Find the electric field from a ring of charge along its axis without using symmetry I am more than happy understanding how to find the equation for the electric field for a ring of charge along its centre axis (see Electric Field: Ring of Charge). But in every explanation I have seen they use the symmetry of the problem to cancel out one of the components. I understand why they do this, it makes the calculations much easier, but I would like to understand how mathematically it cancels out. Any help would be appreciated. Thank you.
 A: Let's set up the problem. There is a ring of charge of radius $a$ and we want to compute the electric field a height $z$ above the centre of the ring. Let's call $r=\sqrt{a^2+z^2}$ and the angle subtended by a radius at the point of interest $\theta = \arcsin(a/r)$.
Now, in the plane of the ring draw a ray going out from the centre. Any point on the ring can be described by the angle $\phi$ it makes with this ray. The field at height $z$ due to the charge at this point is given by Coulumb's law:
$$ d\vec{E} = \frac{dQ}{4\pi\epsilon_0 r^3} \vec{r}$$
(Where dQ is the charge in the part of the ring being considered, so $dQ = (2\pi)^{-1}Qd\phi$). Now we want to work out how the different components of this depend on the angles $\theta,\phi$. The $z$ component doesn't care about $\phi$ and is:
$$dE_z = f(r) \cos(\theta)$$
whilst the $x,y$ components do care about $\phi$. The $x$ component, for example is:
$$dE_x = -f(r) \sin(\theta) \cos(\phi)$$
Now we care about the total field, i.e. the sum from all points $\phi \in [0,2\pi)$ on the ring. We do this sum as an integral, the relevant point is that:
$$E_x \sim \int_0^{2\pi}d\phi \cos(\phi) = 0 $$
since $\cos(x)$ averages to zero over a period.
The same thing happens with $E_y$ but not with $E_z$ where the integral just gives a constant factor of $2\pi$.
A: Suppose we set origin at center of the ring, then we can parameterize the ring as:
$$ R(t) = r \cos t \hat{i} + r \sin t \hat{j}$$
Let the point we want be located at $P=z \hat{k}$, now, by definition, the Electric field is given as:
$$ \vec{E} = \int \vec{dE}= \int \frac{k \lambda dl}{|\vec{s}|^3} \hat{s}$$
Where $\vec{s}$ is the vector joining a point on disc to the point on the axis, $\lambda$ is linear charge density and $k = \frac{1}{4\pi \epsilon_o}$
We can simplify the RHS as:
$$ \vec{E} = k \lambda \int_{t=0}^{2\pi}  \frac{|\frac{dR}{dt}| dt }{(z^2 +r^2)^{\frac32}}\left[ -(r \cos t \hat{i} + r \sin t \hat{j})+ z \hat{k} \right] = k\lambda \frac{r}{(z^2 +r^2)^{\frac32}} \left[  \hat{k}\int_{t=0}^{2\pi} z dt -r \underbrace{\int_{t=0}^{2\pi} \cos t \hat{i} + \sin \hat{j} } \right]$$
Notice that the underbraced term is zero because sines and cosines integrated over a whole period nets a total of zero signed area. Ultimately we only have a component in the $z$ direction.
