1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric field along the y-axis of a charged semicircle

  1. Apr 5, 2013 #1
    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. jcsd
  3. Apr 5, 2013 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

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

    ehild
     
  4. Apr 5, 2013 #3
    Alright, thanks. Do you have any tips on how to solve the integral to get E_y?
     
  5. Apr 5, 2013 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

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

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electric field along the y-axis of a charged semicircle
Loading...