Nonconducting Rod bent into Circle

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SUMMARY

The discussion focuses on calculating the electric field generated by a uniformly charged nonconducting rod bent into a circle with radius R and total charge Q. The electric field at the center (z = 0) is zero due to symmetry, while at infinity (z = ∞), the field approaches zero. The maximum electric field occurs at a specific positive value of z, which can be derived from the equation E = k_e Q z / (z² + R²)^(3/2). For the given values of R = 2.00 cm and Q = 4.00 μC, the maximum electric field can be calculated using this formula.

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Question

A thin nonconducting rod with a uniform charge 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. What is the magnitude of the electric field due to the rod at (a) z = 0 (b) z = ∞ (c) In terms of R, at what positive value of z is that magnitude maximum? (d) If R = 2.00 cm and Q = 4.00 μC, what is the maximum magnitude?

Relevant Equations

dE = (kedq/r2)[itex]\hat{r}[/itex]
λ = Q/Length

Solution Attempt

I'm not really sure how to even approach this problem. I know that E = kzQ/(z2+R2)3/2 but I'm not quite sure how to reach this.

Help is much appreciated.
 
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Part (a) you should be able to figure out by inspection, from symmetry considerations.

For part (b), consider an arbitrary point on the z-axis, and find an expression for the field due to an infinitesimal charge element on the circle. Then integrate. Again, keep symmetry in mind.
 

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