Two charges of # 9 C # and # 2 C# are positioned on a line at points # 4 # and # -4 #, respectively. What is the net force on a charge of # 3 C# at # -2 #?

1 Answer
Aug 4, 2017

Answer:

#vecF_(n e t)=6.76*10^9# #N# (to the right)

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 #8.99xx10^9Nm^2//C^2#, sometimes referred to as Coulomb's constant.

Here is a quick diagram of the situation:

enter image source here

To find the net force on the #3C# charge, we consider the force exerted on it by the #9C# and #2C# charges. Let's call the #3C# charge #q_1#, the #9C# charge #q_2# and the #2C# charge #q_3#. Then:

#vecF_(n e t)=sumvecF=vecF_(2 on 1)+vecF_(3 on 1)#

Recall that like charges repel, while opposite charges attract. Each of our charges are positive. Therefore, the force vectors can be drawn in as follows:

enter image source here

23 should say 21 in the above diagram

where #21 (23)# is the force of #q_1# on #q_2# and #31# is the force of #q_3# on #q_1#

So we can calculate #vecF_(31)# and subtract from that #vecF_(21)# to find the net force on the #q_1# charge.

#|vecF_(31)|=k*(|q_1||q_3|)/(r_(31)^2)#

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

#=1.35*10^(10)# #N#

#|vecF_(21)|=k*(|q_1||q_2|)/(r_(21)^2)#

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

#=6.74*10^9# #N#

Therefore, we have:

#vecF_(n e t)=6.76*10^9# #N# (to the right)