Two point charges q₁ = 4.0 μC and q₂ = -8.0 μC are placed along the x-axis at x₁ = 0 m and x₂ = 0.20 m, respectively. What is the electric potential energy of this system of charges?

Question

Quizplus · Accepted Answer

The electric potential energy of a system of charges is given by the equation:$$U = k\frac{q_1q_2}{r}$$where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.In this case, q1 = 4.0 μC, q2 = -8.0 μC, and r = 0.20 m. Substituting these values into the equation, we get:$$U = (9.0 	imes 10^9 \frac{Nm^2}{C^2})\frac{(4.0 	imes 10^{-6} C)(-8.0 	imes 10^{-6} C)}{(0.20 m)}$$$$U = -1.44 J$$Therefore, the electric potential energy of this system of charges is -1.44 J.

Two Point Charges Q1 = 4

Two Point Charges Q₁ = 4