Question

The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these charges if Q1 is located at (2, 5, 1) and Q2 is located at (3, 2, 3).

Answer 1

F = 3.86 x 10⁻⁶ N

Step-by-step explanation:

First, we will find the distance between the two particles:

`r = \sqrt{(x_(2)-x_(1))^2+(y_(2)-y_(1))^2+(z_(2)-z_(1))^2}\\`

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

`r = √((3-2)^2+(2-5)^2+(3-1)^2)\\r = 3.741\ m\\`

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

`F = (kq_(1)q_(2))/(r^2)\\`

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

`F = ((9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^(-8)\ C)(3\ x\ 10^(-7)\ C))/((3.741\ m)^2)\\`

F = 3.86 x 10⁻⁶ N