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 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 = -14.5 nC , is located at x1 = -1.660 m ;the second charge, q2 = 35.0 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 = 52.0 nC placed between q1 and q2 at x3 = -1.160 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.

Coulomb'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 = -14.5 nC , is located at x1 = -1.660 m ;the second charge, q2 = 35.0 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 = 52.0 nC placed between q1 and q2 at x3 = -1.160 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.

Coulomb'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 = -14.5 nC , is located at x1 = -1.660 m ;the second charge, q2 = 35.0 nC , is at the origin (x=0.0000).

Explanation / Answer

Fnet = F13 + F23

F13 = kq1q3d^2

d = 0.5 m

q1 = -14.5 nC = -14.5 x 10^-9 C

q3 = 52 nC = 52 x 10^-9 C

F13 = -2.7144 x 10^-5 N

F23 = kq2q3/d^2

d = 1.160 m

q2 = 35 nC = 35 x 10^-9 C

F23 = 1.2767 x 10^-5 N

Fnet = F13 + F23

Fnet = -1.437665 x 10^-5 N = -1.44 x 10^-5 N

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