# Potential difference between Earth's surface and 2 meters above

Assuming Earth is a charged sphere of radius $R = 6400\times10^3$ m with uniform surface charge density $\sigma = -10^{-9}$ C/m2 and with $\epsilon_0 = 8.85\times10^{-12}$ F/m

I find that

$$V(R+2)-V(R) = \int_{R}^{R+2}E(r)dr = \frac{\sigma R^2}{\epsilon_0}(-\frac{1}{R+2} + \frac{1}{R}) = 226 V$$

If this is correct, why don't we feel that potential difference ?

• feynman wrote about it in Physics lectures, see here: peaceone.net/basic/Feynman/V2%20Ch09.pdf Answer seems to be that any conductor you bring into air will change that field. So there is no 300V between your head and feet. – aaaaa says reinstate Monica Mar 3 '15 at 21:13
• May I ask where the value for the surface charge density comes from? – Steeven Mar 4 '15 at 0:05
• @Steeven it was given in the exercise, but it seems correct since Feynman mentions a similar potential difference (~200V). – mwa1 Mar 4 '15 at 9:04