Question

# Coulomb's law for the magnitude of the force F between two particles with charges Q and...

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=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) 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.

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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