91影视

The surface charge density, $蟽=\text{1.0e}/{\text{m}}^{\text{2}}=鈭�1.6脳{10}^{\text{-19}}\text{C}/{\text{m}}^{\text{2}}$.

At infinity the potential, _$V=0$

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

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,

_{$V=\frac{1}{4蟺{蔚}_o}鈰�\frac{q}{R}$}

The density is defined by

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

The electric field is,

_{$E=\frac{蟽}{{蔚}_0}$}

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

The electric potential:

_{$V=\frac{1}{4蟺{蔚}_o}鈰�\frac{q}{R}$}

Substitute_$4蟺R^2蟽$for q in the above equation.

$\begin{array}{c}V=\frac{4蟺R^2蟽}{4蟺{蔚}_{o}R}\\ =\frac{R蟽}{{蔚}_o}\end{array}$

Substitute known values in the above equation.

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

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

The electric field outside of the earth is given as,

_{$\begin{array}{c}E=\frac{蟽}{{蔚}_0}\\ =\frac{鈭�1.6脳{10}^{\text{-19}}\text{C}/{\text{m}}^{\text{2}}}{8.85脳{10}^{鈭�12}\text{F}/\text{m}}\\ =鈭�0.18脳{10}^{鈭�7}\text{N}/\text{C}\\ =鈭�\text{18n}鈰�\text{N}/\text{C}\end{array}$}.

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

As,_{$E=鈭�\text{18nN}/\text{C}$},

Therefore, direction of the field must be inward.