A point charge of $+ 3.00 \mu \mathrm { C }$ and a second charge $Q$ are initially very far apart. If it takes $29.0 \mathrm {~J}$ of work to bring them to a final configuration in which the $+ 3.00 - \mu \mathrm { C }$ charge is at the point $x = 1.00$ $\mathrm { mm } , y = 1.00 \mathrm {~mm}$, and the second charge $Q$ is at the point $x = 1.00 \mathrm {~mm} , y = 3.00 \mathrm {~mm}$, find the magnitude of the charge $Q . \left( k = 1 / 4 \pi \varepsilon _ { 0 } = 8.99 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The work done in bringing the charges together is equal to the potential energy of the final configuration. 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 problem, we have:$$q_1 = 3.00 \mu \mathrm { C } = 3.00 \times 10^{-6} \mathrm { C }$$$$x_1 = 1.00 \mathrm { mm } = 1.00 \times 10^{-3} \mathrm { m }$$$$y_1 = 1.00 \mathrm { mm } = 1.00 \times 10^{-3} \mathrm { m }$$$$x_2 = 1.00 \mathrm { mm } = 1.00 \times 10^{-3} \mathrm { m }$$$$y_2 = 3.00 \mathrm { mm } = 3.00 \times 10^{-3} \mathrm { m }$$$$U = 29.0 \mathrm { J }$$Substituting these values into the equation for potential energy, we get:$$29.0 \mathrm { J } = (8.99 \times 10^9 \mathrm { N } \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }) \frac{(3.00 \times 10^{-6} \mathrm { C })Q}{(1.00 \times 10^{-3} \mathrm { m })^2 + (2.00 \times 10^{-3} \mathrm { m })^2}$$Solving for $Q$, we get:$$Q = 2.15 \mu \mathrm { C }$$Therefore, the magnitude of the charge $Q$ is $2.15 \mu \mathrm { C }$.

A Point Charge Of +3.00μC+ 3.00 \mu \mathrm { C }+3.00μC

A Point Charge Of $+ 3.00 \mu \mathrm { C }$