Question

Suppose that earth has a surface charge density of 1 electron/m*2.calculate earth's potential and electric field just outside earth's surface.Radius of earth 6400km.

Answer 1

$\int{EdA}=\frac{q}{\epsilon_0}\to E=\frac{4\pi R^2\cdot e}{4\pi r^2\cdot \epsilon_0}=\frac{R^2\cdot e}{r^2\cdot \epsilon_0}$

$E=-\frac{d\phi}{dr}\to \phi=-\int{\frac{R^2\cdot e}{r^2\cdot \epsilon_0}}dr\to \phi=\frac{R^2\cdot e}{ \epsilon_0}\cdot \frac{1}{r}$

If $r=R$

$E=\frac{R^2\cdot e}{r^2\cdot \epsilon_0}=\frac{e}{\epsilon_0}=\frac{1.6\cdot10^{-19}}{8.85\cdot 10^{-12}}=1.81\cdot10^{-8}V/m$

$\phi=\frac{R\cdot e}{ \epsilon_0}=\frac{6400000\cdot 1.6\cdot10^{-19}}{ 8.85\cdot 10^{-12}}=0.116V$