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How far has the proton been deflected?

  1. Jan 23, 2005 #1
    A proton travelling at v= 1.0*10^5 m/s enters the gap btwn the plates of a 2.0-cm-wide parallel-plate capacitor. The surface charge densities on the plates are +-1.0*10^6 C/m2. How far has the proton been deflected sideways when it reaches the far edge of the capacitor?

    I figured that the horizontal velocity would be constant so I need to find a vertical v to find how far the proton has been deflected. After that I can find the answer.

    So, I need to find the acceleration to find the final vertical v by using a= eE/m.

    The problem is I'm not sure how to find E in N/C w/ the given information. They give 1.0*10^6 C/m2 but I'm not sure how to convert that into E.

    Any hints/tips would be greatly appreciated.
     
  2. jcsd
  3. Jan 23, 2005 #2

    dextercioby

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    I believ the electric field is
    [tex] E=\frac{\sigma}{\epsilon_{0}} [/tex]

    Gauss theorem could prove that...

    Daniel.
     
  4. Jan 23, 2005 #3
    Thanks for the reply, but if you don't mind, can you clarify a bit?

    I know what epsilon_0 is but not sigma. My class hasn't reached that part yet. If sigma is given 1.0*10^-6 C/m^2, it still doesn't make sense on how E = sigma/epsilon_0; I tried but the units don't work out right.

    BTW, if you don't mind, can you teach me how to get those epsilon, sigma, et cetera images? I tried copying your [tex] ... [/tex] but those images don't show up, only the texts were italicized and bold.
     
  5. Jan 23, 2005 #4

    dextercioby

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    1.Sorry,the units have to match...
    [tex] \nabla\cdot (\epsilon_{0}\vec{E})=\rho [/tex]

    is Gauss law for vacuum...Your formula results immediately.

    2.sigma is te electric charge surface density on the plates of the capacitator.U could have figured that after the units (C/m^{2}).

    3.Yes,u mean TeX editing.Dld the file which teaches u how to write the code lines.Click on any of the formulas and then another click in the small window that pops up.It's a PDF file.

    Daniel.
     
  6. Jan 23, 2005 #5
    Daniel, I'm lost here. I still don't get what can I use sigma for exactly? Sigma is the electric charge/area. Only the length is given, how can I find the electric field from sigma then when I have no clue what the area is?

    Also, in the above reply, I see that you use rho in the equation, but rho is not given at all. I'm really confused; if you don't mind, can you clarify a bit?

    Also in the above reply, you used an upside down delta, what is that thing?
     
  7. Jan 23, 2005 #6

    krab

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    (in words) the electric field is equal to the surface charge density divided by the permittivity of free space. You have both numbers, so just divide them.
     
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