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 = -10.0 nC , is located at x1 = -1.695 m ; the second charge, q2 = 31.0 nC , is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 50.5 nC placed between q1 and q2 at x3 = -1.090 m ?

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

Explanation / Answer

Consider force due to charge q1 on q3 is F1.

and due to charge q2 on q3 is F2.

for F1 -

q1 = -10 x 10^-9 C

q3 = 50.5 x 10^-9 C

d1 = 1.695 - 1.090 = 0.605 m

so, F1 = k*q1*q3 / d1^2 = - (9.0x10^9x10 x 10^-9x50.5 x 10^-9) / 0.605^2 = - 1.242 x 10^-5 N

for F2 -

q2 = 31x10^-9 C

q3 = 50.5 x 10^-9 C

d2 = 1.090 m

so, F2 = (9x10^9x31x10^-9x50.5 x 10^-9) / 1.09^2 = 1.186 x 10^-5 N

So, the resultant force on q3 -

F = F1 + F2 = - 1.242 x 10^-5 + 1.186 x 10^-5 = -0.056 x 10^-5 N = -5.60 x 10^-7 N

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