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Potential due to a line charge

  1. May 10, 2017 #1
    1. The problem statement, all variables and given/known data

    A line charge with uniform charge distribution and linear charge density ##\lambda## lie along z-axis from ##(0,0,-d)## to ##(0,0,+d)##. Find electric field due to this charge along z, x axis and potential at ##(0,0,2d)##, ##(\sqrt{3}d, 0,0)## and ##(3d/2, 0 , d)##.

    2. Relevant equations


    3. The attempt at a solution

    ##\displaystyle dE_z = \dfrac{dQ}{R^2} = {\lambda dR \over R^2}##

    Hence, ##\displaystyle E_z = \int^{z+d}_{z-d} {\lambda dR \over R^2} = {2\lambda d \over z^2 - d^2}##

    Where ##z## is a point along z-axis.


    Now for potential
    ##\displaystyle \phi_z = -\int^z_0 E_z dz = -\int^z_0 {2\lambda d \over z^2 - d^2} dz = -\lambda\ln \left| d^2 (z-d) \over z + d\right|##

    Now for x-axis,

    Physics potential.png

    ##\displaystyle dE_x = {dQ \cos \theta \over R^2} = {\lambda dz \cos \theta \over R^2} = {\lambda dz \cos \theta \over x^2 + z^2} = {\lambda x \over (x^2 + z^2)^{3/2}}dz##

    So, ##\displaystyle E_x = \int^{d}_{-d}{\lambda x \over (x^2 + z^2)^{3/2}} dz = {2\lambda d\over x(d^2 + x^2)^{1/2}}##

    Now for potential,

    ##\displaystyle \phi_x = -\int^x_0 {2\lambda d \over x(x^2 + d^2)^{1/2}} = -\lambda \ln\left| \sqrt{d^2 +x^2} - d \over \sqrt{x^2 + d^2 } + d\right|##

    Now we need to find potential at ##(0,0,2d)##, ##(\sqrt{3}d, 0,0)## and ##(3d/2, 0 , d)##,

    Plugging ##(0,0,2d)## in ##\phi_z## I got ##\lambda \ln 3##.

    Plugging ##(\sqrt{3} \times d,0,0)## in ##\phi_x## I got ##\lambda \ln 3##.

    So far so good,

    For potential at ##(3d/2, 0 , d)##, I plugged this coordinate in

    ##\displaystyle \phi (x,0,z) =\phi_x + \phi_z = -\lambda \ln\left| \sqrt{d^2 +x^2} - d \over \sqrt{x^2 + d^2 } + d\right| + -\lambda\ln \left| d^2 (z-d) \over z + d\right|##

    Now as you see for ##z = d## the expression on the far right is undefined because ##\ln 0## is undefined.

    What to do now ? I got every answer correct except this one. The given answer is ##\lambda\ln 3##.

    What did I do wrong ?
     
    Last edited: May 10, 2017
  2. jcsd
  3. May 10, 2017 #2

    haruspex

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    Why?
    Your φx is shorthand for φ(x,0,0). Similarly φz.
     
  4. May 10, 2017 #3
    I guess because ##\phi## is scalar, so it adds up like a scalar.

    ##\displaystyle \phi = \int \vec E \cdot d\vec s = \int E_x dx + E_ydy + E_z dz = \phi_x + \phi_z##

    I took ##\phi_y = 0## because we are not considering it.

    Is this correct ?
     
    Last edited: May 10, 2017
  5. May 10, 2017 #4

    haruspex

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    It is a potential, so adds up like a potential. The potential at B is the potential at A plus the potential difference from A to B.
    Don't know what you mean about its being scalar. You seem to be assuming it is linear, i.e. φ(x+x',y,z)=φ(x,y,z)+φ(x',0,0).
    (Edited... wrote φ(x',y,z) by mistake.)
     
    Last edited: May 10, 2017
  6. May 10, 2017 #5
    So ##\displaystyle \phi(x, 0, z) = \phi_x + \int_{(x, 0 , 0 )}^{(x,0,z)} \vec E \cdot d\vec s## is correct ?
     
    Last edited: May 10, 2017
  7. May 10, 2017 #6

    haruspex

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    Yes.
     
  8. May 10, 2017 #7
    But now how I am going to evaluate this ? I am confused a bit,

    Should I do ##\displaystyle \int E_x dx + E_y dy + E_z dz = \int E_z dz = \phi_z## ?
     
  9. May 10, 2017 #8

    haruspex

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    Your notation confuses me, and it might be confusing you too.
    Ex etc. usually mean x component of the field at a general point, i.e. Ex(x,y,z). But in some of your equations you have used Ex to mean Ex(x,0,0) and Ez to mean E(0,0,z).
    If you take the potential at the origin to be 0 then ##\phi(x,y,z)=\int_SE_x dx + E_y dy + E_z dz## where S is any path from the origin to (x,y,z). In particular, you could take the path (0,0,0)-(x,0,0)-(x,0,z) to reach (x,0,z), so
    ##\phi(x,0,z)=\int_0^x E_x(u,0,0) du + \int_0^z E_z(x,0,w) dw##.
    But first you need an expression for Ez(x,0,z).

    An alternative approach is to consider the potential at (x,0,z) due to some element of the line of charge and integrate along the charge.
     
  10. May 10, 2017 #9
    Lol :H:H , you are correct, I confused myself with my notation. I was adding potetial compoenent wise, what an idiot.
     
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