I'm trying to do this physics problem and I'm messing up the integral somewhere. The problem:

Suppose there is a ring of radius $a$ with a uniform charge distribution and a total charge of $Q$. Calculate $\vec{E}$ for a point $P$ equidistant from all points on the ring and distance $x$ from the center of the ring.

My work:

Let the center of the ring be the origin, let $P = x \hat{i}$, and let $\theta$ be the angle at $0$ between $a\hat{k}$ and a selected point on the ring.

$$d\vec{E} = \frac{k_edQ}{(x^2 + a^2)^{\frac{3}{2}}}(x\hat{i} + a\sin \theta \hat{j} +a \cos \theta \hat{k}) $$ $$\vec{E} = \int_0^{2 \pi}\frac{k_eQ}{(x^2 + a^2)^{\frac{3}{2}}}(x\hat{i} + a\sin \theta \hat{j} +a \cos \theta \hat{k})d \theta $$ $$= \frac{k_eQ}{(x^2 + a^2)^{\frac{3}{2}}} \int_0^{2 \pi}(x\hat{i} + a\sin \theta \hat{j} +a \cos \theta \hat{k})d \theta$$ $$= \frac{k_eQ}{(x^2 + a^2)^{\frac{3}{2}}} \bigg[x \theta\hat i -a\cos \theta \hat j +a \sin \theta \hat k \bigg]_0^{2 \pi} $$ $$= \frac{k_eQ}{(x^2 + a^2)^{\frac{3}{2}}} (2 \pi x \hat i) $$

I feel like this is wrong. If so, what did I mess up?

  • $\begingroup$ It seems right, but a visual would help. Also, try checking out the Physics SE. $\endgroup$ – gen-z ready to perish Sep 21 '17 at 13:27

The $2\pi$ shouldn't be there. You can start by saying that all $dE_y$ will cancel out because of symmetry. Hence, you are left with only $dE_x$.

This means that

$$ dE_x = \displaystyle\frac{k}{r^2}\cos \theta dQ,$$

where $r^2 = x^2 + y^2$ and $\cos \theta = \frac{x}{r}$.

This should mean that

$E_x = \displaystyle\int \displaystyle\frac{kx}{(x^2+a^2)^{3/2}}dQ$ which is a whole lot easier to integrate.

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  • $\begingroup$ The problem is not the math, but the physics. Your math is correct, but the first step in interpreting the problem is incorrect. You are integrating by all small charges ($dQ$), not by coordinates $d\theta$. I hope this helps. $\endgroup$ – cgo Sep 21 '17 at 13:32
  • $\begingroup$ If I understand the problem correctly, though, I'm integrating a series of charges over increasingly small arc lengths. Would I get the correct answer perhaps if my equation was $\vec{E} = \int_0^{2 \pi} \frac{k_e Q \theta}{2 \pi(x^2 + a ^2)^{\frac{3}{2}}}(x \hat i + a \sin \theta \hat j + a \cos \theta \hat k)d \theta$? $\endgroup$ – Adam Gluntz Sep 21 '17 at 14:33
  • $\begingroup$ Yes, the $2\pi$ at the bottom cancels the top $2\pi$ in your answer. But where does the bottom $2\pi$ come from? $\endgroup$ – cgo Sep 23 '17 at 8:32

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