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Line of charge, Feild on the origin

  1. Jul 5, 2008 #1
    1. The problem statement, all variables and given/known data
    A line of charge with a uniform density of 35.0 nC/m lies along the line y = -12.6 cm, between the points with coordinates x = 0 and x = 44.2 cm. Calculate the electric field it creates at the origin. Find the x- and y-components of the feild.

    2. Relevant equations
    I beleive i have to use an integrated formula but im not sure which one.

    3. The attempt at a solution
    I'm not sure if i have the problem straight in my head, but if i do, then i dont know how to approach it. I know that i have to get the x and y components from the magnitude, and i know how to do that, but i don't really know how to find the magnitude.
  2. jcsd
  3. Jul 5, 2008 #2
    if this was a point charge, you would figure out the electric field by using E = kQ/r^2, right? So for a continuous charge distribution, dE = k*dQ/r^2, which you will have to integrate.
  4. Jul 5, 2008 #3
    so after integration, the formula is:

    E = k*Q/(x^2 + a^2) right?
  5. Jul 5, 2008 #4
    no, that cant be right. If it were, that would mean the magnitude was only 659 N/C which is way to low. I must not be integrating properly?
  6. Jul 5, 2008 #5
    well no, thats actually more like the equation before integration (with dE and dQ). Once you get [tex]\int k \frac{dQ}{x^2 + a^2}[/tex] there's a problem because you're integrating with respect to Q (hence the dQ) but you want to integrate with respect to x. So replace dQ with [tex]\lambda dx[/tex]. Now you're integrating x with respect to x.
  7. Jul 5, 2008 #6
    so then that leaves me with E = [tex]\frac{-k\lambda}{x} - \frac{k\lambda}{a}[/tex]
    Last edited: Jul 5, 2008
  8. Jul 5, 2008 #7
    hmm i dont think so, i dont remember off the top of my head what this integral would evaluate to but i'm relatively sure it involves a trig substitution.
  9. Jul 5, 2008 #8
    Hm. Ok, ill have to put this one down for a bit and come back to it. Doing the integral is what is most confusing to me.
  10. Jul 5, 2008 #9


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    You have to set up the differential electric fields in the x and y directions properly first. You can't just integrate with respect to Q because that doesn't take into account the geometrical setup of the problem.

    Now, you can see by means of a diagram (draw it) that [tex]dE_y = \cos \theta dE[/tex], where [itex]\theta[/itex] ranges from 0 to some value and is basically the angle in between the y-axis and the imaginary radial line connecting the differential charged length as it sweeps from x=0 to x=44.2 cm.

    So we have [tex]dE = \frac{\lambda dx}{4 \pi \varepsilon r(x)^2}[/tex]. I wrote r(x) instead of r because r is not a constant. Find an expression for r, the length from the origin to the charged dx line segment. Now when you're done, plug in [itex]dE_y[/tex] into the formula for [itex]dE[/itex] above. You still have to express either cos theta in terms of x or all the x values in the dE expression as functions of theta before performing the integration. I recommend doing the former since it's easier to express cos theta in terms of x and some other constants. Once, you're done simply integrate over the length of the charged line.

    The expression for [itex]E_x[/itex] can be found in a similar manner.
  11. Jul 5, 2008 #10
    So doing the integral of:
    dE = \frac{\lambda dx}{4 \pi \varepsilon r(x)^2}

    E = \frac{-\lambda}{4 \pi \varepsilon x}

  12. Jul 5, 2008 #11


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    You're supposed to be integrating to find either [tex]E_y \ \mbox{or} \ E_x[/tex]. But more importantly you haven't found r(x) yet. r(x) as I wrote it means r as a function of x, not r*x, as you appear to have written it.
  13. Jul 5, 2008 #12
    OK, so just that i have this correct.. i neet to integrate two equations one for Ey and one for Ex. Those would be:
    dEx = dECos[tex]\theta[/tex]
    dEy = dESin[tex]\theta[/tex]

