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Homework Help: Make up test question

  1. Feb 23, 2004 #1
    Ok I got two questions dealing with Charge and Current.

    1) Consider three identical metal spheres . Initally , sphere A contains a charge of 2uC and B contains a charge of 10uC. Sphere C has a charge of -4uC. Spheres A and B touch each other, and then are seperated. Then Sphere B and C touch and then seperate. What is the charge on C now.

    I had gusses positive and he marked .5 off soo i assume i need a number with that. Not sure what to do with it.

    2)If a metal wires carries a current of 80.mA, how long does it take for the 3.00 x10^20 electrons to pass a given corss-sectional area of the wire.

    I had use I= deltaQ/delta t and had gotten 2 seconds. his resposne was "correct, but now solve for delta t.... i throught i just sloved for it.

    Thanks for the help.
  2. jcsd
  3. Feb 24, 2004 #2


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    It's just an accounting problem. You have to make the assumption (I'm assuming) that the contact between the spheres lasts long enough for the charge to reach equilibrium. Do it in two steps.

    Ya, I agree with you. I can't imagine what he was thinking about. (I did not check to see if 2 s is correct.)
    Last edited: Feb 24, 2004
  4. Feb 24, 2004 #3
    Alirhgt awosome, also so what you were saying in my first question. THe charges equalize and then you take that charge and add it to the negative one... correct.
  5. Feb 25, 2004 #4


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    "that charge?" I don't know to what you are referring. A and B join together to share all of their charge. Since they are identical metal spheres (and since we are making previously said assumption), they will share the charge half and half. But don't forget that they separate, after which, you will have a new distribution of charge among the three spheres (not 2, 10, -4 any more). Then, according to this new distribution, you do step 2.

    One way to check to see if you made a mistake is to add all the charges and see if it is the same as what you would get if you add all the charges in the original distribution. You can use this to check each step, because charge is conserved.
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