Coulomb\'s law for the magnitude of the force F between two particles with charg

Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q^prime separated by a distance d is

|F|=Krac{|Q Q^prime|}{d^2},

where K= rac{1}{4 pi epsilon_0}, and epsilon_0 = 8.854 imes 10^{-12};{ m C^2/(Ncdot m^2)} is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q_1 = -16.5 nC, is located at x_1 = -1.655 m; the second charge, q_2 = 34.0 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q_3 = 46.5 nC placed between q_1 and q_2 at x_3 = -1.050 m?

Your answer may be positive or negative, depending on the direction of the force.

Express your answer numerically in newtons to three significant figures.

Explanation / Answer

q1 = -16.5*10^-9 C at x1 = -1.655 m q2 = 34.0*10^-9 C at x2 = 0 q3 = 46.5*10^-9 C at x3 = -1.050 m Seperation between q1 and q3, r13 = 1.655 - 1.050 = 0.605 m Seperation between q2 and q3, r23 = 1.050 m Force on q3 due to q1, F1 = Kq1q3/(r13)^2 (- i)along the -ve X axis                                                                                = - 18865.51*10^-9 i N Force on q3 due to q2, F2 = Kq2q3/)r23)^2 ( - i)                                                                                    = -12906.122*10^-9 N i along the -ve X axis Net force on q3, F = F1 + F2                                                      = - 31771.63*10^-9 i N Magnitude F = - 31771.63*10^-9 N Magnitude F = - 31771.63*10^-9 N

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