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 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?

Answer 1

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

$\Sigma F = F_{E}-W = 0$ (Eq. 1)

Where:

$F_{E}$ - Electrostatic force exerted on human, measured in Newton.

$W$ - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

$q\cdot E-m\cdot g = 0$

$q = \frac{m\cdot g}{E}$ (Eq. 2)

$E$ - Electric field, measured in Newtons per Coloumb.

$m$ - Mass, measured in kilograms.

$g$ - Gravity acceleration, measured in meters per square second.

$q$ - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that $m = 60\,kg$, $g = 9.807\,\frac{m}{s^{2}}$ and $E = -150\,\frac{N}{C}$, the charge that a 60-kg human must have to overcome weight is:

$q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }$

$q = -3.923\,C$

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

$F = \kappa \cdot \frac{q^{2}}{r^{2}}$ (Eq. 3)

Where:

$\kappa$ - Electrostatic constant, measured in Newton-square meter per square Coulomb.

$q$ - Electric charge, measured in Coulomb.

$r$ - Distance between two people, measured in meters.

If we know that $\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$, $q = -3.923\,C$ and $r = 100\,m$, then the force of repulsion between two people is:

$F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]$

$F = 13.851\times 10^{6}\,N$

The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.