Suppose the rubber rod holds -3 pico-Coulomb at x=0 and the uncharged ball has +2 femto-Coulomb at x= 50mm, as well as -2 femto-Coulomb at x=55mm distance. What would be the electric Force applied to the ball?

I used the equation F=q1q2/d^2 and did two separate calcuations, one with -3picoC and 2 femtoC and another with 2femtoC at 50 and -2femtoC at 55. I added them togther and I got the value 2.3x10^-14 N which apparently isn't correct.
Can someone help me figure out what I'm doing wrong?

To find the electric force applied to the ball, you correctly used the equation F = q1q2/d^2, where F is the force, q1 and q2 are the charges, and d is the distance between the charges. However, when adding the forces, you need to consider both the magnitude and the direction.

Let's break down the problem step by step:

1. Calculate the force between the -3 pico-Coulomb charge on the rubber rod at x=0 and the +2 femto-Coulomb charge on the ball at x=50mm:
- Convert -3 pico-Coulomb to Coulombs: -3 pico-Coulomb = -3 × 10^-12 Coulomb
- Convert 50mm to meters: 50mm = 0.050 meters
- Substitute the values into the equation:
F1 = (-3 × 10^-12 C)(+2 × 10^-15 C) / (0.050 m)^2
F1 = -1.2 × 10^-19 N

2. Calculate the force between the +2 femto-Coulomb charge on the ball at x=50mm and the -2 femto-Coulomb charge at x=55mm:
- Convert +2 femto-Coulomb to Coulombs: +2 femto-Coulomb = +2 × 10^-15 Coulomb
- Convert 55mm to meters: 55mm = 0.055 meters
- Substitute the values into the equation:
F2 = (+2 × 10^-15 C)(-2 × 10^-15 C) / (0.055 m)^2
F2 = -1.309 × 10^-17 N

3. To obtain the total force, you need to add the forces considering their signs:
F_total = F1 + F2
F_total = (-1.2 × 10^-19 N) + (-1.309 × 10^-17 N)
F_total = -1.3212 × 10^-17 N

Therefore, the electric force applied to the ball is approximately -1.3212 × 10^-17 N.