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Homework Help: Electric Force on a charge effected by a non-uniform linear charge

  1. Jan 14, 2010 #1
    Ok, so here's the question. I've pretty much got it except I'm having trouble with one part of constructing the integral. The problem is...

    An insulator which lies between the positions -d[tex]\hat{z}[/tex] and d[tex]\hat{z}[/tex] has a nonuniform linear charge density [tex]\lambda[/tex] = [tex]\lambda_{o}[/tex][tex]\frac{z}{d}[/tex] . Find the force on a charge 'q' located at 3d[tex]\hat{z}[/tex].

    2. Relevant equations

    [tex]\int[/tex] d[tex]\vec{E}[/tex] = [tex]\int[/tex][tex]\frac{k dq}{r^{2}}[/tex][tex]\hat{z}[/tex]

    dq = [tex]\lambda[/tex]dz

    [tex]\lambda[/tex] = [tex]\lambda_{o}[/tex][tex]\frac{z}{d}[/tex]

    3. The attempt at a solution

    What I ended up getting when constructing the integral is...

    [tex]\int[/tex][tex]^{ d}_{-d}[/tex]k[tex]\frac{\lambda_{0} z dz}{d r^{2}}[/tex][tex]\hat{z}[/tex]

    Now I'm having trouble with the r[tex]^{2}[/tex] part in the denominator...I know it should be something like...

    (r [tex]\pm[/tex] something)[tex]^{2}[/tex]

    ...but I'm just not sure what it is...any help would be greatly appreciated.

    Thanks!
     
  2. jcsd
  3. Jan 14, 2010 #2
    what is d?
     
  4. Jan 14, 2010 #3
    d is the length of the insulator (from -d to d), as stated in the question.
     
  5. Jan 14, 2010 #4
    okay I see what you want I think. Both the point that you are trying to find the force on, and the insulator are parallel?
    And does the sign on the charge change at the origin? Or is it just the magnitude of the distance from the origin that matters for the charge?
     
  6. Jan 14, 2010 #5
    not exactly, the point and the insulator are both on the same axis...the insulator spans (-d) to (d) and the point is at 3d all along the z axis (z-hat)

    also yes, the charge density is non uniform so it changes at the origin as you stated...lambda z/d changing sign as a function of it's position on the z axis.
     
  7. Jan 14, 2010 #6
    i would do two seperate integrals. the first being from -d --> 0.

    your integral looks good. I would just put on the denominator (4d + z)^2

    and when it goes from 0 ----> d, put on the bottom (3d -z)^2

    but makes sure from -d ---> 0 you have the -z value subbed in.
     
  8. Jan 14, 2010 #7
    whoops it seems i have done that wrong. (3d-z)^2 will work for it all. forget about (4d + z). So I might still do the two integrals seperately, but they should have the same denominator.
     
  9. Jan 14, 2010 #8
    ok, i think i got it now...

    [tex]\int^{d}_{-d}[/tex][tex]\frac{k\lambda_{o}zdz}{d(3d-z)^{2}}[/tex]

    and then removing constants...

    [tex]\frac{k\lambda_{o}}{d}[/tex][tex]\int^{d}_{-d}[/tex][tex]\frac{zdz}{(3d-z)^{2}}[/tex]

    but still the integral doesn't turn out nicely...should i be using the length of the vector r

    |[tex]\overline{r}[/tex]| = [tex]\sqrt{(3d)^{2} - (z)^{2}}[/tex] ?

    if i run it through all the way with the |[tex]\overline{r}[/tex]| i end up with a nice integral that i can substitute and the resulting field at 3d ends up being zero...it seems reasonable enough but is it right...i don't know...?

    i'm relatively happy with that answer so if you see something fundamentally wrong with it you can let me know...otherwise thanks for you help!!
     
  10. Jan 15, 2010 #9
    it cant be zero if i understand the charge density properly. Did you try two integrals, one from -d to 0 and one from 0 - d?
     
  11. Jan 15, 2010 #10
    it cant be zero. do the integral with (3d - z)^2 on the bottom. i got k*lambda/d(3/4 + ln(1/2))
     
  12. Jan 15, 2010 #11
    ok, yeah got it now...thanks
     
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