Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

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

    Gauss theorem could prove that...

  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


    User Avatar
    Science Advisor
    Homework Helper

    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.

  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


    User Avatar
    Science Advisor

    (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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook