# I Electric field of a ring:integrate over dq or ds?

1. Sep 7, 2017

### Tiago3434

Hi people, I was reading on my textbook on electromagnetism (Halliday) about using integral to field the electric field of a charged ring (1D) at a point P located on an axis perpendicular to the ring's plane. The ring is uniformly charged. The book (and my professor) both breakdown each element of charge dq into its linear charge density times an element of length, or λds. I'm ok with the math from that point on, but I noticed that you can get the same result just integrating over dq, given that the integral of dq is q. Is my idea wrong, or is there a reason for adding that extra step?
Thanks!

2. Sep 7, 2017

### Staff: Mentor

Suppose the charge density were not uniform. Integrating over s, you would let λ be some function of s, λ(s). Then your integral would contain λ(s)ds. How would you integrate directly over dq in that case?

3. Sep 8, 2017

### Tiago3434

oh, ok I got you point. Thanks, jtbell

4. Sep 8, 2017

### Tiago3434

Sorry if the question is stupid, but thinking a bit further about this, wouldn't the overall charge just add up to q again (even if all the dq weren't equal)?

5. Sep 14, 2017 at 6:22 PM

### Tiago3434

Or (sorry if this is a calculus question) can you only integrate over a differential dx if dx is constant? Is that the case?

6. Sep 14, 2017 at 7:25 PM

### pixel

The integral of λ(s)ds would be q, the total charge. But you are trying to calculate the electric field, E, at a point, P, on the axis, not the total charge. λ(s) can't be taken out of the integral over s in that case.

7. Sep 15, 2017 at 10:38 PM

### Tiago3434

I'm sorry but I don't get it. Wouldn't the answer just end up being the same, as dq=λ(s)ds?

8. Sep 15, 2017 at 11:21 PM

### Tiago3434

Like, if dq=λds, I don't understand how come the integrals ∫dq and ∫λds could end up being different.

9. Sep 16, 2017 at 12:43 AM

### NFuller

Those integrals are not different but those are not the integrals for the electric field. The integral for the electric field also contains a directional element $\hat{r}$ which may weight the electric field more in one direction than the other.

In general
$$\mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\int\frac{dq}{r^{2}}\hat{r}$$

As an example, lets assume you are looking for $\mathbf{E}$ at the center of the ring, then $\hat{r}=-\text{cos}\theta\hat{x}-\text{sin}\theta\hat{y}$. If the ring has a radius of $R$ then
$$\mathbf{E}=\frac{-1}{4\pi\epsilon_{0}}\int\frac{dq}{R^{2}}(\text{cos}\theta\hat{x}+\text{sin}\theta\hat{y})$$
Now lets assume the ring has some varying linear charge density $\lambda=\alpha\text{cos}\theta$ then
$$dq=\lambda Rd\theta=\alpha\text{cos}\theta\;Rd\theta$$
The electric field is now
$$\mathbf{E}=\frac{-1}{4\pi\epsilon_{0}}\int_{0}^{2\pi}\frac{\alpha\text{cos}\theta\;Rd\theta}{R^{2}}(\text{cos}\theta\hat{x}+\text{sin}\theta\hat{y})$$
$$\mathbf{E}=\frac{-\alpha}{4\pi\epsilon_{0}R}\left[\int_{0}^{2\pi}\text{cos}^{2}\theta\;d\theta\hat{x}+\int_{0}^{2\pi}\text{cos}\theta\text{sin}\theta\;d\theta\hat{y})\right]$$
$$\mathbf{E}=\frac{-\alpha}{4\pi\epsilon_{0}R}\left[\pi \hat{x}+0\hat{y})\right]=\frac{-\alpha}{4\epsilon_{0}R}\hat{x}$$
Now if we consider the case of constant charge density $\lambda$
$$\mathbf{E}=\frac{-1}{4\pi\epsilon_{0}}\int_{0}^{2\pi}\frac{\lambda Rd\theta}{R^{2}}(\text{cos}\theta\hat{x}+\text{sin}\theta\hat{y})$$
$$\mathbf{E}=\frac{-\lambda}{4\pi\epsilon_{0}R}\left[\int_{0}^{2\pi}\text{cos}\theta\;d\theta\hat{x}+\int_{0}^{2\pi}\text{sin}\theta\;d\theta\hat{y})\right]$$
$$\mathbf{E}=\frac{-\lambda}{4\pi\epsilon_{0}R}\left[0 \hat{x}+0\hat{y})\right]=0$$
These two answers are different because including the direction $\hat{r}$ selects which parts of the charge density will add together in some direction and which will cancel out in another direction.