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 Fbetween two particles with charges Q and Qseparated by a distance d is

|F|=K|QQ|d2,

where K=140, and 0=8.854×1012C2/(Nm2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1 = -10.0 nC , is located atx1 = -1.750 m ; the second charge,q2 = 40.0 nC , is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 54.0 nC placed between q1and q2 at x3 = -1.135 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

coulombs constant K = 9*10^9 N-m^2/C^2

Force between q1 and q3 is F1 = k*q1*q3/r1^2 = (9*10^9*10*10^-9*54*10^-9)/(1.75-1.135)^2 = 1.28*10^-5 N along -x-axis

Force between q1 and q2 is F2 = k*q1*q2/r2^2 = (9*10^9*10*10^-9*40*10^-9)/(1.135^2) = 2.79*10^-6 N along -X-axis

net force is F1+F2 = (2.79*10^-6)+(1.28*10^-5) = 1.559*10^-5 N along -X-axis

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