Question

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150\, N/C$, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be :
$\left[\text{Given}\, \epsilon_{0} = 8.85 \times 10^{-12} \,C^{2}/N-m^{2},\,R_{E} = 6.37 \times 10^{6} m\right]$

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

Answer 1

Answer: (C)

$E=\frac{1}{4\pi\varepsilon_{0}} \frac{\theta}{R^{2}}=\frac{\sigma}{\varepsilon_{0}} \Rightarrow \sigma=\varepsilon_{0}E$
$=8.85\times10^{-12}\times 150$
$Q=\varepsilon_{0}E\times 4\pi R^{2}$
$=6.76\times 10^{5}\times 10^{-12}\times 10^{+12}$
$=680\,kC$
for inward will be negative.