Question 308 of 311

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 : [Given : ${\in }_{\text{O}}=\text{8.85}\times 1{0}^{-12}{\text{C}}^{2}/{\text{N-m}}^{2}\text{,}{\text{R}}_{\text{E}}=\text{6.37}\times 1{0}^{6}\text{m}$]

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

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Answer: (A) $-680kC$

$E=150N/C$

We know that $\phi =^{}\vec{\text{E}}^{}\cdot ^{}\vec{\text{A}}$

$=\text{E}\times 4\pi {}^{}{\text{r}}^{2}$

$=150\times 4\pi \times {\left({}_{}\right)}^{}{\left({\text{R}}_{\text{e}}\right)}^{\left({}_{}\right)}$

Now, $\phi =\dfrac{{}_{}}{}\dfrac{{\text{Q}}_{\text{in}}}{{}_{}}$



${\text{Q}}_{\text{in}}{}_{}=\phi {}_{}{\in }_{\text{o}}{}_{}=150\times 4\pi {\left({}_{}\right)}^{}{\left({\text{R}}_{\text{e}}\right)}^{\left({}_{}\right)}{}^{}\times \text{8.85}\times 1{}^{}{0}^{}{}^{} \text{C}$

$=150\times 4\pi {\left\{\right\}}^{}{\left\{\text{6.37}\times 1{}^{}{0}^{6}\right\}}^{\left\{{}^{}\right\}}{}^{}\times \text{8.85}\times 1{}^{}{0}^{}$

$=600\pi \times \text{6.37}\times \text{6.37}\times \text{8.85}C$

$=680 \text{kC}$

As Electric field is inward, $Q$ will be negative