Answer:
a) -537 N/C
, b) - 327.8 N/C and c) 723.7 N/C
Explanation:
The electric field is a vector magnitude, so we must add them as vectors. The electric field equation is
E = k q / r²
Where k is the Coulomb constant that you are worth 8.99 10⁹ N m²/C², that the magnitude of the load and r the distance between the load and the test load
Let's find the field created by each charge at the point x = 0.200 m
charge 1.
This charge is at the origin, the distance is
x₁ = 0.200 m
q₁ = 3 nC = 3 10-9 C
E1 = k q₁ / x₁²
E1 = 8.99 10⁹ 3 10⁻⁹ / 0.2²
E1 = 674.25 N / C
Charge 2
This load is the point a = 0.800 m, so the distance to the test charge at 0.200 m
x₂ = 0.800 - 0.200
x₂ = 0.600 m
q₂ = -5.5 nC = -5.5 10-9 C
E2 = 8.99 10⁹ 5.5 10⁻⁹ / 0.600²
E2 = 137.35 N / C
We already have the value of each field, add them vectorially, remember that if the charges are of the same sign they repel and if they are of the opposite sign they attract, the field E1 is directed to the left and the field E2 is directed to the right
Et = -E1 + E2
Et = -674.25 + 137.25
Et = -537 N / C
The field is headed to the left
b) we perform the same procedure for another distance value
Charge 1
x = 1.20
E1 = 8.99 10⁹ 3 10⁻⁹ / 1.2²
E1 = 18.73 N / C
Chage 2
x = 0.8 - 1.2 = -0.4 m
E2 = 8.99 10⁹ 5.5 10⁻⁹ / 0.4²
E2 = 309.03 N / C
Total field, E1 is on the left and E2 goes on the left
Et = -E1 -E2
Et = -18.73 - 309.03
Et = - 327.8 N / C
The field is headed to the left
c) point at x = -0.200 m
charge 1
x = -0.200 m
E1 = 8.99 10⁹ 3 -10⁻⁹ / (-0.2)²
E1 = 674.25 N / C
charge 2
x = 0.800 - (-0.2) = 1,000 m
E2 = 8.99 10⁹ 5.5 10⁻⁹ / 1²
E2 = 49.45 N / C
Ei goes to the right and E2 goes to the right
Et = E1 + E2
Et = 674.25 +49.45
Et = 723.7 N / C
The field is headed to the right
