At what distance along the central axis of a ring of radius R and uniform charge is the magnitude of the elect

Question

What Distance? At what distance along the central axis of a ring of radius R and uniform charge is the magnitude of the electric field due to the ring's charge maximum? Answer in terms of R.

Answer 1

This can get quite complicated, but I will make a stab at it, but please check my work for errors.

The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is

dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2)

where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis.

However, cos(T) = x/sqrt(x^2 + R^2)

so the equation becomes

dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2)

dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5

Integrating around the ring you get

E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5

E = (R/2*e0)*x*(x^2 + R^2)^-1.5

we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is

dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x}

dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5}

to find the maxima set this = 0, giving

(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0

mult both side by (x^2 + R^2)^2.5 to get

(x^2 + R^2) - 3*x^2 = 0

-2*x^2 + R^2 = 0

-2*x^2 = -R^2

x = (+/-)R/sqrt(2)

I hope I didn't make any algebra or calculus errors. pls check

Answer 2

Check this out
It is the field at the axis of a charged ring

Answer 3

to find the maxima why to set it to zero?

Answer 4

points!!

Answer 5

i don't know but it sounds important? but thanks for the two points!