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

the answer is 3.1 his answer is wrong mines is right

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