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.
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
$$E_y = \frac{2kQ}{\pi}\frac{1}{b^2-a^2}$$
Best Answer
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 $\sin\theta.$