$\mathrm { A } + 5.0 - \mathrm { nC }$ charge is at the point $( 0.00 \mathrm {~m} , 0.00 \mathrm {~m} )$ and a $- 2.0 - \mathrm { nC }$ charge is at $( 3.0 \mathrm {~m} , 0.00 \mathrm {~m} )$. What work is required to bring a $1.0 - \mathrm { nC }$ charge from very far away to point $( 0.00 \mathrm {~m} , 4.0 \mathrm {~m} )$ ? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } \right.$ $= 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$ )

Question

Quizplus · Accepted Answer

The work required to bring a charge from infinity to a point in an electric field is given by the potential energy difference, $W = qV$, where $q$ is the charge being moved and $V$ is the electric potential at the point. The electric potential due to a point charge is given by $V = k \frac{q}{r}$, where $k$ is Coulomb's constant, $q$ is the charge creating the potential, and $r$ is the distance from the charge to the point of interest. For the $5.0 \, \text{nC}$ charge at $(0,0)$, the distance to $(0,4)$ is $4 \, \text{m}$, and for the $-2.0 \, \text{nC}$ charge at $(3,0)$, the distance to $(0,4)$ is $5 \, \text{m}$ (using the Pythagorean theorem). The total potential at $(0,4)$ is the sum of potentials due to each charge:$$V_{\text{total}} = k \left( \frac{5.0 \times 10^{-9}}{4} + \frac{-2.0 \times 10^{-9}}{5} \right)$$$$V_{\text{total}} = 9 \times 10^9 \left( \frac{5}{4} \times 10^{-9} - \frac{2}{5} \times 10^{-9} \right)$$$$V_{\text{total}} = 9 \times 10^9 \left( 1.25 - 0.4 \right) \times 10^{-9}$$$$V_{\text{total}} = 9 \times 10^9 \times 0.85 \times 10^{-9} = 7.65 \times 10^{-9} \, \text{J/C}$$The work required to bring a $1.0 \, \text{nC}$ charge is:$$W = qV = 1.0 \times 10^{-9} \times 7.65 \times 10^{-9} = 7.65 \, \text{nJ}$$Rounding to two significant figures gives $7.7 \, \text{nJ}$, making choice B correct.

A+5.0−nC\mathrm { A } + 5.0 - \mathrm { nC }A+5.0−nC

$\mathrm { A } + 5.0 - \mathrm { nC }$