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Electric charge

  1. Jun 29, 2006 #1

    tony873004

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    Figure 19-38 shows an electron entering a parallel-plate capacitor with a speed of v = 5.25 106 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d = 0.626 cm at the point where the electron exits the capacitor.
    [​IMG]

    (a) Find the magnitude of the electric field in the capacitor.
    b) Find the speed of the electron when it exits the capacitor.


    Part A:
    E=F/q, but q is not given
    F=ma, but m is not given
    F=kQq/r2
    I don't know where to start.

    Part B:
    Compute time:
    [tex]
    \begin{array}{l}
    t = \frac{d}{v} \\
    \\
    t = \frac{{0.025m}}{{5.25 \times 10^6 m/s}} = 4.2857 \times 10^{ - 8} s \\
    \end{array}
    [/tex]

    Compute acceleration:
    [tex]
    \begin{array}{l}
    y_f = y_i + v_i t + 0.5at^2 \\
    \\
    a = \frac{{y_f - y_i - v_i t}}{{0.5t^2 }} \\
    \\
    a = \frac{{0.626m - 0 - 0t}}{{0.5t^2 }} \\
    \\
    a = \frac{{0.00626m}}{{0.5\left( {4.2857 \times 10^{ - 8} s} \right)^2 }} = 6.81644 \times 10^{12} m/s^2 \\
    \end{array}
    [/tex]

    Compute velocity:
    [tex]
    \begin{array}{l}
    v_{y,f} = v_{y,i} + at \\
    \\
    v_{y,f} = 0 + 6.81644 \times 10^{12} \times 4.2857 \times 10^{ - 8} \\
    \\
    v_{y,f} = 292133.33 \\
    \\
    v_f = \sqrt {v_{y,f}^2 + v_x^2 } \\
    \\
    v_f = \sqrt {(292133.33m/s)^2 + (5.25 \times 10^6 m/s)^2} \\
    \\
    v_f = 5.258122 \times 10^6 m/s \\
    \end{array}
    [/tex]
    But this answer is wrong.
     
    Last edited: Jun 29, 2006
  2. jcsd
  3. Jun 29, 2006 #2

    Kurdt

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    What are the question sof part A and B?
     
  4. Jun 29, 2006 #3

    tony873004

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    Oops. Maybe that's why I'm confused. I don't know what they're asking :).

    I edited the post to include the questions. Thanks for pointing that out.
     
  5. Jun 29, 2006 #4

    Kurdt

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    1.)

    F=ma where m is the mass of an electron.
    E=F/q where q is the charge of the electron

    I'll have a look at 2.)
     
  6. Jun 29, 2006 #5

    tony873004

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    Thanks, that's what I was thinking, but I can't trust my answer since I seem to have gotten acceleration wrong in part b, unless my mistake in part b comes after my computation of acceleration.
     
  7. Jun 29, 2006 #6

    Kurdt

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    To be honest part b looks ok to me. Do you have the real answer and perhaps I could work backwards? Probably the fact its past midnight here is not helping me think straight ;)
     
  8. Jun 29, 2006 #7

    Kurdt

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    Is the answer 6x10^6 m/s? If so I'll tell you how I arrived at that.
     
  9. Jun 29, 2006 #8

    tony873004

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    I don't know what the answer is. The online grading system gave me the red x on the answer I put. In fact, I'm out of guesses, so I can't try 6e6. But I'd still like to know how to do it, and I still need a good value for acceleration for the f=ma part.

    Just curious. If you think it's 6e6, then where did we depart in methods?

    Thanks for your help so far.
     
  10. Jun 29, 2006 #9

    Kurdt

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    To calculate the velocity i used:

    V^2=u^2+2as

    That was all really as far as acceleration was concerned I concur with yourself. The final answer was 5.99....... some other bunch of numbers so i rounded it up.
     
  11. Jun 30, 2006 #10

    tony873004

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    What is u and what is s?

    For part A, it gives the units for the answer as N/C.

    I used my acceleration value of 6.81644e12 with the mass of an electron, 9.11e-31, and the charge of an electron 1.6e-19 C.

    6.81644e12 m/s2* 9.11e-31 kg / 1.6e-19 C
    -38.11 N/C

    But it says this is wrong too. This problem looks so simple.
     
  12. Jun 30, 2006 #11

    Doc Al

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    Redo this calculation. It's off by a factor of 10.
     
  13. Jun 30, 2006 #12

    tony873004

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    Thanks for catching that. It's off by more than that. Now I get
    4.76E-09

    I'll blame it on the calculator :)
     
  14. Jun 30, 2006 #13

    Kurdt

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    u is initial velocity and s is displacement. With the numbers for part a that you've quoted i get 38.81 N/C. I fear perhaps there are just a few computational errors. As far as I can see the method for A is fine just computational errors perhaps (remember it says magnitude aswell) and if you do part b with the formula i used that should be alright.
     
  15. Jun 30, 2006 #14

    Kurdt

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    Oh well there you are how did I not spot that!
     
  16. Jun 30, 2006 #15

    tony873004

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    Typos galore in this one! Where I wrote 0.025 should read 0.0225, which is the number I used in the calculator. So I am off by exactly 1 magnitude like Doc Al said, and not more like I thought I was.

    The answer for part A is 3881. This is exactly 2 magnitudes off of the answer we were getting, which is not surprising since time was off by 1 magnitude and it gets squared.

    Thanks Kurdt and Doc Al. Now I understand :)
     
  17. Jun 30, 2006 #16

    Kurdt

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    Was doc al mainly I have to say but its weird cos I tried the time numbers myself and because it was an exact order of magnitude out I did not notice. Apologies.
     
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