Apositive charge q1,-45.5 nc is located at the origin. a negative charge q2 =-5.5 nc is located on the positive x-axis, p = 14 cm from the origin.
choose the correct relation to indicate the continuous region on the x-axis that includes a point where the electric field is zero.

 r = 0.1045 cm

Explanation:

The electric field of a point charge is given by the expression

          E = k q / r²

Let's look for the field created by each load

Q₁ charge

           E₁ = k Q₁ / (r-0)²

           E₁ = - 8.99 10⁹ 45.5 10⁻⁹ / r²

           E₁ = - 409.05 / r²

Q₂ charge

           E₂ = k Q₂ / (r-0.14)₂

           E₂ = - 8.99 10⁹ 5.5 10⁻⁹ / (r-0.14)²

           E₂ = -49.45 / (r-0.14)²

The electric field is a vector quantity, so we must use the sum of vectors to get the total field

           E_total = E₁ + E₂

The burden of proof is always considered positive as the two charges are negative the strength is attractive, consider three regions of interest.

- Left of the two charges in this case the two forces are directed to the right and therefore the field is not canceled

- to the right of the two charges, the forces go to the left and the field is not canceled for any distance

- Between the two charges, in this case each force goes in the opposite direction and there is a point for the field to cancel

             0 = E₁ - E₂

             E₁ = E₂

            -409.05 / r² = - 49.45 /(r-0.14)²

              (r-0.14)² = 49.45 / 409.05 r²

             r² - 2 0.14r + 0.14² = 0.12 r²

             r² (1-0.12) - 0.28 r + 0.0196 = 0

             0.88 r² - 0.28 r + 0.0196 = 0

             r² - 0.318 r + 0.0223 = 0

We solve the second degree equation

            r = [0.318 ±√(0.318² - 4 0.0223)] / 2

            r = [0.318 ± 0.109] 2

            r₁ = 0.209 / 2 = 0.1045 m

            r₂ = 0.2135 m

We see that the result in the area of interest between charges is

             r = 0.1045 cm

- There are zones on the sides of the charge, but the force in these zones has a component that takes the test load to the part between the loads, therefore the field is never canceled

the bal will not change its velocity.

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