Maximum electric field in a circle I have a question as follows:
"Show that $E_x $ on the axis of a ring charge of radius $a$ has its maximum value at $x = \pm a /\sqrt{2} $. Sketch $E_x$ versus x for both positivve and negative values of $x$"
This is a bit more abstract than my other questions "Given some givens, find the electric field at this point". As I understand I need to find a function $E_x (x)$ along all x from $-a$ to $a$, and confirm the maxes by taking its derivative that $E_x '(x) = 0$ at the given maxes.
I take a point on $x$ axis, I call it $x$. Considering all $dQ$ along the circle, all of the electric field should be along the x axis due to symmetry. That's nice I guess. The line towards $dQ$ forms an unknown length I'll call $w$. The line projected down from this point on $dQ$ I'll call $u$, and the point from the end of the projection back to $x$ is $v$. I don't have the rep to draw a pretty picture and share it here unfortunately.
The direction I'm obviously going is some trigonometry trickery, but I have no idea where to go. I have $u,v,w$ as triangle sides and are unknowns and $w$ is needed for the distance part of the coulumb's equation. there is the triangle with $a,x,w$ but isn't a right triangle and I'm unsure of any useful properties to work with there. Perhaps there's some actual physics I have overlooked. Thanks for any help. 
 A: You're very much on the right track, well done.
You won't need too much trickery: try thinking in terms of electrostatic potentials rather than fields. Then any point on the $x$ axis is equidistant from all points on the ring; this constant distance, by Pythagoras's theorem is simply $\sqrt{x^2 + a^2}$. So your electrostatic potential as a function of $x$ for a point on the axis is:
$$V(x) = -\frac{Q}{4\,\pi\,\epsilon_0} \frac{1}{\sqrt{x^2 + a^2}}$$
where $Q$ is the total charge Now differentiate once wrt to $x$ to find $E_x$ (recall $\mathbf{E} = - \nabla V$) and then maximize this latter expression for the electric field.
If you must use electric field rather than potential, then the $x$-component from the charge is $|\mathbf{E}|^2 \cos\theta$ where $\theta$ is the angle between the line joining the charge and the point on the $x$-axis in question and the $x$-axis itself. So you use the inverse square law to find $|\mathbf{E}|$ and then $\cos\theta = \frac{x}{\sqrt{x^2 + a^2}}$.
I'm sure you find arguments like those often given in class "by symmetry" without further explanation annoying, because, as you say, you're then "guessing" or using ones "intuition": I certainly did. Here's a way to tighten such an argument and make it more precise so that hopefully you find it a bit more satisfying: there are many isometric transformations that we could do on this system and still leave it the same relative to any particular Cartesian co-ordinate system: rotation through any angle about the $x$ axis, rotation through a half turn about the $y$ axis, rotation through half a turn about the $z$ axis. So, you conclude, the system's behavior must be the same even after any sequence of these transformations are imparted to it. So, to argue rigorously, you assume that there is a $E_y$ or $E_z$ component along the $x$-axis and impart the $z$ and $y$ rotations to see what happens to the system's description. You can soon prove that there can be no $E_y$ or $E_z$ because then the system's description would otherwise change after these half turns are imparted to it.
