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Consider two point charges located on the x axis: one charge, q1 = -11.5nC , is located at x1 = -1.685m ;the second charge, q2 = 40.0nC , is at the origin (x=0.0000).What is the net force exerted by these two charges on a third charge q3 = 55.0nC placed between q1 and q2 at x3 = -1.195m ?

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usamakld99

Answer:

[tex]F_{net} = 3.75\times10^{-5} Newton towards negative axis[/tex]

Explanation:

q1 = -11.5nC   , x1 = -1.685m

q2 = 40.0nC ,  x=0.0000m

q3 = 55.0nC  , x3 = -1.195m

Force on charge 3 due to charge 1 (attractive)

By applying coulomb law

[tex]F_{13}=K\frac{q_{1}q_{3}}{r^{2} }[/tex]

r = 1.685-1.195

r =  0.49 m

[tex]F_{13}=9\times10^{9}\times \frac{-11.5nC \times -55.0nC}{0.49^{2}m } \times -\iota[/tex]

[tex]F_{13}= 2.37\times10^{-5} (-\iota) N[/tex]

Force on charge 3 due to charge 2 (repulsive)

[tex]F_{23}=K\frac{q_{2}q_{3}}{r^{2} }[/tex]

[tex]F_{23}=9\times10^{9}\times \frac{-40nC \times -55.0nC}{1.195 ^{2}m }[/tex]

[tex]F_{23}=1.38\times 10^{-5}\times -\iota N[/tex]

[tex]F_{net} =F_{13} +F_{23}[/tex]

[tex]F_{net} = 3.75\times10^{-5} Newton   towards negative axis[/tex]

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