Question

A particle with charge q1 = +4.0 µC is located at x = 0, and a second particle with charge q2 = −2.1 µC is located at x = 15 cm. Determine the location of a third particle with charge q3 = +4.6 µC such that the net electric field at x = 25 cm is zero.

Answer 1

Answer:

 x₃ = 0.0725 m

Explanation:

The force between the particles is electrical

          F = $k \frac{q_1q_2}{r_{12}^2}$

          F = q₀ E_total

don q₀ is a positive test charge at the point of interest, therefore we can calculate the electric field at the point of interest x₀ = 0.25 cm

          F = 0   → E_total = 0

          E_total = E₁ + E₂ + E₃

           E_total = $k \frac{q_1}{r_{1o}^2} + k \frac{q_2}{r_{2o}^2} + k \frac{q_3}{r_{3o}^2}$

          E_total = k ( $\frac{q_1}{r_{1o}^2} + \frac{q_2}{r_{2o}^2} + \frac{q_3}{r_{3o}^2}$ )

let's look for the distances

           

          r₁₀ = (x₀ - 0)

          r₁₀ = 0.25 m

          r₂₀ = √(x₀ - x₂) ²

          r₂₀ = √ (0.25 - 0.15) ²

          r₂₀ = 0.10 m

          r₃₀ = √ (0.25 - x₃) ²

we substitute

           0 = 9 10⁹ (4 / 0.25² - 2.1 / 0.1² + 4.6 / (0.25-x₃)² )

            4.6 / (0.25-x3)² = -4 / 0.25² + 2.1 / 0.1²

             4.6 / (0.25-x3) ² = -146

             (0.25 - x3) ² = - 4.6 / 146 = - 0.0315068

              0.25 - x3 = 0.1775

              x₃ = 0.25 - 0.1775

              x₃ = 0.0725 m