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

1 Answer
Dec 30, 2015

Answer:

# F_3=8.219*10^9N#

Explanation:

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Consider the figure. Let the charges #2C, -3C# and #4C# be denoted by #q_1,q_2# and #q_3# respectively.
Let the positions at which charges are placed be in the units of meters.

Let #r_13#be the distance between the charges #q_1 and q_3#.
From figure
#r_13=-2-(-9)=-2+9=7 m#

Let #r_23#be the distance between the charges #q_2 and q_3#.
From figure
#r_23=2-(-2)=2+2=4 m#

Let #F_13# be the force due to charge #q_1# on the charge #q_3#
#F_13=(kq_1q_3)/r_13^2=(9*10^9*(2)(4))/7^2=1.469*10^9N#
This force is repulsive and is towards charge #q_2#.

Let #F_23# be the force due to charge #q_2# on the charge #q_3#
#F_23=(kq_2q_3)/r_23^2=(9*10^9*(3)(4))/4^2=6.75*10^9N#

This force is attractive and is towards charge #q_2#.

The total force or net force on charge #q_3# is the sum of above two forces.
Since the above two forces #F_13# and #F_23# are in same direction therefore they can be added directly.
Let #F_3# be the total force on the charge #q_3#.
#implies F_3=F_13+F_23=1.469*10^9+6.75*10^9=8.219*10^9N#
#implies F_3=8.219*10^9N#
Since #F_13 and F_23# are toward the charge #q_2# therefore the force #F_3# is also toward the charge #q_2#.