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 × 10⁹N ∙ m²/C²)

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Quizplus · Accepted Answer

The work done in bringing the charges together is given by:$$W = \frac{kq_1q_2}{r_f} - \frac{kq_1q_2}{r_i}$$where:* k is Coulomb's constant* q_1 and q_2 are the charges* r_f is the final distance between the charges* r_i is the initial distance between the chargesIn this case, the initial distance between the charges is extremely far apart, so we can assume that r_i = ∞. The final distance between the charges is 6.0 mm, or 0.006 m.Plugging in the values, we get:$$W = \frac{(8.99 	imes 10^9 N \cdot m^2/C^2)(-3.0 	imes 10^{-6} C)(-9.0 	imes 10^{-6} C)}{0.006 m} - \frac{(8.99 	imes 10^9 N \cdot m^2/C^2)(-3.0 	imes 10^{-6} C)(-9.0 	imes 10^{-6} C)}{\infty}$$$$W = 40 J$$Therefore, the work done in bringing the charges together is 40 J.

A -30-μC Point Charge and a -9