A ring with charge Q and radius 'r'

[squelch]

Homework Statement

Show that $\vec{E}_x$ on the axis of a ring charge [I'm assuming they meant "of charge Q"] of radius "r" has its maximum value at $x=\pm \frac{r}{\sqrt{2}}$

Homework Equations

Linear charge density $\lambda=\frac{Q}{2\pi R}$
$dQ=\lambda ds = \frac{Qd\theta}{2\pi}$
$\vec{E}=\frac{KQ}{R^2}$

The Attempt at a Solution

Let the distance between an infinitesimal sector of the ring with charge $dQ$ and a point $P$ be $a$. Then $a=\sqrt{x^2+r^2}$ where r is the radius of the ring. Note that to save some typing, I will use $a$ and $\sqrt{x^2+r^2}$ interchangeably.

Let $\theta$ be the angle formed by a and the x axis, then:
$cos\theta=\frac{x}{a}=\frac{x}{\sqrt{x^2+r^2}}$ and $sin\theta=\frac{r}{a}=\frac{r}{\sqrt{x^2+r^2}}$

Now, $d\vec{E}_y=0$, because all forces in the y-component cancel, leaving $d\vec{E}_{total}=d\vec{E}_x=d\vec{E}cos\theta$

$\therefore d\vec{E}=\frac{KdQ}{a^2} cos\theta$

Because $dQ= \frac{Qd\theta}{2\pi}$ and $cos\theta=\frac{x}{\sqrt{x^2+r^2}}$
$d\vec{E}=\frac{KQx}{2\pi a^3}d\theta$

Now, integrating from 0→2∏ to find $\vec{E}$
$$\vec{E}=\frac{KQx}{2\pi a^3} \int_{0}^{2\pi} d\theta = \frac{KQx}{a^3} = \frac{KQx}{(x^2+r^2)^{3/2}}$$

From here, I'm not sure how to use this to "prove" that $\vec{E}_x$ has a maximum value at $x=\pm \frac{r}{\sqrt{2}}$

 

[gneill]

How do you go about finding the maximum or minimum of a function of x?

 

[squelch]

First derivative test?

 

[gneill]

First derivative test?

You've sort of got the right idea. Set the first derivative w.r.t. x to zero and solve for x. Then test the second derivative at x to see if you've found a maximum or a minimum. It's a Calculus 1 topic.

 

[squelch]

You've sort of got the right idea. Set the first derivative w.r.t. x to zero and solve for x. Then test the second derivative at x to see if you've found a maximum or a minimum. It's a Calculus 1 topic.

Right, but I don't feel like I have my $\vec{E}$ right. I keep seeing that $\sqrt{x^2+r^2}$ and thinking that I missed that I was supposed to do trig substitution somewhere along the way.

 

[ehild]

You formula for E is correct. You need to find x where the function $\frac{x}{(x^2+r^2)^{3/2}}$ has its extrema.

ehild