Forces in a Three-Charge System Coulomb\'s law for the magnitude of the force F

Question

Forces in a Three-Charge System

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 = -17.0 nC , is located at x1 = -1.660 m ; the second charge, q2 = 39.5 nC , is at the origin (x=0.0000)

Part A

What is the net force exerted by these two charges on a third charge q3 = 51.0 nC placed between q1 and q2 at x3 = -1.165 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

magnitude of electric force is given by

F = kQ1Q2/d^2

Force on Q3 due to Q1 = kQ1Q3/d^2 = 9*10^9*17*10^(-9)*51*10^(-9)/(1.66 - 1.165)^2 = 31.84*10^(-6) N

and direction is Q1 to Q3

Force on Q3 due to Q2 = kQ2Q3/d^2 = 9*10^9*39.5*10^(-9)*51*10^(-9)/(1.165)^2 = 13.36*10^(-6) N

and direction is Q3 to Q2

Net force exerted on third charge = F = 31.84*10^(-6) + 13.36*10^(-6) = 4.52*10^(-5) N

and direction is +x axis.

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