Homework Help: Electric field along the y-axis of a charged semicircle

1. Apr 5, 2013

dumbperson

1. The problem statement, all variables and given/known data

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/...z9K2R-JDyhgQrU0CSl08tKghH-4LTTamtYNjY-w7FNfPA

2. Relevant 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)
3. 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

2. Apr 5, 2013

ehild

Check the sign of the electric field. Otherwise it looks correct.

ehild

3. Apr 5, 2013

dumbperson

Alright, thanks. Do you have any tips on how to solve the integral to get E_y?

4. Apr 5, 2013

ehild

It can not be written in closed form, I am afraid. Writing up the potential is easier, but it also involves elliptic integral.

ehild