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dumbperson
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Homework Statement
Find the electric field along the y-axis of a charged semicircle (with the center of the circle in the origin) with radius R, with a uniform charge distribution $$ \lambda$$ . https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcSzwZUlOz9K2R-JDyhgQrU0CSl08tKghH-4LTTamtYNjY-w7FNfPA
Homework Equations
$$ \vec{E} = \frac{1}{4\pi \epsilon_0} \int \frac{dq}{|\vec{r}|^2} \hat{r} $$
$$ k = \frac{1}{4\pi \epsilon_0} $$
(r is the vector pointing from a charge dq to your ''test charge'' on the y-axis)
The Attempt at a Solution
$$ dq= R\lambda d\theta $$
The positionvector of the charge dq $$ =\vec{r_1} = R \cos(\theta)\hat{x} + R \sin(\theta)\hat{y}$$
The positionvector of your ''test charge'' along the y-axis : $$\vec{r_2}=y\hat{y} $$
so the vector $$ \vec{r} = R \cos(\theta)\hat{x} + (R \sin(\theta)-y)\hat{y}$$
$$ |\vec{r}|= \sqrt{R^2\cos(\theta)^2+R^2\sin(\theta)^2+y^2-2yR\sin(\theta))}=\sqrt{R^2+y^2-2yR\sin(\theta)}$$
so $$\hat{r}= \frac{\vec{r}}{|\vec{r}|} = \frac{ R \cos(\theta)\hat{x} + (R \sin(\theta)-y)\hat{y}}{\sqrt{R^2+y^2-2yR\sin(\theta)}} $$
So the integral becomes
$$ \vec{E} = kR\lambda \int_0^\pi \frac{ R \cos(\theta)\hat{x} + (R \sin(\theta)-y)\hat{y}}{(R^2+y^2-2yR\sin(\theta))^{\frac{3}{2}}} d\theta $$
So $$ E_x = kR^2\lambda \int_0^\pi \frac{\cos(\theta)}{(R^2+y^2-2yR\sin(\theta))^{\frac{3}{2}}} d\theta$$
$$ E_y = kR\lambda \int_0^\pi \frac{R\sin(\theta)-y}{(R^2+y^2-2yR\sin(\theta))^{\frac{3}{2}}} d\theta $$
Am I doing this correctly? I managed to solve the integral for E_x and the result is zero, as expected. But I have no clue how to solve E_y, is this integral set up correctly? Well I could solve it for y=0 pretty easily, but that is not the assignment.
Thanks
