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) A)40 J B)81 J C)27 J D)6.8 J

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Quizplus · Accepted Answer

The work done in bringing the 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 the Coulomb constant* q_1 and q_2 are the charges* r is the distance between the chargesIn this case, the charges are -3.0 μC and -9.0 μC, and the distance between them is 6.0 mm. Substituting these values into the equation, we get:$$U = (8.99 	imes 10^9 N \cdot m^2/C^2)\frac{(-3.0 	imes 10^{-6} C)(-9.0 	imes 10^{-6} C)}{6.0 	imes 10^{-3} m}$$$$U = 40 J$$Therefore, the work done in bringing the charges together is 40 J.

A -30-μC Point Charge and a -9