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Potential difference between two large plate changing with time

  1. Jul 11, 2012 #1

    AGNuke

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    The potential difference between two large parallel plates is varied as V=at; a is a positive constant and t is time. An electron starts from rest at t=0 from the plate which is at lower potential. If the distance between the plates is L, mass of electron m and charge on electron -e then find the velocity of the electron when it reaches the other plate.


    I attempted the question by equating the relation between time & velocity and time & distance.

    [tex]A=\frac{Vq}{x}=\frac{eat}{mL}[/tex]
    [tex]\int_{0}^{v}dv=\int_{0}^{t}{Adt} \Rightarrow v=\frac{eat^{2}}{2mL}[/tex]
    [tex]\int_{0}^{L}dx=\int_{0}^{t}vdt+\int_{0}^{t}atdt \Rightarrow L=\frac{ea}{2mL}\int_{0}^{t}t^{2}dt+\frac{ea}{mL}\int_{0}^{t}t^{2}dt=\frac{eat^3}{2mL}[/tex]

    Now dividing the cube of first equation with square of second equation, to eliminate the time, I got [tex]v=\left ( \frac{eaL}{2m} \right )^{1/3}[/tex]

    Now my problem is that the answer stated is [tex]v=\left ( \frac{9eaL}{2m} \right )^{1/3}[/tex]

    Any problems as for what I have done??? :confused:
     
  2. jcsd
  3. Jul 11, 2012 #2

    tiny-tim

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    Hi AGNuke! :smile:

    I don't understand why you're integrating adt …
    why not just use x = ∫vdt ? :confused:
     
  4. Jul 11, 2012 #3

    AGNuke

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    Oops... my fault. I thought that since acceleration is also changing, I better write it too, but I forgot that I already made it up for it when writing v.

    Thanks tim, I got my 9 in the place.
     
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