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 Qseparated by a distance d is

|F|=K|QQ|/d2,

where K=1/40, and 0=8.854×10^(12)xC^(2)/(Nm2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1 = -18.0 nC , is located at x1 = -1.690 m ; the second charge,q2 = 32.0 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 atx3 = -1.205 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

Force exerted by q1 on q3 is,

F31 = - K |q1q3|/r132 (q1 will attract q3 hence force is in negative x direction)

Force exerted by q2 on q3 is,

F32 = - K |q2q3|/r232 (q2 will repel q3 hence force is in negative x direction)

So, total force on q3 is,

F = F31 + F32 = - K |q3| (|q1|/r232 + |q2|/r232)

Now, r13 = (1.690 - 1.205) m = 0.485 m, r23 = 1.205 m

Substituting values, we get,

F = -4.21 * 10-5 N

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