Question

Suppose that a planet has a surface charge density of $1$ electron $/m^2$, given radius of the planet is $8850km$, then calculate potential of the planet due to this charge :

• A. -0.16 V
• B. -160 V
• C. 0.16 V
• D. 160 V

Answer 1

Answer: (A)

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
$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi R^2\sigma }{R}=\frac{R\sigma }{{\epsilon }_0}$
$V=\frac{8850\times 10^3\times \left(-1.6\times 10^{-19}\right)}{8.85\times 10^{-12}}$
$=-0.16$ volt