A tiny object carrying a charge of +3.00 μC and a second tiny charged object are initially very far apart. If it takes 29.0 J of work to bring them to a final configuration in which the +3.0.0-μC object i is at x = 1.00 mm, y = 1.00 mm, and the other charged object is at x = 1.00 mm, y = 3.00 mm, find the magnitude of the charge on the second object. (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A) 2.15 μC B) 4.30 μC C) 10.74 μC D) 4.30 nC

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Quizplus · Accepted Answer

The work done in bringing the two charges together is equal to the potential energy of the system. The potential energy of two point charges is given by:$$U = k\frac{q_1q_2}{r}$$where k is Coulomb's constant, q1 and q2 are the charges of the two objects, and r is the distance between them.In this problem, we have:$$q_1 = +3.00 \mu C = 3.00 	imes 10^{-6} C$$$$x_1 = 1.00 mm = 1.00 	imes 10^{-3} m$$$$y_1 = 1.00 mm = 1.00 	imes 10^{-3} m$$$$x_2 = 1.00 mm = 1.00 	imes 10^{-3} m$$$$y_2 = 3.00 mm = 3.00 	imes 10^{-3} m$$$$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 2.00 	imes 10^{-3} m$$$$U = 29.0 J$$Substituting these values into the equation for potential energy, we get:$$29.0 J = (8.99 	imes 10^9 N \cdot m^2/C^2)\frac{(3.00 	imes 10^{-6} C)q_2}{2.00 	imes 10^{-3} m}$$Solving for q2, we get:$$q_2 = 2.15 \mu C$$Therefore, the magnitude of the charge on the second object is 2.15 μC.

A Tiny Object Carrying a Charge of +3