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I want to find the y-component of the electric at a point $b$ on the x-axis due to a charged semicircle centred around the origin.

diagram

I found $dq$ using

$$\frac{dq}{Q} = \frac{dθ}{π}$$

$$dq = \frac{Q}{π}dθ$$

And the distance $r$ from $dq$ to $b$ using cosine law

$$r^2 = a^2 + b^2 - 2abcosθ$$

Subbing this into $dE = \frac{kdq}{r^2}$ gives

$$dE = \frac{k\frac{Q}{π}}{a^2 + b^2 - 2abcosθ}dθ$$

Since I only want the y-component of the electric field, I multiplied the whole thing by sinθ and integrated to get

$$E_y = \int\frac{k\frac{Q}{π}sinθ}{a^2 + b^2 - 2abcosθ}dθ$$

$$E_y = \frac{kQ}{2πab}ln(\frac{a^2+b^2+2ab}{a^2+b^2-2ab})$$

The answer that I should be getting is

$$E_y = \frac{2kQ}{\pi}\frac{1}{b^2-a^2}$$

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2 Answers 2

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So you have the source point $[a\cos\theta,~a\sin\theta]$ and the field point $[b,~0]$ and therefore the vector connecting the two is $\vec r = [b-a\cos\theta,~a\sin\theta].$

You are correct to calculate $\|\vec r\| = \sqrt{a^2 + b^2 -2ab\cos\theta}$ but notice that when we want to calculate the $y$-component we need to form $\hat y\cdot \hat r/\|\vec r\|^2$ with $\hat y,\hat r$ being unit vectors. You seem to have assumed that $\hat y\cdot\hat r=\sin\theta$ but in fact we can see from the above expressions and the definition that $\hat r = \vec r / \|\vec r\|$ that actually,$$f(\theta) = \hat y\cdot\frac{\hat r}{\|\vec r\|^2} = \hat y\cdot\frac{\vec r}{\|\vec r\|^3} = \frac{a\sin\theta}{(a^2+b^2-2ab\cos\theta)^{3/2}}.$$Thus substituting $u=a^2+b^2-2ab\cos\theta,~du=2ab\sin\theta~d\theta$ in the integral yields,$$\int_{-\pi}^0 d\theta ~f(\theta) = \int_{(a+b)^2}^{(a-b)^2}\frac{du}{2b}~u^{-3/2} =\frac1b~\left(\frac1{a+b} - \frac1{a-b}\right)=\frac1b~\left(\frac{a-b}{a^2-b^2} - \frac{a+b}{a^2-b^2}\right).$$ You did your integral right, but you did not set up the right integral because of the hand-wave about $\sin\theta.$

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  • $\begingroup$ The square root should give $b-a$ instead of $a-b$ according to the OP's diagram, showing $b>a$. $\endgroup$
    – velut luna
    Commented Oct 1, 2017 at 17:50
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Assume $b>a$, as shown in your diagram.

First, your given answer should be wrong. Because if we let $a\rightarrow 0$ and fix $b$, then the $y$-component should go to zero. But your answer goes to the Coulomb field of a point charge.

The $y$ component is not obtained by simple multiplication by $\sin\theta$.

You have

$$d\vec{E} = \frac{k\frac{Q}{π}}{(a^2 + b^2 - 2ab\cos \theta)^{3/2}}d\theta((b-a\cos\theta) \hat{i}+a\sin\theta \hat{j})$$

So the $y$ component is $$dE_y = \frac{k\frac{Q}{π}}{(a^2 + b^2 - 2ab\cos\theta)^{3/2}}d\theta a\sin\theta$$

Then $$E_y= kQ/\pi \int_0^\pi \frac{a\sin\theta d\theta}{(a^2+b^2-2ab\cos\theta)^{3/2}} $$ $$=\frac{kQ}{2\pi b} \int_0^\pi \frac{d(a^2+b^2-2ab\cos\theta)}{(a^2+b^2-2ab\cos\theta)^{3/2}}$$ $$=\frac{kQ}{\pi b}\left[(a^2+b^2-2ab\cos\theta)^{-1/2}\right]_\pi^0$$ $$=\frac{kQ}{\pi b}\left[(a^2+b^2-2ab)^{-1/2}-(a^2+b^2+2ab)^{-1/2}\right]$$ $$=\frac{kQ}{\pi b}\left[\frac{1}{b-a}-\frac{1}{b+a}\right]$$ $$=\frac{2kQ}{\pi}\frac{a}{b}\frac{1}{b^2-a^2}$$

If instead you have $a>b$, then the answer is $$E_y=\frac{kQ}{\pi b}\left[\frac{1}{a-b}-\frac{1}{a+b}\right]$$ $$=\frac{2kQ}{\pi}\frac{1}{a^2-b^2}$$

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  • $\begingroup$ Why is r raised to 3/2 in your equation for dE? $\endgroup$ Commented Oct 1, 2017 at 17:10
  • $\begingroup$ Coulomb's law $\vec{E}=\frac{kQ}{r^3}\vec{r}$. $\endgroup$
    – velut luna
    Commented Oct 1, 2017 at 17:13
  • $\begingroup$ Is it not $E = \frac{kQ}{r^2}$? $\endgroup$ Commented Oct 1, 2017 at 17:14
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    $\begingroup$ You answer should be wrong for if you let $a\rightarrow 0$, you should get zero $y$ component, but your answer is finite. $\endgroup$
    – velut luna
    Commented Oct 1, 2017 at 17:20
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    $\begingroup$ The correct answer should go to zero when $a$ goes to zero!! $\endgroup$
    – velut luna
    Commented Oct 1, 2017 at 17:26

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