Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is

|F|=K|QQ′|d2 , where K=14πϵ0 , and ϵ0=8.854×10−12C2/(N⋅m2) 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 = -1.670 m ; the second charge, q2 = 35.5 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 = 51.0 nC placed between q1 and q2 at x3 = -1.080 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.

Answer 1

Force between two charges q and Q separated by a distance r is given by

$\dpi{100} F = \frac{1}{4 \pi \epsilon_{0}} \, \frac{q Q}{r^{2}}$

Where,   $\dpi{100} \frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \, \, Nm^{2}/C^{2}$   

And the direction of force will depend on the polarity of charges :

two opposite charges attract each other while two same charges repel each other.

Position of the charges and the direction of forces are shown in the diagram given below :

Force on $\dpi{100} q_{3}$ due to charge $\dpi{100} q_{1}$ is

  $\dpi{100} F_{13} = \frac{1}{4 \pi \epsilon_{0}} \, \frac{q_{1} q_{3}}{r^{2}}$

  $\dpi{100} = 9 \times 10^{9} \times \, \frac{18 \times 10^{-9} \times 51 \times 10^{-9}}{(1.670 - 1.080)^{2}}$

  $\dpi{100} = 2.373 \times 10^{-5} \, \, N$

In vector notation

  $\dpi{100} \vec{F}_{13}= 2.373 \times 10^{-5} \, \, \hat{i} \, \, N$

Force on $\dpi{100} q_{3}$ due to charge $\dpi{100} q_{2}$ is

  

$\dpi{100} F_{13} = \frac{1}{4 \pi \epsilon_{0}} \, \frac{q_{2} q_{3}}{r^{2}}$

  $\dpi{100} = 9 \times 10^{9} \times \, \frac{35.5 \times 10^{-9} \times 51 \times 10^{-9}}{1.080^{2}}$

  $\dpi{100} = 1.397 \times 10^{-5} \, \, N$

In vector notation

  $\dpi{100} \vec{F}_{23}= 1.397 \times 10^{-5} \, \, \hat{i} \, \, N$

So, net force on $\dpi{100} q_{3}$ is

  $\dpi{100} F_{net} = F_{13 } + F_{23}$

$\dpi{100} = 3.7699 \times 10^{-5} \, \, N$ (in positive x - direction)

For any doubt please comment and please give an up vote. Thank you.