Two charges of # -2 C # and # 7 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
Feb 9, 2016

Answer:

#F_{Net} = F_A + F_B = 50.5687\times10^9 N#

Explanation:

Let us label the charges as: #Q_A=-2C#; #Q_B=+7C# and #Q=+3C#

Note: You have not mentioned the units of the distance. I assume it to be meters.

Let us label the positions of these charges as : #x_A=+5#m, #x_B=-1#m; and #x=+1#m.

The charge #Q# is placed between #Q_A# and #Q_B#. Because #Q# and #Q_A# are charges of opposite signs they interact attractively and #Q# is pulled towards #Q_A#. Since #Q# and #Q_B# are of the same sign they interact repulsively and #Q# is pushed away from #Q_B#. So #Q# feels a net force toward #Q_A#.

Now apply Coulomb's law to get their magnitudes,
#F_A = k\frac{|Q_AQ|}{(x_A-x)^2}#
#\qquad# #=(8.99\times10^9(N.m^2)/C^2)(\frac{|(-2C)(+3C)|}{(5m-1m)^2})#
#\qquad##=3.37125\times10^9 N#

#F_B = k\frac{|Q_BQ|}{(x-x_B)^2}#
#\qquad# #=(8.99\times10^9(N.m^2)/C^2)(\frac{|(+7C)(+3C)|}{(1m-(-1m))^2})#
#\qquad##=47.1975\times10^9 N#

#F_{Net} = F_A + F_B = 50.5687\times10^9 N#