Q. The surface charge density $(in\ C/m^{2})$ of the earth is about
A) $10^{- 9}$
B) $- 10^{9}$
C) $10^{9}$
D) $- 10^{- 9}$
Explanation
(d) The negative charge distributed over the surface of earth is about $5 \times 10^{5}C.$ The surface charge density of earth is
$\sigma = \frac{Q}{4\pi r_{e}^{2}} = \frac{- 5 \times 10^{5}C}{4\pi\left( 6.37 \times 10^{6}m \right)^{2}}\sim - 10^{- 9}C/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.