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 in bringing a point charge from infinity to a point in the electric field is given by the negative of the potential energy at that point. The potential energy of a point charge q at a distance r from a dipole of charge q' and separation d is given by:$$U = \frac{1}{4\pi \epsilon_0} \frac{q q'}{r^2} \left ( 1 + \frac{3cos^2 	heta - 1}{2} \frac{d^2}{r^2} ight )$$In this case, q = 3.0 μC, q' = 5.0 μC, d = 1.0 mm, r = 3.0 mm, and θ = 0°. Plugging these values into the equation, we get:$$U = \frac{(9.0 	imes 10^9 	ext{ N} \cdot 	ext{m}^2/	ext{C}^2)(3.0 	imes 10^{-6} 	ext{ C})(5.0 	imes 10^{-6} 	ext{ C})}{(3.0 	imes 10^{-3} 	ext{ m})^2} \left ( 1 + \frac{3cos^2 0° - 1}{2} \frac{(1.0 	imes 10^{-3} 	ext{ m})^2}{(3.0 	imes 10^{-3} 	ext{ m})^2} ight )$$$$U = 23 	ext{ J}$$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