Question

A 10.0-cm-diameter and a 20.0-cm-diameter charged ring are arranged concentrically (so they share the same axis). Assume both are centered at the origin. The 10 cm ring is charged to 30.0nC, while the 20 cm ring is charged to 60.0nC. Using the library electric field formula for Ering, compute the total electric field strength numerically at:
a, The origin
b. 40.0 cm along the axis of the 10 cm ring.

Answer 1

Answer:

a)     Eₓ = 0  ,   b)    c)    Ex_total = 3.96 10⁻⁶ N

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