    This means that [tex]Ex = \frac{kx}{(x^2 + a^2)^\frac{3}{2}}Q[/tex]
    and Ey is the same thing only y values in place of x values?
  14. Jul 5, 2008 #13


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    What is Q here? You're supposed to set up the integration in terms of [itex]dE_x[/itex], so how did you get that expression? REmember that the final answer for both Ex and Ey is a numerical value, it should not have x and y in it at all. I assume a here is 12.6cm.
  15. Jul 5, 2008 #14
    Q is the charge on the rod. I used circular geometry and the lambda given to get the charge. Since radius is constant, its easy to find the circumfrence of what would be a circle. Then i used the ratio of 56:304 degrees to find the lenght of the rod. Length of rod multiplied by charge/length gives charge.
  16. Jul 6, 2008 #15


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    You're not understanding the technique for doing such problems. First you have to find the differential electric field contribution due to each differential charged segment, then express it in terms of a variable and then integrate it over the geometrical parameters of the problem. I have given you the expression for [itex]dE_y[/itex] and told you to express [itex]\cos \theta[/itex] in terms of x so that the integration can be performed over the length of the charged rod. As as I have said earlier, the expression for [itex]dE_x[/itex] can be found in a similar manner.

    And as for your reply, I think you've confused this thread and another thread of yours on the e-field due to a circular arc of charge. This one concerns a line, and the "radius", which I take to be the distance between the origin and the differential charged segment is clearly not constant.
  17. Jul 6, 2008 #16
    Ok. So the geometrical parameters for this problem only range from 0 to 44.2 cm along the x-axis. this causes the hypotenuse to change. I beleive this is what you meant by saying r(x) as in r as a function of x? Im thinking this would be [tex]r(x) = \frac{x}{\sqrt{x^2 + y^2}}[/tex] .

    [tex] Cos\theta [/tex] expressed in terms of x is simply r(x) as i have it above, is it not?

    If this is correct, then doing the integral still gives:


    E = \frac{-\lambda}{4 \pi \varepsilon x}


    which is to be evaluated from 44.2 to 0, right?
    and yes i did get this problem confused with the other one for that last reply.
    Last edited: Jul 6, 2008
  18. Jul 6, 2008 #17


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    r(x) is simply the distance from the differential charged segment [itex]\lambda dx[/itex] to the origin. Your expression is not correct. Just use pythagoras theorem to get it.

    As for [itex]\cos \theta[/itex], which angle are you using here? You have to be consistent with your choice of angle when you perform the integration later. The one I used is the angle in between the y-axis and the line connecting the origin to the charged segment dx. You could use the other angle as well. Your current expression for cos theta uses the other angle.

    The expression I wrote for [tex]dE_y[/tex] uses the angle I described earlier. And before doing the integration I suggest you post the expression for either [itex]dE_y[/itex] or [itex]dE_x[/itex] first, because your answers don't look look correct and I can't tell what's wrong.
  19. Jul 6, 2008 #18
    r(x) is 46 cm.

    The theta would be 15.9 deg.

    The expression would be:


    Ex = \frac{\lambda x^2}{4 \pi \varepsilon r(x)^3}

    exaluated from 44.2cm to 0?
  20. Jul 6, 2008 #19


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    I don't know how you got that, but that's not right. r(x) is a function of x, not the some numerical distance in itself. But before I go on, I need to know what you already know. Do you know how to evaluate an electric field at a point due to continuous charge distributions such as line of uniform charge and ring of uniform charge?
  21. Jul 6, 2008 #20
    Well yes, but we're given specific equations for those. For example, E=2k lambda/r for an infinite line of continuous charge where r is a distance from that line. Continuous charge of a sphere (and you're looking for a point withing the sphere) is E = kQr/a^3 where a is the radius of the entire sphere and r is the distance from the center of the sphere to the point inside the sphere that you are looking for. This is pretty much the first time i've had to integrate on my own to get the formula for a question.
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