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.

enter image description here

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?

  • 1
    $\begingroup$ One way to think about it is the following. Think what you would do in any change of variable to compute an integral. Say you make a change from $x$ to $y$ with $x=g(y)$, a function of $y$. Then you write $\int f(x)\mathrm{d}x=\int f(g(y)) \frac{d g}{dy} \mathrm{d}y$. Notice how you are integrating different variables on each side and there is no inconsistency. $\endgroup$
    – secavara
    Commented Mar 31, 2018 at 12:02

2 Answers 2


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.

  • $\begingroup$ Thank you everyone. I really need to work on my Calculus a bit more... I think this is the best answer. $\endgroup$
    – 71GA
    Commented Mar 31, 2018 at 12:15
  • $\begingroup$ I don't like the second approach which deals derivatives as fractions. I even think it should be discouraged. :) $\endgroup$
    – 71GA
    Commented Mar 31, 2018 at 12:28
  • 1
    $\begingroup$ @71GA Of course it's a matter of taste, you can see how the two are always going to give the same answer. When you learn more advanced calculus (i.e. differential forms) the second way will even be the more correct method! I personally use the fraction method for all single-variable problems though you're right that it can lead you astray in general. $\endgroup$
    – knzhou
    Commented Mar 31, 2018 at 12:56

$dE_x$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.