Supposing that the earth has a surface charge density of 1 electron/m2; calculate electric field just outside earth's surface.
The electronic charge is $-1.6 \times 10^{-19} C$ and earth's radius is $6.4 \times 10^6 m$. $\left(\varepsilon_0=8.9 \times 10^{-12} C^2 / N-m^2\right)$
Answer & explanation
Correct answer: option 1
Again, the electric field E just outside the earth's surface is same is if the entire charge q were concentrated at this centre. Thus
$E=\frac{1}{4 \pi \varepsilon_0} \frac{q}{R^2}=\frac{1}{4 \pi \varepsilon_0} \frac{4 \pi R^2 \sigma}{R^2}=\frac{\sigma}{\varepsilon_0}$
Substituting the given value:
$E =\frac{\left(-1.6 \times 10^{-19}\right) C / m^2}{8.9 \times 10^{-12} C^2 / N-m^2}$
$= -1.8 \times 10^{-8}$ N/C
The minus sign indicates that E is radially inward.