91Ó°ÊÓ

The surface charge density, $\mathrm{σ}=\text{1.0e}/{\text{m}}^{\text{2}}=−1.6×{10}^{\text{-19}}\text{C}/{\text{m}}^{\text{2}}$.

At infinity the potential, _$V=0$

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

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,

_{$V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}â‹\dots \frac{q}{R}$}

The density is defined by

_{$\begin{array}{l}\mathrm{σ}=\frac{q}{4\mathrm{π}{R}^2}\\ q=4\mathrm{π}{R}^2\mathrm{σ}\end{array}$}

The electric field is,

_{$E=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}$}

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

The electric potential:

_{$V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}â‹\dots \frac{q}{R}$}

Substitute_{$4\mathrm{π}{R}^2\mathrm{σ}$}for q in the above equation.

$\begin{array}{c}V=\frac{4\mathrm{Ï€}{R}^2\mathrm{σ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{o}R}\\ =\frac{R\mathrm{σ}}{{\mathrm{Î\mu }}_o}\end{array}$

Substitute known values in the above equation.

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

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

The electric field outside of the earth is given as,

_{$\begin{array}{c}E=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\\ =\frac{−1.6×{10}^{\text{-19}}\text{C}/{\text{m}}^{\text{2}}}{8.85×{10}^{−12}\text{F}/\text{m}}\\ =−0.18×{10}^{−7}\text{N}/\text{C}\\ =−\text{18n}â‹\dots \text{N}/\text{C}\end{array}$}.

Hence,the electric field outside of the earth is $−\text{18n}â‹\dots \text{N}/\text{C}$.

As,_{$E=−\text{18nN}/\text{C}$},

Therefore, direction of the field must be inward.