Answer:
a) Eₓ = 0 , b) c) Ex_total = 3.96 10⁻⁶ N
Step-by-step explanation:
For this problem we will look for the expression for the electric field of a ring of charge, at a point on its central axis.
From the symmetry of the ring, the resulting field is in the direction of the axis, we suppose that it corresponds to the x and y axis and the perpendicular direction but since the two sides of the annulus are offset.
dE = k dq / r²
The component of this field in the direction to x is
d Eₓ = dE cos θ
the distance of ring to the point
r² = x² + a²
where a is the radius of each ring
the cosine is the adjacent leg between the hypotenuse
Cos θ = x /r
we substitute
d Ex = k / (x² + a²) dq x /√ (x² + a²)
we integrate
Ex = ∫ k / (x² + a²)^{3/2} x dq
Ex = k x/ (x² + a²)^{1.5} Q
the total field is the sum of the fields created by each ring, since they are on the same line (x-axis), you can perform an algebraic sum
Eₓ_total = Eₓ₁ + Eₓ₂
a) at the origin x = 0
in field is zero, since the numerator e makes zero
Eₓ = 0
b) for point x = 40.0cm = 0.400 m
we substitute the values in our equation
Ex_total = k x [Q (x² + 0.1²)^{1.5} + 2Q / (x² + 0.2²2) ^3/2]
Ex_total = k Q x [1 / (x² + 0.01)^1.5 + 2 / (x² + 0.04)^1.52]
let's calculate
Ex_total = 9 10⁹ 30 10⁻⁹ 0.40 [1 / (0.4 2 + 0.01) 3/2 + 2 / (0.4 2 + 0.04) 3/2] 10⁻⁹
Ex_total = 108 [14.2668 + 22.36] 10⁻⁹
Ex_total = 3.96 10⁻⁶ N