A thin nonconducting rod with a uniform distribution of positive charge Q is ben

Question

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of radius R (see the figure). The central perpendicular axis through the ring is a z axis, with the origin at the center of the ring. What is the magnitude of the electric field due to the rod at (a)z = 0 and (b)z = ?? (c) In terms of R, at what positive value of z is that magnitude maximum? (d) If R = 2.30 cm and Q = 4.13 ?C, what is the maximum magnitude?

(a) Number Units N/C or V/mVkVmVV/sV/m A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a circle of radius R (see the figure). The central perpendicular axis through the ring is a z axis, with the origin at the center of the ring. What is the magnitude of the electric field due to the rod at (a)z = 0 and (b)z = ?? (c) In terms of R, at what positive value of z is that magnitude maximum? (d) If R = 2.30 cm and Q = 4.13 ?C, what is the maximum magnitude?

Explanation / Answer

By symmetry all electric field vectors cancel
with the exception of the z component.

Obtained by integration:
E = kQz/(r^2+z^2)^(3/2)
Set the first derivative of this function to zero.
Using the product rule for differentiation find:
dE/dz = kQ[ (r^2+z^2)^-(3/2) - 3z^2(r^2+z^2)^-(5/2)] = 0
multiply both sides of this equation by (r^2+z^2)^(5/2) and divide both by kQ:
dE/dz = (r^2+z^2) - 3z^2 = 0
r^2 +z^2 -3z^2 = 0
r^2 - 2z^2 = 0
r^2 = 2z^2
=>z = (+/-) r/sqrt(2)

Pop z,r, and Q into E = kQz/(r^2+z^2)^(3/2) to find the answer.
If you need more help, let me know.

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