Question

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?

Short answer

1. The electric potential is, _{$V=-\text{0.12Volts}$.}
2. The electric field outside of the earth is, $E=-\text{18n}\cdot \text{N}/\text{C}$_.
3. The direction of the field is inward.

Step-by-step solution

Given data:

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, _{$R=6.4\times {10}^6\text{m}$}

Determining the concept:

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 $8.85\times {10}^{-12}\text{F}/\text{m}$.

(a) Determining the electric potential:

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 .$-\text{0.12Volts}$

(b) Determining the electric field outside the earth:

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 $-\text{18n}\cdot \text{N}/\text{C}$.

(c) Determining the direction of the field:

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

Therefore, direction of the field must be inward.