Suppose there are only two charged particles in a particular region. Particle 1 carries a charge of +q and is located on the positive x-axis a distance d from the origin. Particle 2 carries a charge of +2q and is located on the negative x-axis a distance d from the origin. Required:
Where is it possible to have the net field caused by these two charges equal to zero?

1. At the origin.
2. Somewhere on the x-axis between the two charges, but not at the origin.
3. Somewhere on the x-axis to the right of q2.
4. Somewhere on the y-axis.
5. Somewhere on the x-axis to the left of q1.

x₂ = 0.1715 d

1) false

2) True

3) True

4) false

5) True

Explanation:

The field electrifies a vector quantity, so we can add the creative field by these two charges

             E₂-E₁ = 0

             k q₂ / r₂² - k q₁ / r1₁²= 0

             q₂ / r₂² = q₁ / r₁²

suppose the sum of the fields is zero at a place x to the right of zero

          r₂ = d + x

          r₁ = d -x

we substitute

           q₂ / (d + x)² = q₁ / (d-x)²

we solve the equation

           q₂ / q₁ (d-x)² = (d + x) ²

let's replace the value of the charges

       q₂ / q₁ = + 2q / + q = 2

          2 (d²- 2xd + x²) = d² + 2xd + x²

          x² -6xd + d² = 0

we solve the quadratic equation

          x = [6d ± √ (36d² - 4 d²)] / 2

          x = [6d ± 5,657 d] / 2

          x₁ = 5.8285 d

          x₂ = 0.1715 d

      with the total field value zero it is between the two loads the correct solution is x₂ = 0.1715 d

this value remains on the positive part of the x axis, that is, near charge 1

now let's examine the different proposed outcomes

1) false

2) True

3) True

4) false

5) True