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E-field question

  1. Nov 7, 2003 #1
    [SOLVED] E-field question

    I'm new here but hope thats not a problem. I have an e-field problem that i was wondering about. Now this is how it was presendetd to me.

    A charge of uniform density 4.0 nC/m is distributed along the x axis from x=-2.0 m to x=+3.0m. What is the magnitude of the electric field at the point x=+5.0m on the x axis?

    this is a fealer quesition. i have some outers but i sorta whant to test this system if you will.
  2. jcsd
  3. Nov 7, 2003 #2


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    Hi zath,

    Welcome to PhysicsForums. :smile:

    Can you evaluate an integration of Coloumb's Law over the line of charge?

    - Warren
  4. Nov 7, 2003 #3
    lambda = 4e-9 C/m
    x1=-2, x2=3, x=5 m

    Q=lambda/x = 0.8e-9 C = 800pC

    ...so you have an line charge streched from x1(-2,0) to x2(3,0). that charge produces an uniform electric field with vector x from x1 and x2, and vector y 0. if you want to measure electric field in y(5,0), there is no electric field there, because there is no charge there.

    welcome to the forums
    Last edited by a moderator: Nov 7, 2003
  5. Nov 7, 2003 #4

    You better think about this some more.
  6. Nov 7, 2003 #5


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    Oh? Why would that be?

    It is true that for an infinitely long line of charge, the E-field will have no component parallel to the line, but this line of charge is NOT infinitely long.

    First, let's do a sanity check. Let's say all 20 nC of the charge was concentrated at the origin, and calculate the field due to it 5 m away.

    E = 1 / (4 pi e0) * q / r2

    = 7.19004 N/C = 7.19004 V/m

    So our answer to this question should be roughly this size -- if its end up way bigger or way smaller, we probably made a mistake.

    Let's break the line of charge up into infinitesimal pieces, and sum the fields due to each piece. This is the superposition principle -- the field due to a number of charges is the simple sum of each of the charges considered in isolation.

    Let's call a little piece of the line ds. How much charge, dq, is on that little piece of line? Simple. dq = [lamb] ds.

    How far is that little piece of the line from our measurement location at x = 5? Simple, r = 5 - s.

    What's the field due to the charge dq on that little piece of the line? It's

    dE = 1 / (4 pi e0) * [lamb] ds / (5 - s)2

    Notice that I've subsituted the expressions for dq and r that I discovered above.

    What's the field due to all the little pieces of the line summed together? It's an integration of the field contributed by each piece of the line, over the length of the line, like this:

    E = Integral (from s = -2 to s = 3) of 1 / (4 pi e0) * [lamb] ds / (5 - s)2

    Doing the integral and plugging in the limits of integration gets me 12.8394 N/C = 12.8394 V/m.

    You can assume from symmetry that this field is directed along the positive x-axis, away from the line of charge.

    Does this make sense?

    - Warren
  7. Nov 7, 2003 #6
    omg...sorry, i mis-read. i tought it said 5,0 on y axis.
  8. Nov 7, 2003 #7


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    There'd be a non-zero E field there, too!

    - Warren
  9. Nov 9, 2003 #8
    how if the line charge is developed on the x axis? if volume charge is developed then it wouldn be zero...
  10. Nov 9, 2003 #9
    Even a point charge has a spherical field, right?
  11. Nov 9, 2003 #10
    thank you very much, you got my question answerd very well (sorry that i'm so long in replying). One quesiton for you. When i do the integration i gt et the same 12.8394 and there is no question about that. When i do it long hand i get some funny units that didnt match yours, so i envoked the power of TI and punched the units into the equation as well, and i ether get only V, or i get T/s. i get voults if i leave the (5-s) demenstionless and i get the T/s if i put the (5-s) as m. What am i doing wrong??
  12. Nov 10, 2003 #11


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    Coulomb's law involves the following units: e0 has the units C2/J-m; q has unit C; r has unit m.


    E (N/C) = 1/4 pi e0 [J-m/C2] * q/r2 [C/m2]

    Cancelling the units, you'll see that E is indeed in J/m-C, where a J/m is the same as Newtons. The E field is indeed N/C.

    You can see how N/C = V/m by considering the units of electrical potential. You can see in the expression

    V = (1/4 pi e0) (q / r)

    that potential has units of J/C.

    Furthermore, since force has units of J/m, you can see quickly that N/C and V/m are the same unit.

    - Warren
    Last edited: Nov 10, 2003
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