Two small spheres, each carrying a net positive charge, are separated by 0.400 m. you have been asked to perform measurements that will allow you to determine the charge on each sphere. you set up a coordinate system with one sphere (charge q1) at the origin and the other sphere (charge q2) at x = +0.400 m. available to you are a third sphere with net charge q3 = 3.00×10−6 c and an apparatus that can accurately measure the location of this sphere and the net force on it. first you place the third sphere on the x-axis at x = 0.200 m; you measure the net force on it to be 6.50 n in the +x-direction. then you move the third sphere to x = +0.600 m and measure the net force on it now to be 3.50 n in the +x-direction.

part a) calculate q1.
part b) calculate q2.

a)  q₁ = 15. 28 10⁻⁶C, b)  q₂ = 5.64 10⁻⁶ C

Explanation:

For this exercise we use Newton's second law where force is Coulomb's electric force

Case 1. Distance (x₁ = 0.200 m) from the third sphere

         F₁ = F₁₃ - F₂₃

         F₁ = k q₁q₃ / x₁² - k q₂ q₃ / (0.4 - x₁)²

         F₁ = k q₃ (q₁ / x₁² - q₂ / (0.4- x₁)²

Case2 Distance (x₂ = 0.6 m) from the third sphere

        F₂ = F₁₃ + F₂₃

        F₂ = k q₁q₃ / x₂² + k q₂q₃ / (0.4- x₂)²

        F₂ = k q₃ (q₁ / x₂² + q₂ / (0.4-x₂)²

The distance is between the spheres, in the annex you can see the configuration of the charge and forces

Let's replace the values

        F₁ = 8.99 10⁹ 3.00 10⁻⁶⁶ (q₁ / 0.2² - q₂ / (0.4-0.2)²

        F₂ = 8.99 10⁹ 3.00 10⁻⁶ (q₁ / 0.6² + q₂ / (0.4-0.6)²

        6.50 = 674. 25 10³ (q₁ –q₂)

        3.50 = 26.97 10³ (q₁ / 0.36 + q₂ / 0.04)

We have a system of two equations with two unknowns, let's solve it. Let's clear q1 in the first and substitute in the second

         q₁ = q₂ + 6.50 / 674 10³

         3.50 / 26.97 10³ = (q₂ + 9.64 10⁻⁶) /0.36 + q₂ / 0.04

         1.2978 10⁻⁴ = q₂ / 0.36 + q₂ / 0.04 + 26.77 10⁻⁶

         q₂ (1 / 0.36 + 1 / 0.04) = 129.78 10⁻⁶ + 26.77 10⁻⁶

         q₂ 27,777 = 156,557 10⁻⁶

         q₂ = 156.557 10-6 /27.777

         q₂ = 5.636 10⁻⁶ C

We look for q1 in the other equation

        q₁ = q₂ + 6.50 / 674 10³

        q₁ = 5.636 10⁻⁶ + 9.6439 10⁻⁶

        q₁ = 15. 28 10⁻⁶C

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

the answer to this is proletariats.

hope this : )