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Q.
Suppose that a planet has a surface charge density of $1$ electron $/ m ^{2}$, given radius of the planet is $8850\, km$, then calculate potential of the planet due to this charge :
Solution:
If $R$ be the radius and $\sigma$ the surface charge density of earth than total charge on surface of earth is given by $Q=4 \pi R^{2} \sigma$
Potential on surface of earth
$V =\frac{1}{4 \pi \varepsilon_{0}} \frac{ Q }{ R }=\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \pi R ^{2} \sigma}{ R }=\frac{ R \sigma}{\varepsilon_{0}} $
$V =\frac{8850 \times 10^{3} \times\left(-1.6 \times 10^{-19}\right)}{8.85 \times 10^{-12}} $
$=-0.16$ volt