Question

assume that Earth has a surface charge density of 1 electron per metre^2 . calculate the earth's electric field​

Answer 1

Given:

Earth has a surface charge density of one electron per metre².

To find:

Electrostatic field intensity at Earth surface.

Calculation:

Applying Gauss' Theorem:

$\therefore \displaystyle \: \int E .ds \: = \: \frac{q}{ \epsilon_{0}}$

$= > \displaystyle \:E \int ds \: = \: \frac{q}{ \epsilon_{0}}$

$= > \:E \times 4\pi {r}^{2} \: = \: \dfrac{q}{ \epsilon_{0}}$

$= > \:E \: = \: \dfrac{q}{ 4\pi {r}^{2} \epsilon_{0}}$

$= > \:E \: = \: \dfrac{1}{ \epsilon_{0}} \times \dfrac{q}{4\pi {r}^{2} }$

$= > \:E \: = \: \dfrac{1}{ \epsilon_{0}} \times \sigma$

$= > \:E \: = \: \dfrac{ \sigma}{ \epsilon_{0}}$

$= > \:E \: = \: \dfrac{1.6 \times {10}^{ - 19} }{ \epsilon_{0}}$

$= > \:E \: = \: \dfrac{1.6 \times {10}^{ - 19} }{8.85 \times {10}^{ - 12} }$

$= > \: E = 1.87 \times {10}^{ - 7} \: N {C}^{ - 1}$

So, final answer is:

$\boxed{ \red{ \bold{ \: E = 1.87 \times {10}^{ - 7} \: N {C}^{ - 1} }}}$