a). If Earth had a uniform surface charge density of_{$\mathbf{1}\mathbf{.0}\mathbf{}\mathbf{\text{electron}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$}(a very artificial assumption), what would its potential be? (Set$\mathit{V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$at infinity.) What would be the

(b) magnitude and

(c) direction (radially inward or outward) of the electric field due to Earth just outside its surface?

The surface charge density, $\sigma =\text{1.0e}/{\text{m}}^{\text{2}}=-1.6\times {10}^{\text{-19}}\text{C}/{\text{m}}^{\text{2}}$.

At infinity the potential, _$V=0$

The radius of Earth, _{$R=6.4\times {10}^6\text{m}$}

Potential due to continuous charge distribution. According to continuous charge distribution, when charges are continuously spread over a line, surface, or volume, the distribution is called continuous charge distribution.

Formulae are as follow:

The potential is,

_{$V=\frac{1}{4\pi {\epsilon }_o}\cdot \frac{q}{R}$}

The density is defined by

_{$\begin{array}{l}\sigma =\frac{q}{4\pi R^2}\\ q=4\pi R^2\sigma \end{array}$}

The electric field is,

_{$E=\frac{\sigma }{{\epsilon }_0}$}

Where, $q$ is charge, R is radius, V is potential, E is electric field, and ${\epsilon }_0$ is the Permittivity of free space having a value $8.85\times {10}^{-12}\text{F}/\text{m}$.

The electric potential:

_{$V=\frac{1}{4\pi {\epsilon }_o}\cdot \frac{q}{R}$}

Substitute_{$4\pi R^2\sigma$}for q in the above equation.

$\begin{array}{c}V=\frac{4\pi R^2\sigma }{4\pi {\epsilon }_{o}R}\\ =\frac{R\sigma }{{\epsilon }_o}\end{array}$

Substitute known values in the above equation.

$\begin{array}{c}V=\frac{(6.4\times {10}^6\text{m})(-1.6\times {10}^{\text{-19}}\text{C}/{\text{m}}^{\text{2}})}{8.85\times {10}^{-12}\text{F}/\text{m}}\\ =-1.2\times {10}^{-1}\text{Volts}\\ =-0.12\text{V}\end{array}$

Hence, the electric potential is .$-\text{0.12Volts}$

The electric field outside of the earth is given as,

_{$\begin{array}{c}E=\frac{\sigma }{{\epsilon }_0}\\ =\frac{-1.6\times {10}^{\text{-19}}\text{C}/{\text{m}}^{\text{2}}}{8.85\times {10}^{-12}\text{F}/\text{m}}\\ =-0.18\times {10}^{-7}\text{N}/\text{C}\\ =-\text{18n}\cdot \text{N}/\text{C}\end{array}$}.

Hence,the electric field outside of the earth is $-\text{18n}\cdot \text{N}/\text{C}$.

As,_{$E=-\text{18nN}/\text{C}$},

Therefore, direction of the field must be inward.