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Homework Help: What are the direction and magnitude of the net electric field

  1. Jan 18, 2008 #1
    1. The problem statement, all variables and given/known data
    There is an uneven arrangement of electrons and protons on a circular arc of radius r = 2 cm with an electron (E1) at 0 degrees, a proton (E2) at 30 degrees, an electron (E3) at 80 degrees, an electron (E4) at 130 degrees, and a proton (E5) at 160 degrees.
    What are the direction and magnitude of the net electric field.

    2. Relevant equations
    E = E1 + E2 + E3 + ...
    E = k|q|/r^2 r-hat

    3. The attempt at a solution
    I calculated the amount of force a single particle exerts.
    E1 = k|q|/r^2 = (8.99E9*|1*1.6E-19|)/(0.02^2) = 3.596E-6N
    All particles should exert the same magnitude of force, correct?
    Then I split the force magnitudes into their vector components.
    For the proton E2 this looks like E2 = (3.596E-6cos(30))i + (3.596E-6sin(30))j, and I repeated this for all the charges
    I added all these forces vectorially - and I got a total of 1.644E-6i + 9.324E-6j
    After this step I use the pythagorean a^2 + b^2 = c^2 identity to figure out the magnitude length of the force vector.
    This gives me an answer in N, but I want an answer in N/C, so I divide by (5*1.6E-19), because I have 5 particles and are the same distance from the point charge.
    It's wrong, though, by a gigantic order of magnitude.

    I don't know what to do. We are given an answer that says 160N/C towards P (which is not specific anyways, since there are 3 protons).
    Last edited: Jan 18, 2008
  2. jcsd
  3. Jan 18, 2008 #2
    Firstly, E1 = kq/r^2 gives you the magnitude of the electric field, not the force, so will be in N/C, not N. You'll also want to use the true value of q, not it's magnitude, since you need a direction as well. Assuming the point P is where you're measuring the field, and is at the locus of the arc, then you'd be right to decompose the field from each particle into components and sum them. The answer isn't particularly specific in 'toward P', since the electric field is a vector quantity, I'm guessing it means both components will be negative. You seem to have done mostly the right thing, it looks like dividing by the fundamental charge was your only mistake.

    http://img530.imageshack.us/img530/7130/ringiz6.th.png [Broken]
    Last edited by a moderator: May 3, 2017
  4. Jan 18, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Field not force

    Realize that E is the electric field due to the charge, which is measured in N/C, not N. (To get the force between two charges, you'd need to muliply by the second charge.)

    [Looks like marc1uk beat me to it! :smile:]
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