I have just started a first-year calculus-based physics course about electromagnetism and waves. I am having trouble understanding what calculus notation means in the context of physics. Here is a paraphrased example from Fundamentals of Physics 9th ed. by Halliday et al. that considers finding the electric field at point P (refer to the figure below):

Let $ds$ be the arc length of any "differential" element of the ring. $dq=\lambda~ds$ is the total charge of the ring, where $\lambda$ is linear charge density. The magnitude of the field is given by


Since the x-components of the field are balanced at point P, after the application of geometry, the equation becomes

$$dE\cos{\theta}=\frac{z\lambda}{4\pi\epsilon_0 (z^2+R^2)^{3/2}}ds$$

The author then puts an integral sign in front of both sides and integrates around the circumference of the ring: $$E=\int dE~\cos{\theta} = \frac{z\lambda}{4\pi\epsilon_0 (z^2+R^2)^{3/2}}\int_0^{2\pi R}ds =\frac{z\lambda(2\pi R)}{4\pi\epsilon_0 (z^2+R^2)^{3/2}}$$

In math, I was taught that Leibniz's notation ($dy/dx$) is not a fraction. After a little bit of research, I found that it sometimes acts as a fraction, even though it is not a fraction when defined with limits. Nevertheless, what should I think of when I see $dq$, $ds$, and $dE$? What does "differential" mean as used in the text above? So far it has made sense for me to think of $dq$ as an "infinitesimal" charge (for example), even though I was taught calculus using limits (I know very little about infinitesimals).

Secondly, the way the integral is taken seems informal and confuses me. I understand that it represents the sum of the electric field caused by all the charge around the ring, but how can you just put an integral sign in front of both sides and take the integral with respect to whatever $dE$ or $ds$ that happens to be there? Also, the integral on the left side is indefinite whilst the one on the right is definite. How should I view this kind of material?


I am unsure if this question belongs on the math stack exchange or this one.

  • 3
    $\begingroup$ Related: physics.stackexchange.com/q/92925/2451 and physics.stackexchange.com/q/70376/2451 $\endgroup$
    – Qmechanic
    Commented Jan 16, 2014 at 20:16
  • 1
    $\begingroup$ Actually, you can treat "differential" as a fraction of two very small quantities, but only the FIRST order of derivative. It is very convenient and helpful to understand calculus notions. Leibniz's notation that $dy/dx$ is not a fraction is strict in mathematics, but Newton's notion is easier to understand in physics. $\endgroup$
    – qfzklm
    Commented Jan 17, 2014 at 11:54

1 Answer 1


To answer your question about how to understand differentials I would essentially reiterate my answer on https://physics.stackexchange.com/a/92931/12029

(I'm assuming dealing with scalars here. Assume $E=\hat{n}\cdot \vec{E}$ and $C=\hat{n}\cdot \vec{C}$ for some direction $\hat{n}$)

The integration is a separate story, and I think it's helpful to go back to first principles. Differentials in this case relate changes in $E$ to changes in $s$. Suppose $dE=f(s) ds$. In integrating both sides, the real thing you're interested in is finding total changes from linear approximations of changes (differentials). So, when integrating both sides, you're really only finding $\Delta E$, the total change in $E$ from all the contributions of whatever domain you integrate $f(s)$ over. So, if you had an external electric field $C$ at the point you're interested in (that is, a field which is present when all the other charges you're interested in are not), the same calculation you just made could still justify writing $E=C+\Delta E$.

As for writing $\int dE=\int_\text{A}^\text{B} f(s)ds$, where one integral has proper limits and one does not, think of change of variables in limits. If you substitute a variable of integration, eg u-substitution, you also need to change the limits of integration. So, if $\Delta E=\int dE$, that means, basically, the integral was taken over $\int_{E_0}^{E_0+\Delta E} dE$, which clearly and rigorously evaluates to $\Delta E$. So I think it could be said that the author left out the limits of integration so as to not confuse the reader. If one sets up the necessary amount of background machinery, one can phrase this as a formal change of variables in the integral (and it would be a good exercise to do so). However, you should not have a difficulty with the problem's solution when interpreted as the limit of a Riemann sum.

  • $\begingroup$ In the third sentence, do you mean charges instead of changes? Also, it might be worthwhile to make sure that we don't confuse the vector $\vec{E}$ with its scalar magnitude $E$. $\endgroup$
    – Edward
    Commented Jan 16, 2014 at 21:06
  • $\begingroup$ @Edward fixed technical correctness, you're right I was being sloppy with vectors/scalars. $\endgroup$
    – user12029
    Commented Jan 16, 2014 at 21:12

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