# 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