Question

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150N/C$, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be :
$\left[\text{Given}{ϵ}_0=8.85\times 10^{-12}{C}^2/N-m^2,R_E=6.37\times 10^{6}m\right]$

• A. +670 kC
• B. -670 kC
• C. -680 kC
• D. +680 kC

Answer 1

Answer: (C)

$E=\frac{1}{4\pi {\epsilon }_0}\frac{\theta }{R^2}=\frac{\sigma }{{\epsilon }_0}\Rightarrow \sigma ={\epsilon }_{0}E$
$=8.85\times 10^{-12}\times 150$
$Q={\epsilon }_{0}E\times 4\pi R^2$
$=6.76\times 10^5\times 10^{-12}\times 10^{+12}$
$=680kC$
for inward will be negative.