A 7.0-μC point charge and a $9.0 - \mu \mathrm { C }$ point charge are initially extremely far apart. How much work does it take to bring the $7.0 - \mu \mathrm { C }$ point charge to the point $x = 3.0 \mathrm {~mm}$ $y = 0.0 \mathrm {~mm}$ , and the $9.0 - \mu \mathrm { C }$ point charge to the point $x = - 3.0 \mathrm {~mm}$ $y = 0.0 \mathrm {~mm} ?$ (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Question

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 Coulomb's constant, q_1 and q_2 are the charges, and r is the distance between them.In this case, the charges are initially extremely far apart, so the potential energy is zero. The work done in bringing the charges together is equal to the final potential energy:$$U = k\frac{q_1q_2}{r} = (9.0 	imes 10^9 	ext{ N}\cdot	ext{m}^2/	ext{C}^2)\frac{(7.0 	imes 10^{-6} 	ext{ C})(9.0 	imes 10^{-6} 	ext{ C})}{6.0 	imes 10^{-3} 	ext{ m}} = 95 	ext{ J}$$Therefore, the correct answer is A.

A 70-μC Point Charge and A A 9.0 9.0−μC9.0 - \mu \mathrm { C }9.0−μC

A 70-μC Point Charge and A A 9.0 $9.0 - \mu \mathrm { C }$