Question

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 charge would a human with a mass of 62.0 kg have to acquire to overcome his or her weight by the force exerted by the earth's electric field?

B) What would be the magnitude of the repulsive force between two people each with the charge calculated in part A and separated by a distance of 110 m ?

C) Is use of the earth’s electric field a feasible means of flight? Why or why not?

Answer 1

To solve this problem we will start using the concepts related to the electric field, from there we will find the load exerted on the body. Through this load it will be possible to make a sum of forces in balance to find the load that a human supports. Finally with these values it will be possible to find the repulsive force. We will proceed as follows,

The electric field is

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

Here,

k = Coulomb's Constant

Q = Charge

R = Distance (At this case from the center of mass of the earth to the surface)

Rearranging to find the charge,

$Q = \frac{ER^2}{k}$

Replacing,

$Q = frac{(150)(6.38*10^6)}{8.99*10^9}$

$Q = 6.79*10^5 C$

Since the electric field is directed towards the center of earth, the charge is negative.

PART A) Once the load is found we can proceed to apply the balance of Forces, for which the electrostatic force must be equivalent to the weight, this in order to satisfy the balance, therefore

$F_w = F_e$

$mg = \frac{kQq}{R^2}$

Replacing,

$(62)(9.8) = \frac{(8.99*10^9)(q)(-6.79*10^5)}{(6.38*10^6)^2}$

Solving for q,

$q = -4.056C$

PART B) Finally using the given distance and the values of the found load we can find the repulsive Force, which is

$F =\frac{kq^2}{d^2}$

$F = \frac{(8.99*10^9)(-4.056)^2}{110^2}$

$F = 1.22*10^7N$

PART C) The answer is no. According to the information found, we can conclude that traveling through an electric field is not viable because there is a repulsive force of great magnitude acting on the body.