Question

The earth has a vertical electric field at the surface, pointing down, that averages$100\mathrm{N}/\mathrm{C}$. This field is maintained by various atmospheric processes, including lightning. What is the excess charge on the surface of the earth$?$

Short answer

The excess charge on the surface of the earth${\mathrm{Q}}_{\text{in}}$is$-4.5\times {10}^5\mathrm{C}$.

Step-by-step solution

Formula for electric flux and charge

The electric flux is that the amount of electrical field that flows through a closed surface.

Gauss' law states that the electrical field passing through a surface is proportional to the charge within the surface.

Electric flux is,

${\mathrm{\Phi }}_{\mathrm{e}}=\mathrm{EA}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{ϵ}}_{\mathrm{o}}}$

${\mathrm{Q}}_{\text{in}}={\mathrm{ϵ}}_{\mathrm{o}}\mathrm{EA}$

localid="1648750984200" ${\mathrm{Q}}_{\text{in}}=4{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{Er}}^2...1$

Calculation for excess charge on the surface of the world

The radius of the gaussian surface is up to the radius of the earth's outer surface,

$\mathrm{r}=6.37\times {10}^6\mathrm{m}$

As it enters the world, the electrical field is negative.

$\mathrm{E}=-100\mathrm{N}/\mathrm{C}$

Substitute values in equation localid="1648752090427" $1$,

we get,

${\mathrm{Q}}_{\text{in}}=4{\mathrm{\pi ϵ}}_{\mathrm{o}}{\mathrm{Er}}^2$

$=4\mathrm{\pi }\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)(-100\mathrm{N}/\mathrm{C}){\left(6.37\times {10}^6\mathrm{m}\right)}^2$

$=-4.5\times {10}^5\mathrm{C}$