An electric dipole with ±5.0 μC point charges is positioned so that the positive charge is $1.0 \mathrm {~mm}$ to the right of the origin and the negative charge is at the origin. How much work does it take to bring a $3.0 - \mu \mathrm { C }$ point charge from very far away to the point x = 3.0 mm, y = 0.0 mm? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The work done by the electric field of a dipole on a point charge is given by:$$W = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$$where $q_1$ and $q_2$ are the charges of the dipole, $r$ is the distance between the point charge and the dipole, and $\epsilon_0$ is the permittivity of free space.In this case, $q_1 = 5.0 \mu C$, $q_2 = -5.0 \mu C$, $r = 3.0 mm = 0.003 m$, and $\epsilon_0 = 8.85 \times 10^{-12} C^2/Nm^2$.Substituting these values into the equation, we get:$$W = \frac{1}{4\pi (8.85 \times 10^{-12} C^2/Nm^2)} \frac{(5.0 \mu C)(-5.0 \mu C)}{(0.003 m)^2}$$$$W = 19 J$$Therefore, the correct answer is A.

An Electric Dipole with ±5 1.0 mm1.0 \mathrm {~mm}1.0 mm To the Right of the Origin and the Negative Charge

An Electric Dipole with ±5 $1.0 \mathrm {~mm}$ To the Right of the Origin and the Negative Charge