A) + 670 kC
B) - 670 kC
C) 680 kC
D) + 680 kC
Correct Answer: C
Solution :
[c] Given, |
Electric field E = 150 N/C |
Total surface charge carried by earth q = ? |
According to Gausss law. |
\[\phi =\frac{q}{{{\in }_{0}}}=EA\]or\[q={{\in }_{0}}EA\] |
\[={{\in }_{0}}E\pi {{r}^{2}}.\] |
\[=8.85\times {{10}^{-12}}\times 150\times {{(6.37\times {{10}^{6}})}^{2}}.\] |
\[\simeq 680Kc\] |
As electric field directed inward hence |
q = - 680 Kc |
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