# Electric field and potential problem

1. Nov 10, 2015

### gracy

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. Nov 10, 2015

### gracy

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

3. Nov 10, 2015

### gracy

Am I right here?

4. Nov 10, 2015

### 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.