1/r2 - 1/r1)
    =(9.0 X 10^9 Nm^2/C^2)(3.0 X 10^-5 C)(+2.3 X 10^-3 C)(1/0.15 cm -1/0.65 m)
    =3184.62 J

    Hmm...the answers are almost the same but not exactly...any thoughts?...maybe rounding error
     
  8. Apr 25, 2008 #7
    yes, it is just rounding error, i used the exact value of V1= 415384.62 V and got the answer 3184.62 V, thanks for helping :)
     
  9. Apr 25, 2008 #8
    I guess doing it this way:

    "Using the equation ∆Ee=Kq1q2(1/r2 - 1/r1)

    ∆Ee=Kq1q2(1/r2 - 1/r1)
    =(9.0 X 10^9 Nm^2/C^2)(3.0 X 10^-5 C)(+2.3 X 10^-3 C)(1/0.15 cm -1/0.65 m)
    =3184.62 J"

    is better because i am not using numbers that i calculated from before which could create errors.
     
  10. Apr 25, 2008 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Those two methods are identical. (Compare the equations!) The answers match to 2 significant figures.

    The only reason the answer are different at all is rounding error. Redo the calculations without rounding anything off until the last step. (Don't round off the potentials in part a.)
     
  11. Apr 25, 2008 #10
    im not smart enough to see that each way is identical, i guess by looking at the units and cancelling them out?
     
  12. Apr 25, 2008 #11
    oh u mean by two significant figures the "31" part of the number?
     
  13. Apr 25, 2008 #12
    Thanks again for help.
     
  14. Apr 25, 2008 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Sure you're smart enough!

    From part a, you have:
    V=kq/r
    So:
    ∆V=Kq1(1/r2 - 1/r1)
    And thus:
    W = ∆Vq2 = Kq1(1/r2 - 1/r1)q2

    Which is identical to what you did for the second method.
    Yes. (Although it would round off to "32".)
    You are welcome! :smile:
     
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