- #1
gracy
- 2,486
- 83
Homework Statement
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
Homework Equations
surface charge density of sphere=##Q##/##4πR^2##
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!