oulomb\'s law for the magnitude of the force F between two particles with charge

Question

oulomb's law for the magnitude of the force Fbetween two particles with charges  Q and  Qseparated 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 = -13.0 nC , is located at x1 = -1.690 m ; the second charge, q2 = 37.5 nC , is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 48.0 nC placed between q1and q2 at x3 = -1.145 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

Q1 exerts force in left direction

Substituting values in Coulomb's Law

We get

1.89*10-5 N

Q2 exerts force in left direction too

Substituting values in Coulomb's Law

We get

1.23*10-5 N

Total force=1.89*10-5 + 1.23*10-5 =3.12*10-5 N (left)

or -3.12*10-5 N

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