Question

If two such generic humans each carried 3.0 C coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 700 N weight?

Answer 1

To find the distance between two point charges that have a force of 1.00 N between them, we can use Coulomb's Law.

Step-by-step explanation:

To find the distance between two point charges that have a force of 1.00 N between them, we can use Coulomb's Law. The equation for Coulomb's Law is:

F = k * (q1 * q2) / r^2

where F is the force, k is the electrostatic constant (9 * 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges. Rearranging the equation to solve for r:

r = sqrt((k * (q1 * q2)) / F)

Plugging in the known values into the equation, we have:

r = sqrt((9 * 10^9 N m^2/C^2 * (3.0 C) * (3.0 C)) / (700 N))

simplifying the equation gives:

r = sqrt((27 * 10^9 N m^2/C^2) / (700 N))

r ≈ 1.35 meters

Answer 2

Distance, r = 10757.05 meters

Step-by-step explanation:

Charge on two generic humans, q = 3 C

They have to be for the electric attraction between them to equal their 700 N weight, F = 700 N

Electric force is given by :

`F=(kq^2)/(r^2)`, r is the distance between charges

`r=\sqrt{(kq^2)/(F)`

`r=\sqrt{(9* 10^9* 3^2)/(700)}`

r = 10757.05 meter

So, distance between two charges is 10757.05 meters. Hence, this is the required solution.