Electric field at a point within a charged circular ring

[bob012345]

That's a good approach, but it would not be quite enough to show that the same choice of pairing does not cancel; you would need to show that for this or some other pairing the net field is always the same way.

Thanks, so I take it that's a confirmation that pairings do cancel for the sphere. In either case, the pairings would always be general not based on specific angles.

 

[haruspex]

Thanks, so I take it that's a confirmation that pairings do cancel for the sphere.

It must be the case that well chosen pairings cancel for the sphere, since we know the net field is zero. I have not checked whether this is true of the pairing you illustrate.

In either case, the pairings would always be general not based on specific angles.

Not sure what you mean by that. You would need a relationship between the angles, $\phi=f(\theta)$, such that the band $(\theta, \theta+\delta\theta)$ cancels the band $(f(\theta), f(\theta+\delta\theta))$ .

 

[bob012345]

It must be the case that well chosen pairings cancel for the sphere, since we know the net field is zero. I have not checked whether this is true of the pairing you illustrate.
Not sure what you mean by that. You would need a relationship between the angles, $\phi=f(\theta)$, such that the band $(\theta, \theta+\delta\theta)$ cancels the band $(f(\theta), f(\theta+\delta\theta))$ .

The element from the right side is proportional to $\Large \frac {R^2 sin(\gamma) d\gamma}{s^2}$ and from the left side $\Large \frac {R^2 sin(\beta) d\beta}{s'^2}$ but in opposite directions along the axis. I know that $sin(\beta) = {\Large \frac{s'}{s }}sin(\gamma)$ and I believe $d\beta = {\Large \frac{s'}{s }}d\gamma$ making the two differential elements cancel at all angles. But I am having difficulty proving the relationship between $d\beta$ and $d\gamma$. It is easy to see when the angles are very small and when $s$ and $s'$ are vertical but not easy to see in between. I did precision numerical calculations however to verify that the relation $d\beta = {\Large \frac{s'}{s }}d\gamma$ holds for wiggles around any angle. All that just makes $\large \frac {R^2 sin(\beta) d\beta}{s'^2} → \large \frac {R^2 sin(\gamma) d\gamma}{s^2}$ so the elements cancel.

 

[haruspex]

View attachment 288165

The element from the right side is proportional to $\Large \frac {R^2 sin(\gamma) d\gamma}{s^2}$ and from the left side $\Large \frac {R^2 sin(\beta) d\beta}{s'^2}$ but in opposite directions along the axis. I know that $sin(\beta) = {\Large \frac{s'}{s }}sin(\gamma)$ and I believe $d\beta = {\Large \frac{s'}{s }}d\gamma$ making the two differential elements cancel at all angles. But I am having difficulty proving the relationship between $d\beta$ and $d\gamma$. It is easy to see when the angles are very small and when $s$ and $s'$ are vertical but not easy to see in between. I did precision numerical calculations however to verify that the relation $d\beta = {\Large \frac{s'}{s }}d\gamma$ holds for wiggles around any angle. All that just makes

$$ \large \frac {R^2 sin(\beta) d\beta}{s'^2} → \large \frac {R^2 sin(\gamma) d\gamma}{s^2}$$

Call angle BPO $\alpha$.
Angle PBO= Angle PCO=$\gamma-\alpha$.
$s\delta\alpha=R\delta\gamma\cos(\gamma-\alpha)$
$s'\delta\alpha=R\delta\beta\cos(\gamma-\alpha)$.

 

[bob012345]

Considering the question of the charged ring in the OP, the paired differential elements are proportional to $\Large \frac{R d\gamma}{s^2}$ and $\Large \frac{R d\beta}{s'^2}$ and making use of $d\beta = {\Large \frac{s'}{s }}d\gamma$ gives $\Large \frac{R d\gamma}{s^2}$ and $\Large \frac{R d\gamma}{ss'}$ which do not cancel because $s^2$ is always larger than $ss'$ until the very end when they become equal as both become vertical in the plot. This is also where the integration ends. Also note even though $s$ and $s'$ depend on $\gamma$ and $\beta$ respectively, the product of $ss'$ is a constant $ss' = (R+r)(R-r)$ where $r$ is the distance point $P$ is from the center along the central axis of the circle. I ignored the cosine factors which make the integration messy because it is the same for both paired elements and the whole point of this approach is to answer the original question if the electric field is zero or not while avoiding doing the integration.

