A point charge of +3.0 X 10-7 coulomb is placed 2.0 X 10-2 meter from a second point charge of +4.0 X 10-7 coulomb. What is the magnitude of the electrostatic force on the charges? A. 6.0 X 10-12 N
B. 3.0 X 10-10 N
C. 5.4 X 10-2 N
D. 2.7 N

D. 2.7 N

Explanation:

Applying

F = kq'q/r² Equation 1

Where F = force, k = coulomb's constant, q' = first charge, q = second charge, r = distance between the charge

From the question,

Given: q' = +3.0×10⁻⁷ C, q = +4.0×10⁻⁷C, r = 2.0×10⁻² m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 1

F = ( +3.0×10⁻⁷)(+4.0×10⁻⁷)(8.98×10⁹)/(2.0×10⁻²)²

F = 26.94×10⁻¹ N

F = 2.694 N

F ≈ 2.7 N

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up