Two charges of # 2# # C # and # 4# #C# are positioned on a line at points # -5 # and # -2 #, respectively. What is the net force on a charge of # 3 # #C# at # 1 #?

1 Answer
May 21, 2017

Answer:

The forces on the #3# #C# charge due to each of the other charges can simply be added together.

The net force is #F = (kxx2xx3)/6^2+(kxx4xx3)/3^2 = 1.35 xx 10^10# #C#

Explanation:

The charges in question are all positive, so the forces will be repulsive. The #3# #C# charge will be experiencing a force to the right, if we imagine the number line.

The distance from the #2# #C# charge to the #3# #C# charge is #6# #m# (we are not told the units of distance being used, but metres are the appropriate SI unit). The distance from the #4# #C# charge to the #3# #C# charge is #3# #m#.

The net force, then, is:

#F = (kxx2xx3)/6^2+(kxx4xx3)/3^2#

where #k = 9xx10^9# #Nm^2C^-2#

#F = (kxx2xx3)/6^2+(kxx4xx3)/3^2 = 1.35 xx 10^10# #C#

This is a very large force, but these are very large charges, quite close together, and the constant is large.