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Electric Field in all of space from charges along the Z-axis

  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A charge is distributed evenly along the ##z## axis with a density ##\lambda = 3 \frac{\mu C}{m}##. There is also a cylinder with radius ##r = 2## and on his side there is a superficial density charge of ##\sigma = -\left( \frac{1.5}{4 \pi} \right) \frac{\mu C}{m^2}##. Both distributions are unlimited on the ##z## axis that coincide with the axis of the system.

    2. Relevant equations
    Gauss Theorem:
    ##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda (x, y, z) dl##
    ##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma (x, y, z) dS##

    3. The attempt at a solution
    So I started first by calulating the electric field inside the cylinder with a ##r_0 < r##. Since it is only the charge on the ##z## axis creating the field, we will use ##\lambda## and integrate on the line.
    ##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda (x, y, z) dl##
    ##\int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda dl##
    ##E_0 2 \pi r_0 l = \frac{1}{\epsilon_0} \int_l \lambda dl##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r_0 l} \int_l \lambda dl##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0 l} \int_l dl##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda l}{r_0 l}##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##

    Then I calculated the electric field at ##r_0 = r##. Now there are both the charge on the ##z## axis and the charge on the cylinder's side ##\sigma##.
    For the contribution of the linear charge we have ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r}##.
    For the contribution of the superficial charge we have:
    ##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma (x, y, z) dS##
    ##\int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma dS##
    ##E_0 2 \pi r l = \frac{1}{\epsilon_0} \int_S \sigma dS##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r l} \int_S \sigma dS##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_S dS##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_{0}^{r} 2 \pi l dr##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi l}{r l} \int_{0}^{r} dr##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi r l}{r l}##
    ##E_0 = \frac{\sigma}{\epsilon_0}##
    So ##E = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r} + \frac{\sigma}{\epsilon_0}##

    In the end I calculated the electric field for ##r_0 > r##. Like before, there are both the charge on the ##z## axis and the charge on the cylinder's side ##\sigma##.
    For the contribution of the linear charge we have ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##.
    For the contribution of the superficial charge we have ##\frac{\sigma}{\epsilon_0 r_0} (r_0 - r)##.
    So ##E = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0} + \frac{\sigma}{\epsilon_0 r_0} (r_0 - r)##

    Is it all correct?
     
  2. jcsd
  3. Apr 25, 2016 #2

    TSny

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    Your answer looks correct for the case ro < r.

    I would not try to use Gauss' law for the special case of r0 = r. For this value of r, you are at the location of the surface charge density where the E field is indeterminate.

    For the case r0 > r, you have not found the correct charge inside the Gaussian surface due to the surface charge density σ. There should not be any integration with respect to r. Since σ is constant, you have ##\int_{S} \sigma dS = \sigma S##. Be sure to draw a figure which helps you identify the surface ##S##.
     
  4. Apr 25, 2016 #3
    Why is it indeterminate? Shouldn't ##E## be the maximus on the surface?

    Why? I'm assuming a cylinder with a radius ##r_0 > r## and I want to find the electric field there. Shouldn't it be this the contribution of the surface charge:
    ##E_0 2 \pi r l = \frac{1}{\epsilon_0} \int_S \sigma dS##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r l} \int_S \sigma dS##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_S dS##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_{r}^{r_0} 2 \pi l dr##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi l}{r l} \int_{r}^{r_0} dr##
    ##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi l}{r l} (r_0 - r)##
    ##E_0 = \frac{\sigma}{\epsilon_0 r} (r_0 - r)##
     
  5. Apr 25, 2016 #4

    TSny

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    I think you'll see why E is undefined for points on the surface of the cylinder once you have worked out E for the inner and outer regions.
    Shoudn't the ##r## on the left side be ##r_0##?
    At this point, use the fact that ##\int {dS} = S## and just substitute the correct expression for S.
    It doesn't make any sense to integrate over r. You are integrating over part of the surface of the cylinder where ##r## is a fixed number, not a variable.
     
  6. Apr 25, 2016 #5
    Yeah, you're right. My bad.
    So simply ##S = 2 \pi r_0 l##?
    Then I have ##E_0 = \frac{\sigma}{\epsilon_0}##. Is it right?
     
  7. Apr 25, 2016 #6

    TSny

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    Think about whether you should have ##r## or ##r_0## in this expression.
     
  8. Apr 25, 2016 #7
    Gauss Theorem says that ##\Phi## of the electric field in the void ##E_0## through a closed surface ##S## is the integral of the charge inside ##S## divided by ##\epsilon_0##. The surface we are talking about in this case is the cylinder with radius ##r## while the electric field we want to find out is at radius ##r_0 > r##. So on the left, where there is ##S##, we have ##r_0##, while on the right, where there is the other ##S##, we have ##r##. This is because on the left we have the surface with radius ##r_0## and on the right the surface with the surface charge ##\sigma##, so the one at radius ##r##. So I got it wrong. I should have written ##E_0 = \frac{\sigma r}{\epsilon_0 r_0}##. Am I right?
     
  9. Apr 25, 2016 #8

    TSny

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    Yes, that's right.
     
  10. Apr 25, 2016 #9

    TSny

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    Of course, there is also the contribution from the line charge.
     
  11. Apr 25, 2016 #10
    Which is this ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##, right?
     
  12. Apr 25, 2016 #11

    TSny

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    Yes.
     
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