Answer :
Answer:
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