Question

A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge

Answer 1

31.5N

I have confirmed it is the right answer.

Answer 2

F = 2.49 x 10⁻⁹ N

Step-by-step explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

`F = (kq_1q_2)/(r^2)\\`

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

`F = ((9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^(-19)\ C)(1.4\ C))/((0.9\ m)^2)`

F = 2.49 x 10⁻⁹ N