In my book (University Physics by Young and Freedman), during solving the common example of finding the electric field along the x-axis from a uniformly charged disk, they arrive at this differential expression:

$$dE_x = {1 \over 4\pi\epsilon_0}{2\pi\sigma rx dr \over (x^2+r^2)^{3/2}}$$

Then they say:

To find the total field due to all the rings, we integrate $dE_x$ over $r$ from $r=0$ to $r=R$ (not from $-R$ to $R$):

$$E_x = \int_0^R {1 \over 4\pi\epsilon_0}{(2\pi\sigma r dr)x \over (x^2+r^2)^{3/2}}$$

I am confused as to how they arrived at this line. The right hand side suggests that they integrated both sides of the differential equation with a definite integral, but if that were the case, the left hand side would have been

$$\int_0^R dE_x = [E_x]_0^R = R - 0 = R$$

Instead, it looks like they integrated the left hand side of the differential equation with an indefinite integral, so they would get

$$\int dE_x = E_x$$

Certainly, integrating one side with an indefinite integral and the other side with a definite integral cannot be a valid step because you are not doing the same thing to both sides.

How can their result be rigorously justified and what is wrong with my reasoning?

  • $\begingroup$ You're integrating with dE, not dr, on the left hand side, so the limits of integration can't be the same. $\endgroup$
    – DanDan面
    Commented Sep 2, 2019 at 18:20
  • $\begingroup$ Good point. I guess the limits of integration should be the electric field at point $r=0$ and at point $r=R$? $\int_{E_x(0)}^{E_x(R)} dE_x = E_x(R) - E_x(0)$ This still doesn't make sense to me. $\endgroup$
    – user109923
    Commented Sep 2, 2019 at 18:25
  • $\begingroup$ But then what does $E_x(r)$ represent? The problem only said that $dE_x$ represents the field component at the point due to a ring. $\endgroup$
    – user109923
    Commented Sep 2, 2019 at 19:12
  • $\begingroup$ ...and that $\vec{E}=E_x\hat{i}$. $\endgroup$
    – user109923
    Commented Sep 2, 2019 at 19:19

2 Answers 2


The left hand integral limits should be with respect to the field you get as you add up the contributions to the field due to each charged ring. Therefore, it should start at $0$ and end with the final field. i.e. $$\int_0^{E_x}\text dE_x' = \int_0^R{1 \over 4\pi\epsilon_0}{2\pi\sigma rx\,\text dr \over (x^2+r^2)^{3/2}}$$

Trivially, the integral on the left side is just the total field $E_x$.

Seeing some of your comments, you seemed to be confused as to what $r$ and $R$ actually represent. It looks like $x$ is the constant distance between the center of the disk and the point at which you are calculating the field at. $r$ is the radius of one of the rings of thickness $\text dr$, and $R$ is the radius of the entire disk. Note that you are not determining the field at $r$, since $r$ is not a position in this case.

Your differential relation $$\text dE_x = {1 \over 4\pi\epsilon_0}{2\pi\sigma rx\,\text dr \over (x^2+r^2)^{3/2}}=\alpha(r)\,\text dr$$ is essentially telling you each ring of radius $r$ and thickness $\text dr$ contributes to the field an amount $\alpha(r)\,\text dr$. It makes sense that we just need to add up (integrate) all of these values (right integral) to determine the total field (left integral).

  • $\begingroup$ Isn't that a bit like assuming you already know how the answer is going to turn out? $\endgroup$
    – user109923
    Commented Sep 2, 2019 at 18:45
  • $\begingroup$ As Aaron points out, they are adding contributions to the field from thin rings of charge located in the disk, each of radius r and thickness dr. Each little segment of a ring contributes at an angle from the x axis. For the x component you multiply by x/(x^2 + r^2)^(1/2). The charge in each ring depends on the charge density and area: σ(2 π r dr). The field drops off with the distance squared: (x^2 + r^2). The limits reflect the distribution of charged rings. $\endgroup$
    – R.W. Bird
    Commented Sep 2, 2019 at 19:20
  • $\begingroup$ @user109923 not at all. Just because I know that adding up all of the field contributions gives me the total field it doesn't mean I know what that field is. That is what the right side integral is for. $\endgroup$ Commented Sep 2, 2019 at 19:32
  • $\begingroup$ @user109923 Is there anything that is unclear to you? $\endgroup$ Commented Sep 2, 2019 at 22:01
  • $\begingroup$ I guess I don't really see what it means to have numbers along the $E_x$ axis, as would be necessarily the case for $0$ and $E_x$ to be limits of integration in an integral with respect to $E_x$. $\endgroup$
    – user109923
    Commented Sep 7, 2019 at 2:40

But then what does $E_x(r)$ represent? The problem only said that $dE_x$ represents the field component at the point due to a ring

$E_x(r)$ is a poor notation here since it certainly looks like the $x$ component of the electric field as a function of the radial coordinate $r$. Further, it is clear from the integral that $E_x = E_x(x)$.

Here, $dE_x$ or $dE_x|_r$ is the contribution to the total electric field (at coordinate $x$ on the $x$-axis) due an infinitesimal ring of charge between $r$ and $r + dr$.

If it helps, consider the approximation by finite rings of charge located between $r$ and $r + \Delta r$. Let there be $N$ such rings such that

$$\Delta r = \frac{R}{N}$$


$$r_n = (n-1)\Delta r$$

To find the total electric field due to the contributions $E_{x,n}$ from the $N$ rings making up the disk, we simply sum the contributions (linearity of electric field).

$$E_x = \sum_{n=1}^NE_{x,n}$$

In the limit $N\rightarrow\infty$, the contributions become infinitesimal $E_{x,n}\rightarrow dE_x|_r$, and the sum becomes an integral.


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