What's the force on charge Q3?

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1 Answer
Nov 5, 2017

#394"N"# to the left.

Explanation:

The force between two point charges is given by Coulomb's Law:

#color(blue)(F=1/(4piepsilon_0)*(abs(q_1)abs(q_2))/r^2)#

where #r# is the distance between the charges

To find the net force on #Q_3#, we will need to calculate the force exerted on it by both #Q_1# and #Q_2#.

We have the following information:

  • #Q_1=-50muC#
  • #Q_2=100muC#
  • #Q_3=-20muC#
  • #r_(12)=20.0"cm"#
  • #r_(13)=40.0"cm"#
  • #epsilon_0=8.85 10^(−12)" C"^2//"Nm"^2#

Let's start with #Q_1#.

  • #Q_1# and #Q_3# are both negative charges. Therefore the charges repel, and the force exerted on #Q_3# by #Q_1# points to the right.

#F_(13)=1/(4piepsilon_0)*(abs(-50*10^-6C)abs(-20*10^-6C))/(40.0*10^-2m)^2#

#=>~~56"N"# (to the right)

Now we can calculate the force between #Q_2# and #Q_3#:

  • #Q_2# is a positive charge, whereas #Q_3# is a negative charge. Therefore the charges attract, and the force exerted on #Q_3# by #Q_2# is to the left.

#F_(23)=1/(4piepsilon_0)*(abs(100*10^-6C)abs(-20*10^-6C))/(20.0*10^-2m)^2#

#=>~~450"N"# (to the left)

  • Then, since the charges are all placed along a straight line, we can simply add these values for force together to get the net force.
  • However, note that these are forces and do have directions associated with them. We can already see that the net force will be to the left, as this is is a much greater value than that to the right.
  • We will therefore subtract #F_(13)# from #F_(23)#.

#F_(" on "Q_3)=450"N"-56"N=394"N"#

That is, #394"N"# to the left.

Alternatively, you can define left as the negative direction and make the calculated force net negative. Then upon adding the two forces you would obtain a negative net force, which indicates that the net force is to the left.