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Speed of an electron due to a ring charge

  1. Feb 11, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniformly charged thin ring has radius 13.0 cm and total charge 21.5 nC . An electron is placed on the ring's axis a distance 32.5 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest.

    Find the speed of the electron when it reaches the center of the ring.

    2. Relevant equations

    ∫dE = ∫k*dq*R^2*cos(θ)

    ∫E * dL = -ΔV

    z = distance from electron to center of ring

    r = radius of ring

    R = point on ring to electron

    3. The attempt at a solution
    So at first I started off with changing ∫dE into a function of only one changing part.

    = kdq/R^2*cos(θ)

    = k*dq*z/R^3

    = k*Q*z/(z^2 + r^2)^(3/2))dz

    = k*Q/((r^2 + z^2)^.5) = ΔV ≈ 8.85*10^-17 = ΔKE = .5mv^2

    (8.85*10^-17)*2/(9.11*10^-31) ≈1.9*10^7


    I've tried this problem a couple of times, and have gotten anywhere between 1.0-2.0*10^7. I know it's suppose to be around this, but if any errors could be pointed out that would be great. I don't want to lose anymore points on this problem. Thanks!

    Also, as a completely off related topic, is there anywhere here that gives a guideline on how to use the templates for powers and whatnot? I'm not quite sure how to use the whole , and would like to learn how to make posts more readable.
     
  2. jcsd
  3. Feb 11, 2017 #2

    blue_leaf77

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    That doesn't look correct. The potential difference is equal to the difference between its value at the starting point of the particle and that at the center of the ring.
    Let alone the correctness for ##\Delta \textrm{KE}## for now, check again if there should be 2 somewhere and whether the power is 2 or 1/2.
    We use Latex here.
     
  4. Feb 11, 2017 #3

    Simon Bridge

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    It would help if you annotated your working.
    You seem to have used cylindrical-polar coordinates with the z axis pointing along the ring axis.
    Your strategy appears to be to apply the work energy relation ... ##\frac{1}{2}mv^2 = -\int F(z)\; dz## ... ##\vec F = -e\vec E## ... but I think you could just have used Coulomb's law directly.

    OK - so set up the coordinates so the initial position of the electron is z=-a=-32.5cm ... so the +z axis points in the direction of the motion we are interested in.
    Net force on the electron will be in the +z direction - define ##\vec F = F\hat k##
    - let the radius of the ring be r and the total charge on the ring be Q.

    The charge element on the ring between ##\theta## and ##\theta+d\theta## is given by ##dq = (Q/2\pi)\; d\theta## which contributes $$dF = \frac{ke\; dq}{r^2+z^2}\frac{z}{\sqrt{r^2+z^2}}$$ ... to the net force. Add them up by taking the integral over ##\theta \in [0,2\pi)##
    $$F(z) = keQ\frac{z}{(z^2+r^2)^{3/2}}$$
    From the work-energy relation: $$v^2 = -\frac{2keQ}{m} \int_{-a}^0 \frac{z}{(z^2+r^2)^{3/2}}\; dz$$
    Evaluating the integral gives... $$v^2 = \frac{2keQ}{m}\left[ \frac{1}{r} - \frac{1}{\sqrt{r^2+a^2}}\right]$$ ... something like that? (I didn't see a definite integral in your working.) [edit: I see I was beaten to it.]
    Now you can check my working: did I make a mistake? ;)

    Mostly I just want you to see how to annotate your working to help the reader follow what you did. This is great in assignments because it helps the marker give you more marks.
     
  5. Feb 11, 2017 #4

    I do see an error in what I wrote for the original equation. It should be:

    ∫dE = kQ/(r^2 +x^2)^(3/2) where the limits are from 0 to .325m (distance from starting point to center of ring) which I believe should give the change in potential.
     
  6. Feb 11, 2017 #5

    blue_leaf77

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    I didn't check the numerical value but the expression you got for potential energy difference is not correct. See Simon's post above.
     
  7. Feb 11, 2017 #6

    Ah, I see. I completely forgot about the charge of the electron in my post. I was still factoring it into my initial work, but I just didn't write it in, which was sloppy of me. Should be e ∫ E • dL = -ΔV
     
  8. Feb 11, 2017 #7

    Your work is entirely correct. My mistake was not doing a direct integration instead of indirect, so I lost the $$ \frac {1}{r}$$

    I really appreciate your help!
     
  9. Feb 11, 2017 #8

    Simon Bridge

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    That's easy to do: you see a zero in the limits of the integral and automatically ignore one of the terms because that is what usually happens right?
    I usually make a mistake in cancelling minus signs :)
     
  10. Feb 12, 2017 #9

    You forgot a negative sign when you evaluated the integral. $$\int { \frac {zdz} {\sqrt [3] {z^2 + r^2}}}$$

    Gives:

    $$\frac {-1} {\sqrt{z^2 + r^2}}$$

    Which, when evaluated at the limits

    $$ \int_0^{-a}$$

    Gives the answer of what you got, just with an extra negative sign to counteract the negatively charged value of the electron. Sorry I didn't point out the mistake to you in the first place. I'll make sure to do that next time you ask me to check your work. :)

    Once again, thank you very much for your help!
     
