Three charges are placed as follows along the x- and y-axis of an xy-coordinate system: q₁ = 2.00 μC at x₁ = 0 m, q₂ = -4.00 μC at x₂ = 3.00 m, and q₃ = 6.00 μC at y = 4.00 m. 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:$$U = k\frac{q_1q_2}{r_{12}} + k\frac{q_1q_3}{r_{13}} + k\frac{q_2q_3}{r_{23}}$$where k is Coulomb's constant, q₁, q₂, and q₃ are the charges of the three particles, and r₁₂, r₁₃, and r₂₃ are the distances between the particles.In this problem, we have:$$q_1 = 2.00 \mu C = 2.00 imes 10^{-6} C$$$$q_2 = -4.00 \mu C = -4.00 imes 10^{-6} C$$$$q_3 = 6.00 \mu C = 6.00 imes 10^{-6} C$$$$r_{12} = 3.00 m$$$$r_{13} = \sqrt{4.00^2 + 3.00^2} m = 5.00 m$$$$r_{23} = 4.00 m$$Substituting these values into the equation for U, we get:$$U = (9.00 imes 10^9 N m^2/C^2)\frac{(2.00 imes 10^{-6} C)(-4.00 imes 10^{-6} C)}{3.00 m} + (9.00 imes 10^9 N m^2/C^2)\frac{(2.00 imes 10^{-6} C)(6.00 imes 10^{-6} C)}{5.00 m} + (9.00 imes 10^9 N m^2/C^2)\frac{(-4.00 imes 10^{-6} C)(6.00 imes 10^{-6} C)}{4.00 m}$$$$U = -40.2 mJ$$Therefore, the electric potential energy of this system of charges is -40.2 mJ.

Three Charges Are Placed as Follows Along the X- and Y-Axis