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 +3.00 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)

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Quizplus · Accepted Answer

The electric potential energy between 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* r is the distance between the two objectsIn this problem, we are given that U = 29.0 J, q1 = +3.00 μC, x1 = 1.00 mm, y1 = 1.00 mm, x2 = 1.00 mm, and y2 = 3.00 mm. We can use the distance formula to find the distance between the two objects:$$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$$$r = \sqrt{(1.00 	ext{ mm} - 1.00 	ext{ mm})^2 + (3.00 	ext{ mm} - 1.00 	ext{ mm})^2}$$$$r = 2.00 	ext{ mm} = 0.002 	ext{ m}$$Now we can solve for q2:$$U = k\frac{q_1q_2}{r}$$$$29.0 	ext{ J} = (8.99 	imes 10^9 	ext{ N} \cdot 	ext{m}^2/	ext{C}^2)\frac{(3.00 	imes 10^{-6} 	ext{ C})q_2}{0.002 	ext{ m}}$$$$q_2 = \frac{29.0 	ext{ J} \cdot 0.002 	ext{ m}}{(8.99 	imes 10^9 	ext{ N} \cdot 	ext{m}^2/	ext{C}^2)(3.00 	imes 10^{-6} 	ext{ C})}$$$$q_2 = 2.15 	imes 10^{-6} 	ext{ C} = 2.15 \mu	ext{C}$$Therefore, the magnitude of the charge on the second object is 2.15 μC.

A Tiny Object Carrying a Charge of +3