In the figure, +4.0-μC and -4.0-μC point charges are located as shown. Now an additional +2.00-µC point charge is placed at point A. What is the electric potential energy of this system of three charges, relative to infinity? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The electric potential energy of a system of charges is given by the formula: 
U = k*q1*q2/r, where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

First, we need to calculate the electric potential energy between the +4.0-μC and -4.0-μC charges. Their distance is 0.08 m, and their charges have opposite signs, so the electric potential energy is:

U1 = (8.99 × 10^9 N∙m^2/C^2) * (4.0×10^-6 C) * (-4.0×10^-6 C) / (0.08 m) = -0.1798 J

Next, we need to calculate the electric potential energy between the +2.00-µC point charge and each of the other charges. The distance between the +2.00-µC charge and the +4.0-μC charge is 0.04 m, and their charges have the same sign, so the electric potential energy is:

U2a = (8.99 × 10^9 N∙m^2/C^2) * (2.0×10^-6 C) * (4.0×10^-6 C) / (0.04 m) = +0.8996 J

The distance between the +2.00-µC charge and the -4.0-μC charge is also 0.04 m, but their charges have opposite signs, so the electric potential energy is:

U2b = (8.99 × 10^9 N∙m^2/C^2) * (2.0×10^-6 C) * (-4.0×10^-6 C) / (0.04 m) = -0.8996 J

Now we can calculate the total electric potential energy of the system:

Utotal = U1 + U2a + U2b = -0.1798 J + 0.8996 J - 0.8996 J = -0.1798 J

Since the problem asks for the energy relative to infinity, we need to subtract the electric potential energy of the +2.00-µC charge when it is at infinity. At infinity, the electric potential energy is zero, so:

Uinfinity = 0 J

The final answer is:

Ufinal = Utotal - Uinfinity = -0.1798 J - 0 J = -0.1798 J ≈ -264 mJ (rounded to three significant figures)

Therefore, the best choice is B (-264 mJ) as the answer.

In the Figure, +4