Please help me with this problem. Thank you!? The following constants can be use

Question

Please help me with this problem. Thank you!?

The following constants can be used:

ke = 8.99×109 Nm2/C2

qe = 1.6×10–19 C

mp = 1.67×10–27 kg

me = 9.11×10–31 kg

eps0 = 8.85×10–12 C2/(Nm2)

g = 9.80 m/s2

Calculate the magnitude and direction of the Coulomb force on the left charge shown in the figure to the right. The charge of the middle charge is 2.49 ?C. Let forces to the right be positive and forces to the left be negative.?

Defined symbols:
   q2 charge of the middle charge in ?C
   q1 charge of the left charge in ?C
   q3 charge of the right charge in ?C
   d12 distance between q1 and q2 in m
   d23 distance between q2 and q3 in m
   The answer is F1 magnitude of the force on the left charge in N
Hints:
    This problem is complicated by the necessity to add the forces from multiple charges. We will only arrange charges along straight lines in this class, so you won't need to resolve the forces into components; however, you do need to be careful about the signs of the forces.
    Draw a free-body diagram with the direction of the forces shown and put absolute values of the charges in the formula for Coulomb's Law. You don't need to put the signs of the charges in these equations, as the signs only indicate if the force is attractive and repulsive, and you've already taken that into consideration by drawing the arrows in the free-body diagram.
    Be careful to use the correct distances in this problem.
    Range of answers: –45 N to –140 N 6.00 -2.00 AC 3.00 cm 2.00 cm

Explanation / Answer

Apply Columb's law between the charges

The coulmb force between 6 uC and 2.49uC is

F1 = 9 * 10^9 ( 6 * 10^-6)( 2.49 * 10^-6)/(0.03)^2 = 149.4 N

The coulmb force between 6uC and 2uC is

F2 = 9 * 10^9 ( 6 * 10^-6)( 2 * 10^-6)/(0.05)^2 = 43.2 N

The coulmb force between 2uC and 2.49uC is

F3 = 9 * 10^9 ( 2.49 * 10^-6)( 2 * 10^-6)/(0.02)^2 = 112 N

force on 6 uC charge is

F6uc= F1- F2 = 106.2 N to the left

F2.49 uC = F1 + F3 = 261.2 N to the right

F(-2uC) = F2 + F3 =155.2 N to the left

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