At what point on the x-axis is the field greatest?

[mugzieee]

a circular ring of wire radius r lies in a plane perpendicular to the x-axis and is centered at the origin. The ring has a positive electric charge spread uniformly over it. The electric field in the x-direction, E, at the point x on the x-axis is given by E= kx/(x^2 + r^2)^(3/2) for k>0. At what point on the x-axis is the field greatest? Least?

After taking the first derivative i ended up with:
E' = k(-2x^2 + r^2 - 3xr)/(x^2 + r^2)^(5/2)
i know after the derivative I am supposed to find the critical points then classify them and find the global min and global max, but for critical points i end up with only a zero, what does it look like is wrong here?

 

[Naeem]

The Important thing to know is that , The charge on a conductor ( in this case the circular ring ) resides on its outer surface. Therefore, electric field E inside the ring is zero.

2nd, For a point on the charged spherical conductor or outside it, the charge may be assumed to be concentrated at its centre.

Based on this least E which is zero, inside the ring.

The greatest is outside, the ring.

I hope this helps.

I don't know if you really need to take the derivative in this question. This seems more like a conceptual problem.

 

[wais]

The critical point found from taking the derivative is correct, as it is the only point where the derivative is equal to zero. However, this does not necessarily mean it is the only critical point. To determine if it is a maximum or minimum, you need to take the second derivative and evaluate it at the critical point. If the second derivative is positive, then the critical point is a minimum, and if it is negative, then it is a maximum. In this case, taking the second derivative would result in a positive value, indicating that the critical point is a minimum. Therefore, the point on the x-axis where the electric field is greatest is at x = 0, and the point where it is least is at the point where the electric field is equal to zero, which can be found by setting the original equation equal to zero and solving for x.