A 7.0-?C 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 = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²)

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Quizplus · Accepted Answer

The work done in bringing the charges together is given by the potential energy of the system:$$U = k\frac{q_1q_2}{r}$$where $k$ is Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.In this case, $q_1 = 7.0 \mu C = 7.0 \times 10^{-6} C$, $q_2 = 9.0 \mu C = 9.0 \times 10^{-6} C$, and $r = 6.0 mm = 6.0 \times 10^{-3} m$.Substituting these values into the equation, we get:$$U = (9.0 \times 10^9 N \cdot m^2/C^2)\frac{(7.0 \times 10^{-6} C)(9.0 \times 10^{-6} C)}{6.0 \times 10^{-3} m} = 95 J$$Therefore, the work done in bringing the charges together is 95 J.

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

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