 

[bob012345]

Call angle BPO $\alpha$.
Angle PBO= Angle PCO=$\gamma-\alpha$.
$s\delta\alpha=R\delta\gamma\cos(\gamma-\alpha)$
$s'\delta\alpha=R\delta\beta\cos(\gamma-\alpha)$.

Thanks. I finally understand your derivation for $d\beta\ = \Large\frac{s'}{s} \large d\gamma$. The difference between $B$ and $B'$ and $C$ and $C'$ is greatly exaggerated.
.

 

[rude man]

Anyway, the answer is obviously no. Going close to the ring - what is the (simple) expression relating charge density to E field on the surface of a dielectric?

I

You could try writing down the integral for the net field at P and look for extrema as the distance of P from the centre of the hoop varies.

I think that would involve elliptical integrals which are best left alone here.

 

[haruspex]

I think that would involve elliptical integrals which are best left alone here.

Since only extrema are of interest, it might not be necessary to solve the integral.

 

[bob012345]

Since only extrema are of interest, it might not be necessary to solve the integral.

Since the OP only asked if it is zero or not at one point along the radius, you would not have to actually do any integrals but just set the problem up. I found the formula worked out here*. The answer does involve Elliptical integrals of the first and second kind. If it wasn't enough to just look at the formula and see it is not zero, one can evaluate the elliptical integrals by lookup table at that point and see the value is not zero. I still think none of this is necessary based on the previous strategy.

$$E_r = \frac{2\pi R \lambda}{ 4πε_0} \frac{2}{ ^{\large \pi q^\frac{3 }{2}} (1−µ) }\frac{1 }{µ} (2R(1−µ) K(k)−(2R−µ(r +R))E(k))$$

where $~~q = r^2 + R^2 +2rR ~~; k = \sqrt\mu$ and $\mu = \large \frac{4rR}{q}$ and $~~K(k) , E(k)$ are the complete elliptical integrals of the first and second kind.

This can be simplified by using $a = \large\frac{r}{R}$ and noticing $q = (R + r)^2 = R^2(1 + a)^2$

we get finally;

$$E_r = \frac{2 \lambda}{ 4πε_0 R} \left[\frac{1}{a} \left(\frac{K(k)}{(1+a)} - \frac{E(k)}{(1-a)}\right)\right],~~~a>0$$

or in the form with $k_e =\large \frac{1}{4πε_0}$ and $Q = 2\pi R \lambda$;

$$E_r = \frac{ k_e Q}{ \pi R^2} \left[\frac{1}{a} \left(\frac{K(k)}{(1+a)} - \frac{E(k)}{(1-a)}\right)\right],~~~a>0$$

For $a = \large \frac{2}{3}$, the ratio$\frac{ \Large E_r R^2}{\Large k_e Q}$gives ~-0.693 which is not zero.

since $a =\Large \frac{r}{R}$ and $k = \Large \frac{2\sqrt a}{1+a}$ as $a → 0, E_r → 0$ as $a →1$ $E_r$ diverges.

then as $a, r →∞, k → 0, K(k), E(k) → \frac{\Large\pi}{2}$ and $E_r → \frac{ \large k_e Q}{ \ \Large r^2}$

Here is a plot of what the field looks like. The black curve is the equivalent plot of a point charge equal to the ring for reference. The $x$ axis is distance of the point $P$ from the center of the ring in units of $R$.

*http://www.mare.ee/indrek/ephi/efield_ring_of_charge.pdf