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Magnitude of elec field to balance electron?

  1. Dec 4, 2005 #1
    Hi all,

    Im doing a question in the homework, its #13 and Im struggling on how to get started.

    The question is: what are the magnitude and direction of the electric field that will balance the weight of (a) an electron and (b) a proton?

    There is an answer in the book, for the electron "55.8 pN/C down"

    This is day 2 of my trying to figure this one out, since the electron has a simple single charge of "e" how to get the force to balance -1.602 E-19 C ? Im searching the book to find out how to convert this into N/C

    Ive used the examples on the books companion site, move the charge around and it shows the vector and resulting charge, neat. So Im guessing since the weight is so small, one only has to make a field that counters one charge, but 55.8 pN/C??

    I already found a typo in the book, it was showing a right triangle, the 2 sides = .1 and the long side was square root of 2 times .1? so now I dont trust the book, and cannot begin to see how they got the 55 number there, I was going to answer just -1.602 E-19 because one electron should balance the other?? no thats probably not it?

    well thanks in advance
  2. jcsd
  3. Dec 4, 2005 #2
    The force needed to balance the particle can be found by coulomb's law.
    The electric field is [itex] E = F/q_0 [/itex]. This will give units of N/C.
  4. Dec 5, 2005 #3
    ok I found out that the direction of the E-field must be vertically downward. This will generate an upward electrical force since the electron has a negative charge -e.

    For the magnitude of the E field, solve

    m g = e E

    E = g m /e where g is the acceleration of gravity. now the number comes out correctly.
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