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Beam of particles shot through slits

  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A beam of positively charged particles with radius ##R## (current ##I##) is shot from a region of potential ##-V## through a potential difference of ##+V## to a grounded plate. The beam passes through a slit on this plate then passes through another region of length ##L## meters and passes through a slit on a second plate before hitting a target. The proton density is ##p## protons per cubic meter.

    a. What is the speed, ##v## of the particles when the beam reaches the first slit?

    b. What is the electric field as a function of the radius within and outside the beam?

    c. What is the electric potential as a function of the radius if the potential at the center of the beam is ##0##?

    d. Due to particle interactions, some of beam diverges over the distance ##L##. If the radial field is constant and ##1/2## the maximum field, estimate how far a proton would move radially due to this field over the distance ##L##.

    2. Relevant equations

    ##U=qV##
    ##K=\frac{1}{2}mv^2##
    ##\rho = \frac{Q}{Volume}##
    ##E=\frac{Q}{4\pi ε_0 R^2}##
    ## V =\frac{Q}{4 \pi ε_0 R}##

    3. The attempt at a solution

    a. The initial electric potential energy of the particles is converted to kinetic energy as the particles reach the first plate, so ##qΔV=\frac{1}{2}mv^2 \rightarrow v = \sqrt{\frac{2qV}{m}}##

    b. The volume of the beam over the distance ##L## is ##V= (\pi)(R^2)(L)## so there are ##p\pi R^2 L## protons and the total charge is## (1.6*10^{-19})p\pi R^2 L = Q##. So the electric field within the beam at a distance ##r<R## is ##E=\frac{Qr}{4\pi ε_0 R^3}## and for ##r>R##, ##E=\frac{Q}{4\pi ε_0 r^2}##

    c.For ##r<R##, the electric potential is ##V=\frac{Q}{4\pi ε_0 R}## and for ##r>R, V=\frac{Q}{4\pi ε_0 r}##

    d. No idea
     
  2. jcsd
  3. Jan 29, 2014 #2

    BvU

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    Don't understand the step from b) answer to c) answer for r<R. Wasn't E the the divergence of V ?In the OP they state that V=0 at r = 0.
    Your V for r<R comes from what assumption about the charge distribution in the cross section of the beam ?
    Very quick reply, may need to edit some more...

    The equations you give are for a point charge. The beam is probably more like a line charge!
     
    Last edited: Jan 29, 2014
  4. Jan 29, 2014 #3

    lightgrav

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    a) is ok.

    b) you need the Area of the cylinder shell, length L and radius r.

    in c), they wanted V =0 in the beam center (not at infinity) , so you integrate E from inside, outward.

    d) how long (time) do the protons take to travel L?
    where is the maximum E-field? (r=R) ... presume that ½ this value will provide constant acceleration outward.
     
  5. Jan 29, 2014 #4
    Ok so this beam is essentially a cylinder of charge? But the circular faces aren't pierced since it's continuous.

    For part B, with ##r## being the distance I use Gauss' law to calculate the electric field for ##r<R## so the enclosed charge ##q## is then ##Q\frac{\pi r^2 L}{\pi R^2 L} = Q\frac{r^2}{R^2}##, so using Gauss' law, ##E(2\pi rL)= \frac{Qr^2}{ε_0 R^2} \rightarrow E = \frac{Qr}{2\pi ε_0 R^2 L}## for ##r<R##. Thus for ##R<r, E(2\pi rL) = \frac{Q}{ε_0} \rightarrow E = \frac{Q}{2\pi ε_0 rL}## Where ##Q = \pi R^2 L(p)(1.6*10^{-19})## and the field is maximum when ##r=R##
     
    Last edited: Jan 29, 2014
  6. Jan 29, 2014 #5

    lightgrav

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    b) the E-field is radial outward, so NONE of it pierces either end-cap (r+L) →L; otherwise ok.
    c) straightforward integral, now that E is right.
    d) trying to use non-constant acceleration as the beam expands would be a nightmare.
    so just use constant-acceleration approx (ok, if they're going fast)
     
    Last edited: Jan 29, 2014
  7. Jan 29, 2014 #6
    so for part C, I would integrate E from ##-\frac{L}{2}## to ##\frac{L}{2}##. Not going to show all the work but does this integral look correct: ##\int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{Q}{2\pi ε_0 rL}\,dr## I'm just not sure if I'm using the correct equation for the electric field. since the field is zero at the center, I used the electric field for r>R

    Then for part D, I calculated the acceleration from ##v^2 = 2aL \rightarrow a = \frac{qV}{mL}##. to calculate the distance a proton moves radially, do I need to relate this to the work done by the electric force?
     
    Last edited: Jan 29, 2014
  8. Jan 29, 2014 #7

    lightgrav

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    c) E is RADIAL, so integrate from r=0, outward. The center of the beam (r=0) has V=0.
    what happened to the r on top? how did it arrive on the bottom ... there used to be a R² down there.
     
  9. Jan 29, 2014 #8
    I see what you mean by radial but I have two different equations for electric field. So I need to integrate each one from 0 to R (electric field in the beam) and then from R to infinity (electric field outside beam)?
     
  10. Jan 29, 2014 #9

    lightgrav

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    c) yes, integrate the Qr/R² from r=0 to r=R, then add to that the integral of Q/r from r=R.
    d) no, qV/mL was the acceleration along the beam, before the grounded plate.
    here we want the acceleration outward (radially) of a typical proton (say, at r=R/2),
    while the beam drifts (constant speed, found in (a)) from ground-plate to beam-dump.
     
  11. Jan 29, 2014 #10
    OK so my two integrals are ##\int_0^R \frac{Qr}{2\pi ε_0 R^2 L} \,dr = \frac{Q}{4 \pi ε_0 L}## and ##\int_R^r \frac{Q}{2\pi ε_0 rL} \,dr = \frac{Q}{2\pi ε_0 L} \ln{\frac{r}{R}}##The second one somehow doesnt seem right though... since r > R, shouldnt potential decrease the farther you get and shouldn't my first equation imply V=0 at the center?

    For part D ok so the radial acceleration would be from ##\frac{v^2}{r}##?
     
    Last edited: Jan 29, 2014
  12. Jan 29, 2014 #11

    lightgrav

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    c) looks good. notice that a uniform field (|| plate capacitor) has V increase _linearly_ with distance.
    d) the protons repel each other, by the E-field you calculated in b) ... F = eE causes mproton a
     
  13. Jan 29, 2014 #12
    Great, and for D, the displacement would then be calculated using ##v^2 = 2 a \Delta x## and the velocity from a)? Thanks so much for your help.
     
  14. Jan 30, 2014 #13

    lightgrav

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    no, the velocity you calculated in part (a) was along the length L; use it to find time-of-flight.
    part (d) wants Δr = ½ a t²
     
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