Potential energy of point-charges: A -3.0-μC point charge and a -9.0-μC point charge are initially extremely far apart. How much work does it take to bring the -3.0-μC charge to x = 3.0 mm, y = 0.00 mm and the -9.0-μC charge to x = -3.0 mm, y = 0.00 mm? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The work done in bringing the charges together is equal to the change in potential energy. The potential energy of the system is given by:$$U = k\frac{q_1q_2}{r}$$where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.Initially, the charges are extremely far apart, so the potential energy is zero. When the charges are brought to a distance of 6.0 mm, the potential energy becomes:$$U = (8.99 	imes 10^9 N \cdot m^2/C^2)\frac{(-3.0 	imes 10^{-6} C)(-9.0 	imes 10^{-6} C)}{6.0 	imes 10^{-3} m} = -40 J$$The negative sign indicates that the work done is done on the system.Therefore, the work done in bringing the charges together is 40 J.

Potential Energy of Point-Charges: a -3