Question

Coulomb's law for the magnitude of the force Fbetween two particles with charges Q and Q′separated by a distance d is

|F|=K|QQ′|/d^2,

where K=1/4πϵ0, and ϵ0=8.854×10^−12C^2/(N⋅m^2) 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 = -1.665 m ; the second charge, q2 = 36.0 nC ,is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 47.5 nCplaced between q1 and q2 at x3 = -1.130 m ?

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

get force on q3 with three significant figures.

Answer 1

q1 exerts a force of attraction on q3

the force eerted by q1 on q3 acts towards q1 ( i.e -x direction )

F13x = -k*q1*q3/((x3-x1)^2 = -(1/4pie0)*(q1*q3/(x3-1)^2)

F13x = -(1/(4*pi*8.854*10^-12))*(19*10^-9*47.5*10^-9/(-1.13-(-1.665))^2)

F13x = -2.834*10^-5 N

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q2 exerts a force of repulsion on q3

the force eerted by q1 on q3 acts -x direction

F23x = -k*q2*q3/((x3-x2)^2 = -(1/4pie0)*(q2*q3/(x3-x2)^2)

F23x = -(1/(4*pi*8.854*10^-12))*(36*10^-9*47.5*10^-9/(-1.13-0)^2)

F23x = -1.204*10^-5 N

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net force F = F13x + F23x

F = -4.037*10^-5 N <<<<<-----------answer