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Homework Help: Speed of an electron in an electric field - trying to find my error.

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

    What is the speed v(final) of the electron when it is 10.0 cm from charge 1?

    2. Relevant equations

    Ek= (mv^2)/2
    U= (k(q1q2))/r

    3. The attempt at a solution

    I used the following equation:

    Ek(f)+U(f) = Ek(i)+U(i)

    (mv(f)^2)/2 + (k(q1q2))/r = (mv(i)^2)/2 + (k(q1q2))/r

    (9.1x10^-31)(vf)^2)/2 + (((9x10^9)(3.45nC)(1.6x10^-19))/0.10m) + (9x10^9)(1.85nC)(1.6x10^-19))/0.40m) = (((9x10^9)(3.45nC)(1.6x10^-19))/0.25m) + (9x10^9)(1.85nC)(1.6x10^-19))/0.25m)

    (9.1x10^-31)(vf)^2)/2 + 5.6x10^-17 = 3.0 x 10^-17

    (9.1x10^-31)(vf)^2)/2 = -2.6 x 10^-17

    Then my issue: I end up needing to take the square root of a negative number...yah.

    If someone could please point me to the error I've made, it would be appreciated. I'm iffy on my "Ek(f)+U(f) = Ek(i)+U(i)". Did I use the correct equations? Please help, thanks.
  2. jcsd
  3. Mar 9, 2009 #2


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    Homework Helper

    You are releasing an electron which carries a negative charge. It looks like you haven't carried that into the calculation of U correctly.
  4. Mar 9, 2009 #3
    That would do it. I end up getting 7.53x10^6 m/s. Seems high - thoughts before I submit?
  5. Mar 9, 2009 #4


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    Homework Helper

    I didn't follow your calculation, but off the cuff with mass at 10-31 a little change in potential should go a long way.
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