Two charges of 6 C and -3 C are positioned on a line at points 7 and 2 , respectively. What is the net force on a charge of -1 C at -2 ?

1 Answer
Dec 30, 2015

F_3=4.13*10^8N

Explanation:

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Consider the figure. Let the charges 6C, -3C and -1C 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_13be the distance between the charges q_1 and q_3.
From figure
r_13=7-(-2)=7+2=9m

Let r_23be the distance between the charges q_2 and q_3.
From figure
r_23=7-2=5m

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*(6)(1))/9^2=0.667*10^9N
This force is attractive and is towards charge q_1.

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)(1))/5^2=1.08*10^9N
This force is repulsive and is towards charge q_3.

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 we have to subtract smaller force from larger force.
Let F_3 be the total force on the charge q_3.
implies F_3=F_23-F_13=1.08*10^9-0.667*10^9=0.413*10^9=4.13*10^8N
implies F_3=4.13*10^8N
Since The larger force F_23 is towards charge q_3.
Therefore the force F_3 is also towards charge q_3.