Practice: Earth's Net Charge

Since we are considering the Electric Field outside the Earth, we can consider Earth to be a point mass having charge $q$ , and taking the Electric Field at a distance of $R$ from this point, we have,

$E = \frac{kq}{R^2}$

Thus, substituting the values of $k$ , $R$ , and $E$ , we get,

$q = 676281.66$

Since the answer has to be an integer,

$q = \boxed{676282}$

$E = \frac{k \times Q}{R^{2}}$ we were looking for Q so,

$Q = \frac{E \times R^{2}}{k}$

$Q = \frac { 150 \times (6370 \times 10^{3})^{2}}{9 \times 10^{9}}$

$Q = 6.76282 \times 10^{5} = 676282 C$

We can use the formula E.d=U initially, because the 150V/m can be used in this case. The distance "d" is 6370000 m. Then, 150 . 6370000 = 955500000. This result is the electrical potential difference. We use it in other formula, V = k . Q/d . We have that 955500000 = 9 . 10E9 . Q/6370000. Isolating Q, we have that Q= 676281.666666. Q = net charge in Coulombs.

• From Gauss Law, $q/ε = E*A$
• A = $4πr^{2}$ $m^{2}$
  C
  
• q = $4πεr^{2} * E$

• E = 150 V/m

• q = $15*637*637*10^{9}$ / $9*10^{9}$ = 676281.667 c