assume that Earth has a surface charge density of 1 electron per metre

1.	assume that Earth has a surface charge density of 1 electron per metre^2 . calculate the earth's electric field
Answer» 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: $\boxed{ \red{ \bold{ \: E = 1.87 \times {10}^{ - 7} \: N {C}^{ - 1} }}}$