1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnitude and direction of an electric field outside of a plate?

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data

    A circular metal plate of radius 0.2 m has 1010 excess electrons uniformly distributed over its surfaces. What is the magnitude and direction of the field just outside the plate near its center?

    2. Relevant equations
    E = kQ/r2



    3. The attempt at a solution

    Starting with E = kQ/r2,

    I solved for Q: (1010) (1.60 x 10-9 C)

    = 16 C

    Plugging into the equation for E:

    E = (16 C) (9.0 x 109 N m2/C2) / (.02 m)2

    I end up with a really high value of 3.6 x 1012 N/C.

    However, I know this is wrong because I should end up with a value of around 720. What am I doing wrong?

    For direction, I know the electron will move towards the plate.
     
    Last edited: Sep 26, 2013
  2. jcsd
  3. Sep 26, 2013 #2
    I like gauss law better.:biggrin:

    Charge of electron is 1.6x10-19

    It is because you are assuming all charge to be at the same place. In this case you will either need to integrate (long and tedious) or use gauss' law.
     
  4. Sep 26, 2013 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Compute the surface charge density σ. Then, if you don't know the formula for the E field just outside the plate, set up a gaussian surface, say a right circular cylinder, with one end-surface inside the plate and the other just outside. Then compute net flux leaving the surfaces. Remember to make the height of the cylinder very short. EDIT: also the area of the end sections.
     
    Last edited: Sep 26, 2013
  5. Sep 26, 2013 #4
    Ok so Gauss's Law is:

    E = 2pikσ

    Q = (1010) (1.6 x 10-19) C

    Q = 1.6 x 10-9 C

    σ = Q/A = 1.6 x 10-9 C/pi (.02)2 = 2.04 x 10-13 C/m2

    E = 2pi (9.0 x 109 N m2/C2) (2.04 x 10-13 C/m2)

    E = .011

    I'm still not sure where I'm going wrong with this..
     
  6. Sep 26, 2013 #5
    What is the total surface area of the plate?
    And Gauss' law does not look like what you wrote. What is k?
     
  7. Sep 26, 2013 #6
    Actually
    ##\int E.ds=\frac{q}{ε_{o}}##
    ##E.A=\frac{q}{ε_{o}}##
    ##E=\frac {σ}{ε_{o}}## As field is perpendicular to the face of cylinder.
    ##E=\frac{σ}{1/4 \pi k}##
    ##E=4 \pi \sigma k##
     
  8. Sep 26, 2013 #7
    Oh whoops, ok, I see I wrote the expression for a conductor, sorry about that! Gauss's Law is otherwise expressed as:

    ∅E = Q /ε0

    Total surface area of the plate, isn't it A = pi*r2?

    So finding Q again:

    Q = (1010) (1.6 x 10-19 C) = 1.6 x 10-9 C



    ∅E = (1.6 x 10-9 C) / (8.85 x 10-12 F m-1) = 181?

    The final answer should be 720. I'm not sure if I'm missing something to this problem...
     
    Last edited: Sep 26, 2013
  9. Sep 26, 2013 #8
    Lets begin at the beginning. What is your Gaussian surface?
     
  10. Sep 26, 2013 #9
    Ok so for a circular plate, will my Gaussian surface be:

    E =4pi*r2?
     
  11. Sep 26, 2013 #10
    That is the equation for a gaussian surface (what gaussian surface I have no idea). Gaussian surface is a 3D closed figure.
    See post #3 by rude man.
     
  12. Sep 26, 2013 #11
    I'm not finding much on how to compute finding a value for the Gaussian surface of a circular plate though. Unless I need to integrate the area of A for what a circular plate would normally be? i.e. A of a circle = TTr2?
     
  13. Sep 26, 2013 #12
    Okay, lets take RM's gaussian surface.
    Let the radius of cylinder be r and height be h.
    Now this cylinder will enclose how much charge?
     
  14. Sep 26, 2013 #13
    So for a cylinder:

    Won't we have, E = kQ/r-h, so to solve for Q, we'd end up with an equation of:

    E(r-h)/k = Q

    Is this correct? Am I thinking about this right?
     
  15. Sep 26, 2013 #14
    Noooo....
    Lets put that equation aside for this sum.
    What is the charge in the cylinder?
    You have the charge of the disc and its area (remember a disc has 2 sides) So find the surface charge density and multiply it with area of the disc our gaussian surface i.e.. cylinder encloses.
     
  16. Sep 26, 2013 #15
    Sorry, its getting a bit late here. I am off to :zzz:
    For your problem:
    After you have got the charge enclosed, just use the Gauss's law expression. Note that you only need the area of the top face of the cylinder, as h and r are very small, we can assume E to be perpendicular to that face.
    :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Magnitude and direction of an electric field outside of a plate?
Loading...