A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of radius R. The central perpendicular axis through the ring is a z-axis, with the origin at the center of the ring. (a) What is the magnitude of the electric field due to the rod at z = 0? N/C

(b) What is the magnitude of the electric field due to the rod at z = infinity? N/C

(c) In terms of R, at what positive value of z is that magnitude maximum? R

(d) If R = 4.00 cm and Q = 9.00 �C, what is the maximum magnitude?N/C

a)  E_{z} = 0, b) E_{z} = 0, c)    z = 1.73 [1 + ], d)   Emax = 9.7  10¹⁰ N / C

Explanation:

For this exercise we use the expression

          E = k ∫ dq / r²

By applying this expression to our problem of a ring of radius R, with a perpendicular axis in the z direction, we can calculate the electric field for a small charge element

          dE = k dq / r²

In the attachment we can see a diagram of the electric field, it is observed that the fields perpendicular to the z axis cancel and the field remains in the direction of the axis

           d= dE cos φ

we substitute

         E_{z} = k∫    cos φ

let's write the expressions

          r² = R² + z²

          cos φ = z / r

we substitute in the integral, where we see that the load differential does not depend on the distance and the value of the total load is + Q

          E_{z} = k   ∫  dq

          E_{z} = k Q    

This is the expression for the electric field in the axis perpendicular to the ring, we analyze this expression to answer the questions

a) the magnitude of the field at z = 0

         E_{z} = 0

b) the magnitude of the field for z = inf

        when z »R the expression remains

          E_{z} = k Q

          E_{z} = k Q

therefore when the value of z = int the field goes to E_{z} = 0

c) In value of z for which the field is maximum.

    We have an extreme point when the first derivative is equal to zero

we solve

    (R² + z²)^{ 3/2} = 3 z² /(R² +z²) ^{1/2}

    (R² + z²)² = 3z²

    r² + z² = √3    z

    z² –1.732 z + R² = 0

we solve the quadratic equation

       z = [1.732 ± ]/ 2 = [1.73 ± 1.73   ] / 2

       z = 0.865 [1 ± ]

Therefore there are two points where the field has an extreme point one, one is a maximum and the other a minimum, as we have already determined a minimum at z = 0 the maximum point must be

  z = 0.865 [1 + ]

d) the value of Emax

         z₁ = 0.865 [1+) = 1.73 [1 + √0.99786]

          z₁ = 1.729 m

         z₂ = 0.865 [1 - √0.99786 ]

         z₂ = 0.0011 m≈ 0

for which the field has a maximum value substituting in equation 1

            = 81 10⁹  

            Emax = 9.7  10¹⁰ N / C

an enlarged virtual image