Calculation of a net electric field for a charged ring - weird integration

Question

I am just reading book "University physics with modern physics 14-th edition (Young & Fredman)". And on page 702 there is an example 21.9 which says:

Charge $Q$ is uniformly distributed around a conducting ring of radius $a$. Find the electric field at point $P$ on the ring axis at a distance $x$ from center.

So author first states that linear charge density $\lambda = Q/2\pi a \rightarrow \lambda = \text{d}Q/\text{d}s$ where $\text{d}s$ is a diferential of the ring length. It is also immediately clear that there won't be any net electric field in the $y$ direction as $\text{d}{E}_y$ cancel out while on the other hand all $\text{d}{E}_x$ sum up. Therefore we can write this in scalar form:

\begin{equation*} \begin{split} \text{d}E_x &= \text{d}E \cdot cos(\alpha)\\ \text{d}E_x &= \frac{1}{4\pi\varepsilon_0}\frac{\text{d}Q}{r^2} \cdot cos(\alpha)\\ \text{d}E_x &= \frac{1}{4\pi\varepsilon_0}\frac{\text{d}Q}{r^2} \cdot \frac{x}{r}\\ \text{d}E_x &= \frac{1}{4\pi\varepsilon_0}\frac{\text{d}Q}{(x^2+a^2)^2} \cdot \frac{x}{x^2 + a^2}\\ \text{d}E_x &= \frac{1}{4\pi\varepsilon_0}\frac{x}{(x^2 + a^2)^{3/2}}\cdot \text{d}Q\\ \text{d}E_x &= \frac{1}{4\pi\varepsilon_0}\frac{x\cdot \lambda}{(x^2 + a^2)^{3/2}}\cdot \text{d}s\\ \end{split} \end{equation*}

This is all fine, but then he integrates over all the ring's length. But if we have equation we have to integrate in a same way on both sides right? So I think integration should look like this:

\begin{equation*} \begin{split} \int_0^{2\pi a}\text{d}E_x\, \text{d}s &= \int_0^{2\pi a}\frac{1}{4\pi\varepsilon_0}\frac{x\cdot \lambda}{(x^2 + a^2)^{3/2}}\cdot \text{d}s \, \text{d}s\\ \int_0^{2\pi a}\text{d}E_x\, \text{d}s &= \frac{1}{4\pi\varepsilon_0}\frac{x\cdot \lambda}{(x^2 + a^2)^{3/2}}\cdot \int_0^{2\pi a} \text{d}s \, \text{d}s \end{split} \end{equation*}

What is weird to me is integral on the right. Well author of the book doesn't even integrate in a same way on both sides of equation. What he writes down is:

\begin{equation*} \begin{split} \int \text{d}E_x &= \frac{1}{4\pi\varepsilon_0}\frac{x\cdot \lambda}{(x^2 + a^2)^{3/2}}\cdot \int_0^{2\pi a} \text{d}s\\ E_x &= \frac{1}{4\pi\varepsilon_0}\frac{x\cdot \lambda}{(x^2 + a^2)^{3/2}}\cdot 2\pi a\\ \end{split} \end{equation*}

Is he alowed to do that? Why?

Answer 1

You're essentially asking why separating and integrating works. Consider solving the equation $$\frac{dy}{dx} = f(x).$$ You can solve this equation by integrating with respect to $dx$ on both sides, $$\int \frac{dy}{dx} \, dx = \int f(x) \, dx.$$ The left-hand side is, by the chain rule (or if you prefer, by $u$-substitution) $$\int \frac{dy}{dx} \, dx = \int dy$$ where the integration variable is now $y$. Therefore we have $$\int dy = \int f(x) \, dx.$$ You could also get this by treating the original derivative as a fraction, splitting it, and integrating both sides, $$\frac{dy}{dx} = f(x) \quad \Rightarrow \quad dy = f(x) dx \quad \Rightarrow \quad \int dy = \int f(x) \, dx.$$ This is pretty intuitive, too: it says the total change in $y$ is found by adding up all the little changes in $y$, which each happen to be equal to $f(x)$ times the little change in $x$. Depending on how you formalize differentials, the second method can also be totally mathematically correct, though it isn't in the way taught in a first calculus course, so it's understandable it looks weird to you.

Your book is using the second method. If it makes you feel better, you can divide both sides by $ds$ and integrate with respect to $s$ on both sides, converting it to the first method. Of course both are equivalent.

Answer 2

$dE_x$ is the $x$-component of the small field $d\vec E$ due to the charges located on a small segment. Thus, $E_x=\int dE_x= \int\frac{1}{4\pi\varepsilon_0}\frac{x\cdot \lambda}{(x^2 + a^2)^{3/2}}\text{d}s$: there is no need to multiply again by $ds$, but you need to integrate (i.e. sum) the contributions of all the small segments.

For a small segment located at $\theta$ (in plane polar coordinates) on the ring, $ds=a d\theta$, so summing the segments of the ring leads to integration of your source distribution from $\theta=0$ to $\theta=2\pi$.

As you’ve set it up, your approach in incorrect in that you are integrating over $dx$, but $x$ is the distance on the axis and so it is something fixed in your problem, i.e. the distance along $\hat x$ between your sources and the point on the axis is constant.