If the electric field intensity in a fair weather atmosphere is 100 V / m, then the total charge on the earth's surface is (radius of the earth is 6400 km )

Question

If the electric field intensity in a fair weather atmosphere is 100 V / m, then the total charge on the earth's surface is (radius of the earth is 6400 km ) (A) 4.55 × 107 C (B) 4.55 × 108 C (C) 4.55 × 105 C (D) 4.55 × 106 C

Tardigrade Admin · Accepted Answer

E=100 ( V / m ) R=6400 km By Gauss law E A=(q/ε0) ⇒ q=E A ε0 ⇒ q=200(4 π(6400 × 103)2) × 8.85 × 10-12 ⇒ q=4.55 × 105 C

Q. If the electric field intensity in a fair weather atmosphere is $100 V / m$ , then the total charge on the earth's surface is (radius of the earth is $6400 km$ )

$4.55 \times 1 0^{7} C$

$4.55 \times 1 0^{8} C$

$4.55 \times 1 0^{5} C$

$4.55 \times 1 0^{6} C$

$E = 100 \frac{V}{m}$
$R = 6400 km$
By Gauss law
$E A = \frac{q}{ε _{0}}$
$\Rightarrow q = E A ε_{0}$
$\Rightarrow q = 200 (4 π (6400 \times 1 0^{3})^{2}) \times 8.85 \times 1 0^{- 12}$
$\Rightarrow q = 4.55 \times 1 0^{5} C$

Q. If the electric field intensity in a fair weather atmosphere is 100V/m, then the total charge on the earth's surface is (radius of the earth is 6400km )

4.55×107C

4.55×108C

4.55×105C

4.55×106C