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Homework Help: Electrons to coloumbs

  1. Feb 19, 2010 #1
    I have a homework question with regards to a charged particle in an electric field.
    The particle has a charge of -4e
    The field is constant at 45000 N C^1

    For this i have been using the equation F_el= q E(r)

    The trouble i have been having is trying to work out -4e in Coloumbs.
    I know -e=1.602 * 10^-19 coloumbs but is it as simple as 4 * 1.602 * 10^-19?
    It just seems to give me a really tiny force in Newtons. I know its only a small particle but something tells me its just not right.

    Any help if im going the right or wrong way would be much appreciated :)
  2. jcsd
  3. Feb 19, 2010 #2
    Yep. The charge of an electron is just that, a unit of charge. If you have 4 times the charge of one electron, you have 4*e of charge.

    The force is small, since the charge is very small.
    To put it in proportion, let's look at the mass of the electron and the acceleration it experiences as a result of the electric field. You should note, that the field we're dealing with here is MASSIVE in terms of laboratory fields.

    Writing out Newton's Second Law for a stationary electron (Ignoring the directions of force, acceleration and field):





    [tex]m=9.1\cdot 10^{-31} kg[/tex]

    [tex]q=1.6\cdot 10^{-19} C[/tex]

    [tex]E=45,000 \frac{N}{C}[/tex]

    [tex]a=7.9\cdot 10^{15} \frac{m}{s^2}[/tex]

    That force doesn't seem too tiny now, does it?
  4. Feb 19, 2010 #3
    Thanks for your quick reply. I agree it is a massive force and am a bit worried about my previous calculations now.

    The field is produced by a fixed charge of 2.00*10^-11C at a diastance of 2mm

    I used the equation for the field
    E(r) = Q/ 4pi*E0*r^2

    which is electric field at distance r is equal to point charge Q divded by 4pi * the value for the quantity of the permittivity of free space (9*10^9) *( 2.00*10^-3)^2

    I arranged to equal 9*10^9 * Q/r^2

    That gave me 45000 N C^-1

    I hope that makes sense :)
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