Two charges of # -6 C # and # -3 C# are positioned on a line at points # -5 # and # 8 #, respectively. What is the net force on a charge of # 4 C# at # 2 #?

1 Answer
Aug 13, 2017

Answer:

The net force is #1.40xx10^9" N"# to the left.

Explanation:

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

#|vecF|=k(|q_1||q_2|)/r^2#

where #q_1# and #q_2# are the magnitudes of the charges, #r# is the distance between them, and #k# is a constant equal to #8.99xx10^9" Nm"^2//"C"^2#, sometimes referred to as Coulomb's constant.

  • Since unspecified, I'll assume radii values to be in meters so that the units work out.

Here is a diagram of the situation:

enter image source here

To find the net force on the #4C# charge, we consider the forces exerted on it by the #-6C# and #-3C# charges.

We have the following information:

  • #|->Q_1=-6"C at " -5#
  • #|->Q_2=4"C at " 2#
  • #|->Q_3=-3"C at " 8#
  • #|->k=8.99xx10^9" Nm"^2//"C"^2#

#color(blue)(vecF_("net on 3")=sumvecF=vecF_(1 on 2)+vecF_(3 on 2))#

Recall that like charges repel, while opposite charges attract. Each of the charges acting on #Q_2# are negative. Therefore, the force vectors can be drawn in as follows:

enter image source here

where #vecF_(1" on " 2)# is the force of #Q_1# on #Q_2# and #vecF_(3" on " 2)# is the force of #Q_3# on #Q_2#

So, we can calculate #vecF_(1" on " 2)# and subtract from the value we obtain from #vecF_(3" on " 2)# to find the net force on the #Q_2# charge.

#abs(vecF_(1" on " 2))=k*(|q_1||q_2|)/(r_(12)^2)#

#=(8.99xx10^9" Nm"^2//"C"^2)*(6C*4C)/(7"m")^2#

#color(blue)(=4.40xx10^(9)" N")#

#abs(vecF_(3" on " 2))=k*(|q_3||q_2|)/(r_(32)^2)#

#=(8.99xx10^9" Nm"^2//"C"^2)*(3C*4C)/(6)^2#

#color(blue)(=3.00xx10^9" N")#

Therefore, we have:

#color(blue)(vecF_(n e t)=-1.40xx10^9" N")#

That is, #1.40xx10^9" N"# to the left.

#- - - - - - - - #

Here is what the forces exerted on charges by other charges of the same or different sign look like:

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