  11. Feb 12, 2017 #10

    Simon Bridge

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    Good to see you looking closely.
    I believe I got that...
    ... only those are not the limits I used ... double check for me?

    Starting from post #3, 3rd equation down - I did: $$ \begin{align}
    v^2 &= -\frac{2keQ}{m}\int_{-a}^0 \frac{z\; dz}{(r^2+z^2)^{3/2}} \tag{1}\\
    &= -\frac{2keQ}{m}\int_{-a}^{0} \frac{z\; dz}{(r^2+z^2)^{3/2}} \tag{2}\\
    &= -\frac{2keQ}{m}\left[ \frac{-1}{\sqrt{r^2+z^2}}\right]_{-a}^{0} \tag{3}\\
    &= -\frac{2keQ}{m}\left( \frac{-1}{\sqrt{r^2+0^2}} - \frac{-1}{\sqrt{r^2+(-a)^2}}\right) \tag{4}\\
    &= -\frac{2keQ}{m}\left( \frac{-1}{r} + \frac{1}{\sqrt{r^2+a^2}}\right) \tag{5}\\
    &= \frac{2keQ}{m}\left( \frac{1}{r} - \frac{1}{\sqrt{r^2+a^2}} \tag{6}\right)
    \end{align}$$
    Notes on LaTeX ...
    ##\int zdz = \int dz^2## sometimes, so put a thin space in there vis: ##\int z\; dz## (don't rely on context).
    Some people go further and use mathrm: ##\int z\; \mathrm{d}z## but that's annoying to type out.
    You can use double-hash marks to put equations inline.
    PDFLaTeX automatically adds the line spacing before and after equations, so you don't need to.
    LaTeX is not used for text, so unneeded empty lines are still typeset.

    Notice: when I did the setup I defined things so I did not need explicit vector calculus (I did the geometry on paper). I did not bother with the cosine step you did - if I had, I would need to include a diagram to define the angle, which I would not have called ##\theta## because that is already used as part of the cylindrical-polar coordinate system.

    BTW: ##\sqrt[3]{x}## represents the cube root of x or ##x^{1/3}##, if you don't like fractions for powers try: ##x^{3/2} = \sqrt{x^3}## or, more generally, ##x^{a/b} = \sqrt[ b ] {x^a}##.
     
    Last edited: Feb 12, 2017
  12. Feb 13, 2017 #11

    Whoops, alright yeah what I did doesn't work out. I flipped the integral limits, which unless I'm making up some sort of fake math technique, allows us to take out a negative sign. So why doesn't that work out this scenario? I understand physically that you integrate from the initial location of the electron towards the center of the ring, which is considered to be zero, but why does that not work out if you consider the zero point to be where the electron starts?

    Also, I'm following the method that my teacher taught us how to do these kind of integration for ring charges, which doesn't use polar coordinates, even if that is a bit more convenient.
     
  13. Feb 14, 2017 #12

    Simon Bridge

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    You have to do work on the electron to move it from the center of the ring, storing energy in the electric field.
    That energy gets released when the electron returns.

    That is the physical reason for the change in sign when you swap the limits of the integral over.

    Your teacher is implicitly using polar coordinates anyway - being explicit makes it easier to talk about but lots of students find their minds rebel against non-cartesian coordinates and they stop learning.

    The full expression for the coulomb force on charge Q due to some distribution of charges is $$d\vec F = \frac{kQ}{r^3}\vec r\; dq$$ ... where ##\vec r## points from the particular dq to the position of Q. The strategy is to evaluate the force for every bit of charge and then add them up. That is what integration is: adding up lots of little bits. Your job in evaluating the integral is to exploit everything you know about the situation and your geometry tricks to make this sum as simple as possible.

    Imagine you want to find the area between the x-axis and the curve y=f(x), between limits x=a and x=b where a<b, then you are evaluating ##A = \int \; dA## between the limits. That is, you are looking for all the little areas dA and adding them up. (Bear with me, I know you would normally jump to the end.)

    You can see that ##\int \;dA = A + c## works for c=0, but that is not helpful ... what you want to know is how f(x) and the limits affect A. So you need to put dA in terms of other stuff you know.

    The element of area at position (x,y) has area dA = dx.dy if we use rectangular coordinates. Depending on f(x) that may not be a good choice but hey ho.
    ... between x and x+dx, the y value varies between y=0 and y=f(x), which suggests the integral you want to evaluate is: $$A = \int_a^b \left[\int_0^{f(x)}\; dy \right]\; dx$$ ... this is not how you are used to doing it - but have a go evaluating the y axis integral and you will see a familiar result.

    That was a simple example where you already knew the result ... the thought process works for more complicated sums.
     
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