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Electron's velocity

  1. Mar 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Two thin cylindrical shells, respectively with radius ##R_1 = 12 cm## and ##R_2 = 20 cm## and height ##D = 100 m##, are concentric and characterized by a charge density ##\lambda_1 = -0.38 \frac{\mu C}{m}##, for the internal cylinder, and ##\lambda_2 = 0.32 \frac{\mu C}{m}##, for the external one. An electron (##m_e = 9.1 \cdot 10^{-31} kg## and a charge ##e = 1.6 \cdot 10^{-19} C##) is released stationary from the internal cylinder's surface; what is the velocity when it reaches the other cylinder?

    2. Relevant equations
    Gauss Teorem
    ##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{Q}{\epsilon_0}##
    and
    ##a = \frac{q}{m}E##

    3. The attempt at a solution
    So, I started by calculating the electric field on both the cylinders.
    ##\Phi_S (\vec E_1) = \int_S \vec E_1 \cdot d\vec S = E_1 2 \pi r D = \frac{\lambda_1 D}{\epsilon_0}##
    so
    ##E_1 = \frac{\lambda_1}{2 \pi r \epsilon_0}##
    Same thing for the other one.
    ##E_2 = \frac{\lambda_2}{2 \pi r \epsilon_0}##

    At this point I should calculate the velocity starting from the acceleration ##a = \frac{e}{m_e}E##, but I'm blocked here. The ##E## in the acceleration is the "total" electric field or it is ##E_2## since the electron is now moving inside the external cylinder?
    I know there is this other equation:
    ##v = \sqrt{\frac{2q(V_1 - V_2)}{m}}##
    But in this case I should calculate the two electric potential, which would be:
    ##\int_A^B \vec E_0 \cdot d\vec l = V(A) - V(B)##
    with ##A## being the internal cylinder and ##B## the other one.
     
    Last edited: Mar 11, 2016
  2. jcsd
  3. Mar 11, 2016 #2

    mfb

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    Working with the potential is the easiest approach here.

    What is the field inside a long cylinder with a constant surface charge? (Just the field from this surface charge)
     
  4. Mar 11, 2016 #3
    It should be ##E = \frac{\sigma R_2}{\epsilon_0 r}##, right?
     
  5. Mar 11, 2016 #4

    mfb

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    Inside the cylinder?
    So at r=0 you have infinite field strength?
     
  6. Mar 11, 2016 #5
    I did a drawing of the exercise, just to be sure I'm getting this right.
    I used Gauss Theorem to get that. I did like this:
    ##\Phi_s (E) = \int_S \vec E \cdot d \vec S = E 2 \pi r D = \frac{\sigma 2 \pi R_2 D}{\epsilon_0}##
    then
    ##E 2 \pi r D = \frac{\sigma 2 \pi R_2 D}{\epsilon_0}##
    ##E = \frac{\sigma 2 \pi R_2 D}{\epsilon_0 2 \pi r D}##
    ##E = \frac{\sigma R_2}{\epsilon_0 r}##
    so I think yes. Unless I got something wrong.
     

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  7. Mar 11, 2016 #6

    mfb

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    For Gauß theorem, you need the charge enclosed by the area you are integrating over. What is the charge in a smaller cylinder inside a charged cylinder?
     
  8. Mar 12, 2016 #7
    I'm guessing it should be smaller? Because the charged cylinder act like some kind of insulator to the other one inside?
    About the area, I should do something like this then: ##E = \int_{R_1}^{R_2} \frac{\sigma R_2}{\epsilon_0 R_1}##
    With ##R_1## and ##R_2## being the two points describing the area.
     
  9. Mar 12, 2016 #8

    mfb

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    Forget about the two cylinders for a moment, the problem is earlier.

    What is the field inside a single spherical shell with a constant surface charge?

    What is the field inside a single cylinder with a constant surface charge?
     
  10. Mar 12, 2016 #9
    ##\vec E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}##
    with ##\sigma = \frac{Q}{4 \pi r^2}## we have then
    ##\vec E = \frac{\sigma}{\epsilon_0}##

    ##\vec E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}##
    with ##\sigma = \frac{Q}{2 \pi r h}## we have then
    ##\vec E = \frac{\sigma h}{2 r \epsilon_0}##

    Is this what you asked for?
     
  11. Mar 12, 2016 #10

    mfb

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    Where does that formula come from and what does it calculate?

    It does not calculate the electric field inside anything. The right side is also not a vector, so writing E as vector at the left side is not right.
     
  12. Mar 12, 2016 #11
    Oh, I'm sorry about the vector thingy.
    I know this formula:
    ##\vec E_0 (\vec r) = \frac {1}{4 \pi \epsilon_0} \int \frac{\sigma(x', y', z')(\vec r - \vec r')}{|\vec r - \vec r'|^3}dS'## which is just a general way(not only for constant surface charge) and I thought that since ##dq = \sigma(x, y, z) dS## it would end up with the one I wrote (##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}##). Is this not correct?
     
  13. Mar 12, 2016 #12

    mfb

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    It is not correct.
    What would r even be there? The position where you want to evaluate the field, or the radius of the sphere?

    Anyway, let's stay at the sphere: if it has a radius R, what is the electric field at a radius r where r<R?
    You can use Gauß law, but don't blindly plug numbers into a random formula. Think what the formula is doing when, and what the meaning of its components is.
     
  14. Mar 12, 2016 #13
    ##r## is the position where I want to evaluate the field.

    Okay, in ##r < R## we have a ##Q = \sigma 4 \pi r^2##
    using Gauß law we have
    ##\Phi = \frac{Q}{\epsilon_0}##
    which becomes
    ##E 4 \pi r^2 = \frac{\sigma 4 \pi r^2}{\epsilon_0}##
    ##E = \frac{\sigma}{\epsilon_0}##
     
  15. Mar 12, 2016 #14

    mfb

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    Where would that charge be? All the charge is at the spherical shell, at radius R. There is nothing at smaller radii.
     
  16. Mar 12, 2016 #15
    Oh! Because we are talking about surface charge and not volume charge. Because if it was a volume charge then it would have a charge inside the sphere and so it would have ##Q = \rho \frac{4}{3} \pi r^3##, right?
     
  17. Mar 12, 2016 #16

    mfb

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    If it would be a volume charge, yes, but it is not.
     
  18. Mar 12, 2016 #17
    So, going back to what you asked me, for ##r < R## the field would be ##E = 0## inside the sphere and ##E = \frac{\sigma R^2}{\epsilon_0 r^2}## on the surface, right?
     
  19. Mar 12, 2016 #18

    mfb

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    Directly at the surface the field is not well-defined, but outside, yes.
    Inside it is zero, right.

    So going back to the cylinders: does the charge from the outer cylinder matter for the field inside?
     
  20. Mar 12, 2016 #19
    No, it doesn't matter. Only the charge inside matter. This means I only have to calculate the field inside the outer cylinder(which it is the field outside the internal cylinder) ##E_1##, but I have the linear charge density ##\lambda_1## and ##\lambda_2##, not the surface charge, right?
     
  21. Mar 12, 2016 #20

    mfb

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    You have the surface charge of the inner cylinder, but a single charged line along its symmetry axis would produce the same field.
     
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