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 dis

|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 = -12.5 nC , is located at x1 = -1.660 m ; the second charge, q2 = 32.0 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 = 48.5 nC placed between q1 and q2 at x3 = -1.050 m ?

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

Explanation / Answer

A)

Force on q3 due to q1:

F1 = k*q1*q3/r1^2

where r1 = (1.66 - 1.05) = 0.61 m

So, F1 = 9*10^9*(12.5*10^-9)*(48.5*10^-9)/0.61^2

So, F1 = 1.47*10^-5 N

Force on q3 due to q2

F2 = k*q2*q3/r2^2

So,

F2 = 9*10^9*(32*10^-9)*48.5*10^-9/1.05^2

So, F2 = 1.27*10^-5 N

So, net force :

Fnet = F1 + F2

So, Fnet = 1.47*10^-5 + 1.27*10^-5 = -2.74*10^-5 N

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