A 6.0 mC charge is placed at the origin and a second charge is placed on the x-axis at x = 0.30 m.If the resulting force on the second charge is 5.4 N in the positive x-direction,what is the value of its charge?

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The force on the second charge is given by Coulomb's law:$$F = k\frac{q_1q_2}{r^2}$$where k is Coulomb's constant, q1 is the charge at the origin, q2 is the charge on the x-axis, and r is the distance between the charges.We are given that F = 5.4 N, q1 = 6.0 mC, r = 0.30 m, and we want to find q2.Substituting these values into Coulomb's law, we get:$$5.4 N = (9.0 	imes 10^9 Nm^2/C^2)\frac{(6.0 	imes 10^{-3} C)q_2}{(0.30 m)^2}$$Solving for q2, we get:$$q_2 = 9.0 	imes 10^{-3} C = 9.0 mC$$Therefore, the charge on the x-axis is 9.0 mC.

A 60 MC Charge Is Placed at the Origin and a a Second