A point charge of +3.00 μC and a second charge Q are initially very far apart. If it takes $29.0 \mathrm {~J}$ of work to bring them to a final configuration in which the $+ 3.00 - \mu C$ charge is at the point x = 1.00 mm, y = 1.00 mm, and the second charge Q is at the point x = 1.00 mm, y = 3.00 mm, find the magnitude of the charge Q. (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 potential energy of the final configuration. 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, and r is the distance between them.In this problem, we have:$$q_1 = +3.00 \mu C = +3.00 	imes 10^{-6} C$$$$x_1 = 1.00 mm = 1.00 	imes 10^{-3} m$$$$y_1 = 1.00 mm = 1.00 	imes 10^{-3} m$$$$x_2 = 1.00 mm = 1.00 	imes 10^{-3} m$$$$y_2 = 3.00 mm = 3.00 	imes 10^{-3} m$$$$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2.00 	imes 10^{-3} m)^2} = 2.00 	imes 10^{-3} m$$$$U = k\frac{q_1q_2}{r} = (8.99 	imes 10^9 N \cdot m^2/C^2)\frac{(3.00 	imes 10^{-6} C)q_2}{2.00 	imes 10^{-3} m}$$$$U = 13.49 q_2 J$$We are given that the work done in bringing the charges together is 29.0 J, so we can set U = 29.0 J and solve for q2:$$13.49 q_2 J = 29.0 J$$$$q_2 = \frac{29.0 J}{13.49 J/C} = 2.15 \mu C$$Therefore, the magnitude of the charge Q is 2.15 μC.

A Point Charge of +3 29.0 J29.0 \mathrm {~J}29.0 J Of Work to Bring Them to a Final Configuration in Q

A Point Charge of +3 $29.0 \mathrm {~J}$ Of Work to Bring Them to a Final Configuration in Q