An electric dipole with ±5.0 ?C 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 = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²)

Question

Quizplus · Accepted Answer

The work done in bringing a point charge from infinity to a point in the electric field of a dipole is given by:$$W = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r_1 - r_2}$$where $q_1$ and $q_2$ are the charges of the dipole, $r_1$ is the distance from the positive charge to the point, and $r_2$ is the distance from the negative charge to the point.In this case, $q_1 = 5.0 \mu C$, $q_2 = -5.0 \mu C$, $r_1 = 4.0 mm$, and $r_2 = 3.0 mm$. Plugging these values into the equation, we get:$$W = \frac{1}{4\pi (8.85 \times 10^{-12} C^2/Nm^2)} \frac{(5.0 \times 10^{-6} C)(-5.0 \times 10^{-6} C)}{4.0 \times 10^{-3} m - 3.0 \times 10^{-3} m}$$$$W = 23 J$$Therefore, the work done in bringing the point charge from infinity to the point x = 3.0 mm, y = 0.0 mm is 23 J.

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