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Homework Help: Gaussian infinite line charge, find force

  1. Jun 9, 2008 #1
    -----problem-------

    an infinite line of charge produces an electric field of 20N/C at a point that is 3 meters from the line. the electric field points toward the line of charge

    what is would be the force (magnitude and direction) on a negative 6 microcoulomb charge that is located 6m from the line?

    -----equations/constants----

    i want to use E = force/q_test but the equation does not factor distance of 6m in

    but if i use the F = kq_1q_2/r^2 equation i don't know the second charge

    what should i do?

    thanks

    electric field line of charge E = lambda/2pi(epsilon_o)r where lambda = 3.34*10^-9C/m, epsilon_o = 8.85*10^-12, r = 6m

    capacitance C = Q/V where q is charge, v is electric potential

    electric field E = F/q_test = kq/r^2 where F is force, k = 9*10^9, r is distance = 6m

    force F = kq_1q_2/r^2

    -----attempt-----

    if i use the E = F/q_test equation it does not allow me to factor in distance/radius r of 6m

    but if i use the F =kq_1q_2/r^2, i don't know what the the second charge is

    what equation should i use?

    thanks
     
    Last edited: Jun 9, 2008
  2. jcsd
  3. Jun 9, 2008 #2

    rl.bhat

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    Since the electric field points toward the line of charge, the charge on the line must be negative. The electric field 20 N/C at 3 m from the line of charge is the vector sum of forces by all the charges. So you can find the field at 6 m from the line by applying inverse square law.
     
  4. Jun 9, 2008 #3

    Dick

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    Use "E = lambda/2pi(epsilon_o)r" to find lambda. You know E when r=3m. Once you have lambda, use the same formula to find E at r=6m. Once you have done that look back and think about how you could have treated this as a ratio problem and skipped actually solving for lambda.
     
  5. Jun 9, 2008 #4
    so the inverse square law states Q/surface area = E area where Q is pt charge, since we are talking about a line charge, the surface area is the electric flux=2piLEr where L is length, E is electric field, r is radius ---> the E's cancel out though, how will i get force?

    E = -6*10^-6/2piLEr where r = 6m...

    http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html
     
    Last edited: Jun 9, 2008
  6. Jun 9, 2008 #5

    Dick

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    E's don't cancel out. L cancels out. What are you talking about? You have a formula "E = lambda/2pi(epsilon_o)r". You just have to figure out how to use it.
     
  7. Jun 9, 2008 #6
    when i did it L remained, the 2pi and r canceled out.

    what would having L, the length do? do i use it as distance/radius to solve for force?

    F = kq_1q_2/r^2 --> do i use L for r? i still don't have the second charge?

    i think i have completely missed something
     
  8. Jun 9, 2008 #7

    Dick

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    "E = lambda/2pi(epsilon_o)r", I keep putting that in quotes because you stated it. Why don't you use it? "F = kq_1q_2/r^2" only applies to point charges. Don't use it. A continuous line charge isn't a point charge.
     
  9. Jun 9, 2008 #8
    i did use that equation, the one in quotes, and set that equal to the E in the inverse square law. there is only one L in the inverse square law, the 2pi and r cancel but the L remains, ultimately you end up with length L.

    i tried using the inverse square law except E is on the left and right side of the equation and would get cancelled out, that is why i use the eq in quotes, which yielded a value for L.

    or i could use F = qE and use E = equation in quotes, because equation in quotes contains r plus i know lambda and epsilon_o

    i think that is the eq to get me the force

    correct?
     
  10. Jun 9, 2008 #9

    Dick

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    Didn't I say the inverse square law doesn't apply to things that aren't point charges? Why are you using it? You don't even need it. Yes, just use "E = lambda/2pi(epsilon_o)r" and F=qE.
     
  11. Jun 10, 2008 #10
    yes you were right, those eq's i got what i was looking for, F = 6.1*10^-5 N, away from the line
     
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