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Homework Help: What's the Current of the Potential Difference?

  1. Sep 28, 2007 #1
    1. The problem statement, all variables and given/known data

    A potential difference is applied between the electrodes in a gas discharge tube. In 2.0 s, 3.3*10^16 electrons and 1.2*10^16 singly charged positive ions move in opposite directions through a surface perpendicular to the length of the tube. What is the current in the tube?

    2. Relevant equations

    I = C/T

    3. The attempt at a solution

    I reason that the current is in the direction of the movement of the positive ions:

    I = { 1.2E16 positive ions * 1.60E-19 C / 1 positive ion } / 2 seconds
    = 9.6E-4 A

    However the above is not the correct solution, and neither is when I use 3.3E16, the number of electrons. I think I'm not understanding a concept here.
  2. jcsd
  3. Sep 28, 2007 #2


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    Homework Helper

    You know, current is "flowing" in the positive charge direction, i.e opposite direction of electron motion. So that the total flow of charge is:

    deltaQ = (1.2*10^16 + 3.3*10^16)*1.602*10^-19 C

    Draw a picture of electrons moving towards one end of the tube, and positive ions towards the other. Then draw arrows that represent the current assoiciated with each "charge-type", then add :)

    Then divide by 2sec; then the average current is: I_avg = 0.0036 [C/s = A]
  4. Sep 28, 2007 #3
    This question is ambiguous.

    When those positive ions hit the cathode, they'll pick up electrons and start moving in the opposite direction. We might presume that since it says they are singly charged, that each ion carries two electrons each trip.

    What is not made clear, is whether the value specified for electrons moving from the cathode to the anode includes the electrons that are carried by negative ions.

    If the value specified for electrons includes those carried by negative ions, then simply consider the electron flow that is specified, and ignore the statement about the ions.

    If the value specified for electrons does not include those carried by negative ions, then double the ion flow (because each ion carries two electrons each trip) and add the electron flow to get the total flow.
    Last edited: Sep 28, 2007
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