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Electric field and potential problem

  1. Nov 10, 2015 #1
    1. The problem statement, all variables and given/known data

    Suppose that earth has a surface charge density of 1 electron/metre^2 .Calculate earth's potential and electric field just outside earth's surface.Radius of earth 6400 km
    2. Relevant equations
    surface charge density of sphere=##Q##/##4πR^2##

    3. The attempt at a solution
    Let's assume earth to be spherical.Then
    surface charge density of sphere=##Q##/##4πR^2##
    ##Q##=-##1.6##×##10^-19##×##4πR^2##
    Electrical field of a sphere at distance r=##E##=##\frac{Q}{4πr^2}##
    Earth's field just outside earth's surface
    We can take r=R
    Therefore Earth's field =just outside earth's surface=##E##=##\frac{Q}{4πR^2}##
    =##E##=##\frac{-1.6×10^-19×4πR^2}{4πR^2}##
    =-1.6×10^-19V/m
    But it is wrong.I want to know what went wrong.Similarly in case of potential difference
    ##V##=##\frac{Q}{4πR}##
    =##\frac{-1.6×10^-19×4πR^2}{4πR}##
    =-1.6×10^-19×R
    =-1.6×10^-19×64×10^5
    =102.4×10^-14 V
    It is also wrong,I want reason.
    Thanks!
    EDIT:I think I have got .There is something called"ε0".
    gracy!
     
  2. jcsd
  3. Nov 10, 2015 #2
    So,just put ##ε0##in formula of ##E##And ##V##.Problem solved.
    Solution:##E##=##\frac{Q}{4πε0R^2}##

    =##\frac{-1.6×10^-19×4πR^2}{4πε0R^2}##

    =##\frac{-1.6×10^-19}{8.854×10^-12}##

    =-1.8×10^-8 volt/m

    Solution:##V##=##\frac{Q}{4πε0R}##

    =##\frac{-1.6×10^-19×4πR^2}{4πε0R}##

    = ##\frac{ 1.6×10^-19×R}{ε0}##

    =-##\frac{1.6×10^-19×64×10^5}{8.854×10^-12}##

    =-0.116 volt
     
  4. Nov 10, 2015 #3
    Am I right here?
     
  5. Nov 10, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    Yes, that's fine. Technically it would be r = R + ε, where ε is an infinitesimal displacement so that you're just barely off the surface of the sphere. But R is so large by comparison that R + ε → R.
     
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