Question

Electric Field of the Earth. The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth’s electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth’s electric field a feasible means of flight? Why or why not?

Short answer

Magnitude and sign of the charge is $q=-3.9C$

Force of repulsion is $1.40\times {10}^{7}N$

Step-by-step solution

Step 1:

E=150N/C and directed downwards to the center of the earth, as the charge acquired by the human body with mass m=60kg to overcome the weight within the same time of exerted force by E.

Thus, the force (F) is equal to the weight

$$\begin{gathered} F=W=mg \\ F=\left|q\right|E \\ \left|q\right|=\frac{F}{E} \\ \left|q\right|=\frac{mg}{E} \end{gathered}$$

Putting the value,

$$\begin{gathered} \left|q\right|=\frac{60\times 9.8}{150} \\ \left|q\right|=3.9C \end{gathered}$$

The charge will be q=-3.9C.

Therefore, the Magnitude and sign of the charge is$q=-3.9C$

Step 2:

For the repulsive force between two human bodies with charge, and separate by the distance r.

$F=\frac{1}{4\pi {\in }_0}.\frac{q}{r^2}$

Putting the values:

$$\begin{gathered} F=\left(9\times {10}^9\right)\times \frac{\left(-3.9\right)}{\left({100}^2\right)} \\ =1.40\times {10}^{7}N \end{gathered}$$

Therefore, force of repulsion is$1.40\times {10}^{7}N$