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Introductory Physics Help

  1. Jan 9, 2014 #1
    The question is:
    What current is flowing in a metal wire if 3.0*10^18 electrons pass a given point in a time of 0,10s?

    I know that 1 electron has a charge of 1.60*10^-19
    I also know that one coulomb is 6.24*10^18
    Lastly I know to find current I need to use the formula Q=I*t, but change it to I=Q/t (I= current in Amps, Q= # of electrons or quantity of charge in coulombs, and t= time.

    I attempted the solution and this is what I got:

    In order to find the current I first need to figure out the quantity of charge (Q).
    Q=the number of electrons.
    So I multiplied the charge of one electron by the number of electrons in the circuit.
    (1.60*10^-19)*(6.24*10^18)= 0.00048 C
    Q= 0.00048 C

    Now I used the formula I=Q/t
    0.00048/10=0.000048.
    I= 0.000048 A

    Therefore the current in the circuit is 0.000048 A.

    All I wanted to know is if I did it correctly and if I did it incorrectly how I could solve it.
    Thank you so much in advance :)
    Panda-Monium
     
  2. jcsd
  3. Jan 9, 2014 #2
    1.60*10^-19 of what?

    6.24*10^18 of what?

    Correct in theory.

    What are all those numbers? How come that multiplying two dimensionless constants gives you coulombs?

    That would have been correct, if 0.00048 C had been correct.
     
  4. Jan 9, 2014 #3
    FOR VOKO

    Sorry for the lack of information. I guess I need more sleep :) By 1.60*10^-19 I meant that is the charge of one electron in coulombs. So 1 electron=1.60*10^-19 C,
    For 6.24*10^18 I meant that this is the amount of electrons that are in one coloumb. So 1 C = 6.24*10^18 .

    The numbers I multiplied to get me coulombs is what I am confused about. I read that Q= no. of electrons*charge of one electron. Is that right?

    Thank you for your help so far.
     
  5. Jan 9, 2014 #4
    So 1.60*10^-19 C is the charge of one electron. The number of electrons given was 3.0*10^18. But instead of that, you multiplied with 6.24*10^18? And somehow got 0.00048 C?
     
  6. Jan 9, 2014 #5
    I am sorry again I made a mistake in the question I stated it but did not show it.
    I multiplied the charge one electron by the number of electrons in in the circuit. So (1.60*10^-19 C)*( 3.0*10^18 ). This is how I got 0.00048 C.
    This is what I meant.
    Sorry about all this.
     
  7. Jan 9, 2014 #6
    Because 10^-19 * 10^18 = 0.1, (1.60*10^-19 C)*( 3.0*10^18 ) = (1.60 C)*( 0.3 ) = 0.48 C.

    How did you get 0.00048 C?
     
  8. Jan 9, 2014 #7
    Well I guess one of my mistakes was I mixed up calculations. So this means the final answer should be
    I=Q/t
    0.48C/10s=0.048A
    So that means 0.048A is the current/answer.
    Right?
     
  9. Jan 9, 2014 #8
    I am not sure what "0,10s" in the original question means. Is that 10 seconds or 0.1 second? If that is 10 seconds, then your answer is correct.
     
  10. Jan 9, 2014 #9
    Your so observant I checked the question it is 0.1s. So it would be
    0.48C/0.1s=4.8A
    And my last question is would you happen to know the reason for this relationship. How Q(Charge in coloumbs or the number of electrons) = # of electrons*charge of one electron?
     
  11. Jan 9, 2014 #10
    Think of charge as of "money". You can split your money into coins, and let's say, for simplicity, all the coins have the same value. The total sum of your money is the sum of all your coins times the value of each coin.

    So electrons are "coins" of charge.
     
  12. Jan 9, 2014 #11
    That makes more sense. But it just leaves me with one question. How does this all relate to columbs?.
    Like how is multiplying the number of electrons by the amount of electrons giving you columbs?
     
  13. Jan 9, 2014 #12
    You don't multiple the number of electrons by the "amount of electrons", just like you don't multiply the number of coins by the "amount of coins". You multiply the number of coins by the value of each coin. And so you must multiply the number of electrons by the "value", i.e., charge of electron.
     
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