Four point charges are located at the corners of a square, 1.00 m by 1.00 m. On each of two diagonally opposite corners are 1.00 µC charges. On each of the other two corners are −1.00 µC charges. What is the magnitude of the force on one of the positive charges? A) 8.2 × 10^-3 N B) 1.3 × 10^-2 N C) 3.2 × 10^-3 N D) 1.9 × 10^-2 N E) 0.00

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Quizplus · Accepted Answer

The force on the positive charge is due to the two negative charges. The force between two point charges is given by Coulomb's law:$$F = k\frac{q_1q_2}{r^2}$$where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.In this case, the charges are q1 = -1.00 µC and q2 = 1.00 µC, and the distance between them is r = 1.00 m.Substituting these values into Coulomb's law, we get:$$F = (9 	imes 10^9 \frac{Nm^2}{C^2})\frac{(-1.00 	imes 10^{-6} C)(1.00 	imes 10^{-6} C)}{(1.00 m)^2}$$$$F = -9.00 	imes 10^{-3} N$$The negative sign indicates that the force is attractive.The magnitude of the force is:$$|F| = 9.00 	imes 10^{-3} N$$Therefore, the magnitude of the force on one of the positive charges is 9.00 × 10^-3 N, which is choice A.

Four Point Charges Are Located at the Corners of a Square