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 separated 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 = -19.0 nC , is located at x1 = -1.715 m ; the second charge, q2 = 30.5 nC , is at the origin (x=0.0000)

What is the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 atx3 = -1.150 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

distance of q3 form q1, d1 = 1.715 - 1.150 = 0.565 m

distance of q3 from q2, d2 = 1.150 m

q1 and q2 both attarct q2 towards them,

k = 1/4pie0 = 9 x 10^9

Fnet = [ (9x 10^9 x 30.5 x 10^-9 x 55 x 10^-9)/(1.150^2)] - [ (9x 10^9 x 19 x 10^-9 x 55 x 10^-9)/(0.565^2)]

Fnet = - 1.805 x 10^-5 N

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