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 [ Given \; \; \epsilon _{0} = 8.85\times 10^{-12}\: \: \: C^{2}/N-m^{2},R_{E}= 6.37\times 10^{6}m\right ]

  • Option 1)

    +670 kC

  • Option 2)

    - 670 kC

  • Option 3)

    - 680 kC

  • Option 4)

    + 680 kC

 

Answers (2)
N neha
S solutionqc

As we discussed in the concept

Infinite Plane parallel sheets of charge -

If

\sigma _{A}=\sigma and \sigma _{B}=-\sigma \rightarrow  E_{p}= E_{R}=0

and 

E_{Q}= \frac{\sigma }{\epsilon _{0}}

-

 

 Electric Field E = 150 N/C

Total surface charge carried by earth q= ?

q=\varepsilon _{0EA} = \varepsilon _{0}E \pi r^{2}

= 8.85\times 10^{-12}\times 150\times (6.37\times 10^{6})^{2}

\simeq 680 KC

As electric field is directed inwards, hence, q=-680 KC

 


Option 1)

+670 kC

Option 2)

- 670 kC

Option 3)

- 680 kC

Option 4)

+ 680 kC

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