If atmospheric electric field is approximately 150 volt/m and radius of the earth is 6400 km, then the total charge on the earth's surface is

Question

If atmospheric electric field is approximately 150 volt/m and radius of the earth is 6400 km, then the total charge on the earth's surface is (A) 6.8 × 105 coulomb (B) 6.8 × 106 coulomb (C) 6.8 × 104 coulomb (D) 6.8 × 109 coulomb

Tardigrade Admin · Accepted Answer

E=150 V / m R=6400 , km4 Using Gauss' Law ⇒ E A=(q/ε0) ⇒ 150(4 π(6400 × 103)2)=(q/ε0) q=150 × 4 π ×(6400 × 103)2 × 8.854 × 10-12 q=6.8 × 105 C

Q. If atmospheric electric field is approximately $150$ volt/m and radius of the earth is $6400 km$ , then the total charge on the earth's surface is

$6.8 \times 1 0^{5}$ coulomb

$6.8 \times 1 0^{6}$ coulomb

$6.8 \times 1 0^{4}$ coulomb

$6.8 \times 1 0^{9}$ coulomb

$E = 150 V / m$
$R = 6400, km 4$
Using Gauss' Law
$\Rightarrow E A = \frac{q}{ε _{0}}$
$\Rightarrow 150 (4 π (6400 \times 1 0^{3})^{2}) = \frac{q}{ε _{0}}$
$q = 150 \times 4 π \times (6400 \times 1 0^{3})^{2} \times 8.854 \times 1 0^{- 12}$
$q = 6.8 \times 1 0^{5} C$

Q. If atmospheric electric field is approximately 150 volt/m and radius of the earth is 6400km, then the total charge on the earth's surface is

6.8×105 coulomb

6.8×106 coulomb

6.8×104 coulomb

6.8×109 coulomb