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Superposition of total electric field

  1. Dec 12, 2009 #1
    2 Cylinders each of length, L, are separated by a distance d. Each has a radius, a. Use the principle of superposition to find the total electric field at a distance, r, from the 1st cylinder.

    What I know so far: For One cylinder
    Applying gauss law, i have E(2([tex]\pi[/tex])(r)(L) = charge enclosed/ [tex]\epsilon[/tex]0
    so solving for E, E = [tex]\sigma[/tex]a/[tex]\epsilon[/tex]0r

    I'm not sure how to apply the principle of superposition for both fields combined. Any advice?

    Attached Files:

    Last edited: Dec 12, 2009
  2. jcsd
  3. Dec 13, 2009 #2
    The principle of superposition just says that to find the field for both cylinders, just add the fields each one would have in isolation.

    Your problem and diagram don't really have enough information--it would help to know whether the cylinders are conductors or have a charge density as a function of r? How much charge total? Also, we're supposed to assume L>>r?

    Here's the problem I'll do to demonstrate the principle of superposition: two infinite parallel line charges with lambda charge per unit length, one at position r=0 and the other at r=d in the x direction.

    Solving for the field around one line charge using Gauss's law,
    [tex]\overrightarrow{E}=\frac{\lambda}{2\pi \epsilon _{0}} \frac{1}{r}\widehat{r}[/tex]

    Now to do the superposition principle, let's superimpose some cartesian coordinates. Call the direction along the line charges the y direction, and let the direction from the left line charge to the right line charge be the x direction. We'll find the field as a function of x and z, since it is independent of y. The left line charge is our origin, i.e., [tex]\overrightarrow{r}(\textrm{left line charge})=\overrightarrow{0}[/tex] Therefore, the position of the right line charge is [tex]\overrightarrow{d}=d\widehat{x}[/tex]. Call the vector from the left line charge to our field point [tex]\overrightarrow{r}=x\widehat{x}+z\widehat{z}[/tex] and the vector from the right line charge to the field point [tex]\overrightarrow{{\mathfrak{r}}}[/tex].

    So superposing these two fields,
    [tex]\overrightarrow{E_{total}}=\overrightarrow{E_{L}}+\overrightarrow{E_{R}}=\frac{\lambda }{2\pi \epsilon _{0}}\frac{1}{||\overrightarrow{r}||}\widehat{r}+\frac{\lambda }{2\pi \epsilon _{0}}\frac{1}{||\overrightarrow{\mathfrak{r}}||}\widehat{{\mathfrak{r}}}=\frac{\lambda }{2\pi \epsilon _{0}} \left (\frac{\overrightarrow{r}}{x^{2}+z^{2}} +\frac{\overrightarrow{r}-\overrightarrow{d}}{(\left x-d \right )^{2}+z^{2}} \right )=\frac{\lambda }{2\pi \epsilon _{0}} \left (\frac{x\widehat{x}+z\widehat{z}}{x^{2}+z^{2}} +\frac{\left ( x-d \right )\widehat{x}+z\widehat{z}}{(\left x-d \right )^{2}+z^{2}} \right )[/tex]
    Kind of a messy form but I'm sure you could massage it a bit.

    Now the field outside a charged cylindrical is equal to that but zero inside the cylinder's radius, so you can easily make that correction for the problem.
    Last edited: Dec 13, 2009
  4. Dec 13, 2009 #3
    Yes, we assume that L>>a (the radius of the cylinders). The one on the left has a (+sigma) density charge and the one on the left has a (-sigma) density charge. Hope this information helps!
  5. Dec 13, 2009 #4
    Oh well I did it in terms of charge per length instead of charge per area. You can easily swap out lambda. For a cylinder radius a,
    [tex]\lambda =2\pi a\sigma [/tex]
  6. Dec 13, 2009 #5
    Ok, so that would be the combined electric force at that point, r, away from the 1st cylinder?
  7. Dec 13, 2009 #6
    Well the answer I gave you is for two infinite line charges. It tells you the field at a vector [tex]\overrightarrow{r}=x\widehat{x}+z\widehat{z}[/tex] away from the left line charge.

    I don't know if you want to treat your finite cylinders as infinite. (Given only L>>a is not good enough, because maybe r>>L and the cylinders could look like point charges.) If you do, that will be the answer provided r is not inside one of the cylinders. If the point r is inside one of the cylinders, the contribution from the line charge associated with that cylinder would disappear.

    Also, if your cylinders are conducting cylinders I think there may be some image charge effects that I'm not really sure of.
    Last edited: Dec 13, 2009
  8. Dec 13, 2009 #7
    They are treated as infinite and the distance r is never inside the other cylinder. the distance of the cylinders apart from each other is d. And we want to find the total e-field at a point r (for r<d) The cylinders are non conducting cylinders well. Does this make more sense now?
  9. Dec 13, 2009 #8
    Yes, it makes a lot more sense now.

    Keeping in mind [tex]\overrightarrow{r}=x\widehat{x}+z\widehat{z}\Rightarrow r=\sqrt{x^2+z^2}[/tex]
    For all points r<d,
    \overrightarrow{E}(x,z)=\frac{a\sigma }{\epsilon _{0}}\left\{\begin{matrix}
    \frac{(x-d)\widehat{x}+z\widehat{z}}{(x-d)^{2}+z^{2}} & r<a\\
    \frac{x\widehat{x}+z\widehat{z}}{x^{2}+z^{2}}+\frac{(x-d)\widehat{x}+z\widehat{z}}{(x-d)^{2}+z^{2}}& a<r<d\\

  10. Dec 13, 2009 #9
    how did u get the answer[tex]\alpha\sigma[/tex]/[tex]\epsilon0[/tex]??

    for a infinite line charge its just E= [tex]\lambda[/tex]/2[tex]\pi\epsilon0\alpha[/tex] for one of the cylinders rite? and then if we apply the superposition principle we get [tex]\alpha\sigma[/tex]/[tex]\epsilon0[/tex]??
  11. Dec 13, 2009 #10
    Here's the equation:

    [tex]\overrightarrow{E_{\textrm{single line charge}}}(\overrightarrow{r}) = \frac{\lambda}{2\pi \epsilon _{0}} \frac{1}{r}\widehat{r} =\frac{a\sigma }{\epsilon _{0}}\frac{1}{r}\widehat{r}[/tex]

    You have one of each of those for the left and right line charge. The left and right line charges have two different r vectors, as I stated above, where I called them [tex]\overrightarrow{r}[/tex] and [tex]\overrightarrow{\mathfrak{r}}[/tex].
    That's what lead to that big formula leading to those messy x and z expressions.

    However, those big x and z expressions give the answer for two LINE charges. To change the answer to be for a cylinder and satisfying all the other conditions you listed, the contribution from the left line charge disappears for r<a.
    Last edited: Dec 13, 2009
  12. Dec 13, 2009 #11
    ook, sounds good, thanks so much!
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