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Electric Field's Direction

  1. Jan 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A negative point charge q_1 = -4.00 nC is on the x-axis at x = 0.60 m. A second point charge q_2 is on the x-axis at x = -1.20 m.

    a. What must the sign and magnitude of q_2 be for the net electric field at the origin to be 50.0 N/C in the +x-direction?

    b. What must the sign and magnitude of q_2 be for the net electric field at the origin to be 50.0 N/C in the -x-direction?



    2. Relevant equations

    E = k*q/r^2

    3. The attempt at a solution


    a. The charge must be negative for subtraction to occur.

    (-q2) O__<____. _>__O (-q1)
    -----E2 = (-)------E1= (+)
    negative diagram
    . = field point (0,0)
    O = particle
    Arrow indicates field direction based on positive test charge

    E1 = (8.988*10^9)*(-4.00*10^-9 C)/(.60 m)^2 = 99.8667 N/C (+)

    99.8667 N/C + E2 = 50 N/C
    E2 = -49.8667 C

    q_2 = r^2*(49.86667)/(8.988*10^9) = -7.9893*10^-9 C ?????




    b. E1 = (8.988*10^9)*(-4.00*10^-9 C)/(.60 m)^2 = 99.8667 N/C (+)

    E2 = -50 N/C – 99.8667 = -149.8667 C

    q_2 = [(r^2)149.8667 N/C]/k = [(r^2)149.8667 N/C]/k = (1.2^2)*149.8667/(8.988*10^9) = -2.40*10^-8 C ??????

    Thanks.
     
  2. jcsd
  3. Jan 12, 2007 #2

    berkeman

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    Staff: Mentor

    For a), why do you say "The charge must be negative for subtraction to occur"? E field points from + charge to - charge, and a) wants a net positive E field at the origin, doesn't it? Maybe I'm misreading something.
     
  4. Jan 12, 2007 #3
    u should know that q1 and q2 exert on the origin ... so there is 2 forces what means there is 2 electric fields..
    the net electric fields should be =50 and in the direction of the force exert by q2 on the origin...u have this condition and the equations ...u should solve it i think
    tr to get the net force on the origin assuming theer is test point charge. then using the condition u have and the values u will have the answer.
     
  5. Jan 12, 2007 #4

    BobG

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    Science Advisor
    Homework Helper

    Part a is correct.

    You can look at this by proportions as a sanity check. q2 is twice as far away as q1, so if it had an equal charge, the strength of its field at the origin would be 1/4 as strong as q1's. Since the field was half as strong as it would be with q1 alone, the charge on q2 has to be about twice as strong as q1.
     
  6. Jan 12, 2007 #5
    hey BobG is what i said true? just to check if i know the concept
     
  7. Jan 12, 2007 #6

    BobG

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    That's a tough question.

    You could insert a test point of some mass (1kg to keep things simple), but you don't really need it. You'd be multiplying everything by your test mass, including your field at the origin.
     
  8. Jan 12, 2007 #7
    Is my answer for Part B incorrect?
     
  9. Jan 12, 2007 #8

    BobG

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    Science Advisor
    Homework Helper

    No, I think it's correct. I just didn't look at it. A quick check comparing the proportions looks right.
     
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