A dipole with ±6.0 μC charges is positioned so that the positive charge is 1.0 mm to the right of the origin and the negative charge is at the origin.How much work does it take to bring a 10.0 μC charge from infinity to a position x = 3.0 mm,y = 0.0 mm? (The value of k is 9.0 $\times$ 10⁹ N·m²/C². )

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Quizplus · Accepted Answer

The work done in bringing a charge q from infinity to a point near a dipole is given by $$W=k\frac{qQ}{r^2}$$where k is the Coulomb constant, Q is the charge of the dipole, q is the charge brought from infinity and r is the distance between the charge and the dipole.In this case, Q = 6.0 μC, q = 10.0 μC, and r = 3.0 mm = 0.003 m. Substituting these values into the equation, we get:$$W=9.0	imes10^9\frac{(6.0	imes10^{-6})(10.0	imes10^{-6})}{(0.003)^2}=90	ext{ J}$$

A Dipole with ±6 ×\times× 109 N·m2/C2

A Dipole with ±6 $\times$ 10⁹ N·m²/C²