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Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance ddd is |F|=K|QQ′|d2|F|=K|QQ′|d2, where K=14πϵ0K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2)ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1q1q_1 = -13.5 nCnC , is located at x1x1x_1 = -1.735 mm ; the second charge, q2q2q_2 = 35.5 nCnC , is at the origin (x=0.0000)

Answer :

Poltergeist

Answer:

[tex]{F_{tot} = -1.092*10^{-2}N[/tex]

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

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

Solution:

The coulomb force is given by the equation

[tex]F =k\dfrac{q_1 q_2}{d^2}.[/tex]

where [tex]d[/tex] is the separation between the charges [tex]q_1[/tex] and [tex]q_2[/tex].

Now, in our case

[tex]q_1=-13.5nC=-13.5*10^{-9}C[/tex]

[tex]q_2=-1.735nC=-1.735*10^{-9}C[/tex]

[tex]q_3= 47*10^{-9}C[/tex]

The separation between charges [tex]q_3[/tex] and [tex]q_1[/tex] is

[tex]d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m[/tex]

Therefore, the force between them is

[tex]F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N[/tex]

and it is directed in the negative x-direction.

The separation between charges [tex]q_2[/tex]  and [tex]q_3[/tex] is

[tex]d_{23} =-1.240mm-0=-1.240*10^{-3}m[/tex]

therefore, the force between them is

[tex]F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N[/tex]

Therefore the total force on charge [tex]q_3[/tex] is

[tex]F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}[/tex]

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