Moving 2.0 coulombs of charge a distance of 6.0 meters from point A to point B within an electric field requires a 5.0-newton force. What is the electric potential difference between points A and B?

15 V

Explanation:

Electric potential is given as the product of Electric field and distance. Mathematically:

V = E * r

Where E = Electric field

r = distance = 6m

Electric field, E, is:

E = F / q

Where F = Electric force = 5N

q = Electric charge = 2C

Therefore, electric potential will be:

V = (F * r) / q

V = (5 * 6) / 2

V = 15V

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explanation:

that would be because of both gravity and friction. so     c