Two +1C charges are separated by 3000m. What is the magnitude of the electric force between them?

1000 N

Explanation:

The Electric Force between two charged particles with charges q1 and q2 respectively, separated by a distance, d, is given as:

F = (k * q1 * q2) / d²

Where k = Coulombs constant

From the question:

Charge of first particle = charge of second particle = q = +1 C

Distance between both particles, d = 3000 m

Therefore, Electric Force is:

F = (9 * 10^9 * 1 * 1) / (3000²)

F = (9 * 10^9) / (9 * 10^6)

F = 1 * 10^3 N = 1000 N

The electric force between both charges is 1000 N.

1000 N

Explanation:

The electric force between two charges is given as

F = kq'q/r² Equation 1

Where F = electric force between the charges, q' = magnitude of the first charge, q = magnitude of the second charge, r = distance between the charges, k = coulombs constant

Given: q' = q = 1 C, r = 3000 m, k = 9×10⁹ Nm²/kg²

Substitute into equation 2

F = 1²×(9×10⁹)/3000²

F = 1000 N.

Hence the magnitude of the electric force between them = 1000 N

c. bronchi

i know this is true mom has it.

mark me branliest only got it 1 time!

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge