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 (approximately):
$[Given{\epsilon }_0=8.85\times {10}^{-12}{C}^2/{N-m}^2,R_E=6.37\times 10^{6}m]$

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

Answer 1

Answer: (D)

$+680kC$

Earth $=150N/C$

${ϵ}_0=8.85\times {10}^{-12}{C}^2/N-m^2$

$R_E=6.37\times {10}^{6}m$

Given, Electric field $E=150N/C$

Total surface charge carried by earth $q=?$

According to Gauss's law

$\varphi =\frac{q}{{ϵ}_0}=EA$

$q={ϵ}_{0}EA$

$={ϵ}_{0}E\pi r^2$

$=8.854\times {10}^{-12}\times 150\times {(6.37\times {10}^6)}^2$

$=680KC$