A spring stretches 2 meters when a 12 N force is acting on it. What is the spring constant (k)? 10 N/m 0.17 N/m 2 N/m 6 N/m
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Answer: magnitude = 169.66N/C
direction = 8.6859°
Explanation:
Given from the question, we have that;
the Length of line of charge L = 0.808 m
Linear charge density λ = 3.35 × 10⁻⁶ C/m
charge q = -7.32 × 10⁻⁷ C
Coulombs force constant K = 1/(4π ε0) = 8.99 × 10⁹ N·m²/C².
NB. The picture uploaded gives a diagrammatic description of the problem.
From Pythagoras theorem we have,
tan Θ = 3.75 / (10.7-1.56)
Θ = 22.3076 °
recall that the Electric field at point P due to the finite wire is;
È = Kλ (L / b(L+b)) Î .............. (1)
where È rep the electric field.
from equation 1, we have that
È = 8.99 × 10⁹
where È rep the electric field.
from equation 1, we have that
È = 8.99 × 10⁹ × 3.35 × 10⁻⁶ (0.808/ 10.7(10.7 – 0.808))
È = 0.229905 × 10³ N/C Î
recall also that the Electric field at point -P due to -q is;
È = (8.99 × 10⁹ × 7.32 × 10⁻⁷) / ((3.75)² + (10.75-1.56)²) = 0.6742 × 10² N/C
where E = -E₁cosθÎ + E₁sinθĴ
E = - 0.62446 ×10²Î + 0.2562 ×10²Ĵ
The Resultant Electric charge Er is given as;
Er = 1.6771 ×10²Î + 0.2562 ×10²
Er = [√(1.6771)² + (0.2562)² ] × 10² = 169.66 N/C
∴ Magnitude = 169.66 N/C
Having gotten the magnitude, let us find the direction;
⇒ Direction = tan Ф = 0.25621/1.6771 = 8.6859°
Direction = 8.6859°
cheers i hope this helps
