1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    "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

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gaussian infinite line charge, find force
Loading...