Answer:
For situation (a)
net charge E = E₊₂ + E₋₅ + E₋₃
E = K(q/d²)
where K = 8.99e9
d = 5.7cm = 5.7e-2m
Therefore,
E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)
E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)
E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)
thus
E = E₊₂ + E₋₅ + E₋₃
= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)
= 7.88e6(x) + 7.88e6(y)
use Pythagorean theorem
I E I = [tex]\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}[/tex] = 1.242e6[tex]\frac{N}{C}[/tex]
∅ = [tex]tan^{-1}[/tex][tex](\frac{7.88e5}{7.88e5} )[/tex] = [tex]tan^{-1}(1)[/tex] = 45°
Thus for (a) net magnitude = 1.115e6[tex]\frac{N}{C}[/tex] @ 45° above +x axis
for situation (b)
net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆
E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)
E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)
E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)
E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)
thus,
E = E₊₄ + E₊₁ + E₋₁ + E₊₆
= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)
= 7.88e5(x) + 7.88e5(y)
use Pythagorean theorem
I E I = [tex]\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}[/tex] = 1.242e6[tex]\frac{N}{C}[/tex]
∅ = [tex]tan^{-1}[/tex][tex](\frac{7.88e5}{7.88e5} )[/tex] = [tex]tan^{-1}(1)[/tex] = 45°
Thus for (a) and (b) the net magnitude = 1.242e6[tex]\frac{N}{C}[/tex] @ 45° above +x axis
Explanation:
I attached a sample image, i hope that corresponds to your question
