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Calculating the current

  1. Jul 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Through a fluorescent tube of a diameter of 3cm cross the transversal secion per second 2x10^18 electrons and 0.5x10^18 positive ions (with charge +e) What is the current that goes through the tube?

    2. Relevant equations

    3. The attempt at a solution
    I have Vd*n for the electrons that is 2x10^18
    I have Vd*n for the electrons that is 0.5x10^18
    I calculate the area as 2pi(3/2)^2
    And I have the charge of an electron
    I know that talking about current you don't have to make a distinction between positive and negative charges, but that's theroy. In practice I don't know whether to add the current generated by the positive charges with the other one, to substract them, or what?
    How am I supposed to proceed in this cases?

    Thanks in advance.
  2. jcsd
  3. Jul 24, 2008 #2
    Conventionally, the flow of current is the flow of positive charges. If you have negative charges flowing against them then, of course, you will need to subtract to get the net current. Try drawing a quick diagram to see if that helps your understanding.
  4. Jul 24, 2008 #3
    Ok. But no, that's wrong. You have to take the absolute value of the charge, so both currents will add. At least that's what Sears and Semansky say.
  5. Jul 24, 2008 #4


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    Are you sure that's what they say? If the electrons are flowing in the same direction as the positive ions, then some of it will cancel out.
  6. Jul 24, 2008 #5


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    In an electrical discharge like this, the electric field will cause the positive and negative charges to flow in opposite directions. So add the currents, don't subtract them.
  7. Jul 25, 2008 #6
    What? Are you sure? That doesn't make any sense.

    [tex]I=\int \mathbf{J} \cdot d\mathbf{a}[/tex]


    [tex]\mathbf{J} = \rho \mathbf{v}[/tex]

    The current densities will create two currents that we can, more or less, think of as vectors. Vectors add, sure, but one of them is negative.

    I don't even want to think about the electric field because SR will make it a mess. I could be convinced otherwise, but it just doesn't seem right.
  8. Jul 25, 2008 #7


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    I think he meant to say was that because although you have 2 charge carriers, electrons and positive ions, the presence of an electric field in one direction only will cause their respective current contributions to add up. My earlier post was directed at #3 which wasn't a clear explanation.
  9. Jul 25, 2008 #8


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    Defennder is right about my meaning. Sorry, I didn't expect what I said would be confusing.

    A discharge lamp has a voltage difference applied between its 2 electrodes. Since there's a voltage difference between two points in space, there is an electric field. I thought this was simple enough stuff.

    What do people think is causing the charges to move in a lamp, if there is no electric field present?
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