# Electric field in two dimensions

1. Jan 29, 2004

### stunner5000pt

I have this problem which im stuck on: Find an expression for the electric field at point P as in this figure
P
|
|
|
|
|
|
+++++++++++++++++++++++++++++++++++++++++++++++++++++++
<----------------------L-------------------------------->
Where the distance from P to the positive charged rod is Y.

All i know is that you need to break the positive charged rod into little differentials such as dL and integrate someting like this -
Ey = (integrate)-L to +L Q/(4pi Epsilon(o) d^2)dq

2. Jan 30, 2004

### himanshu121

Why dont u apply Gauss Th

3. Jan 31, 2004

### UpQuark

Well this one can be tricky. Now with your picture there and then your claim I am confused, is the point located perpendicular to the end of the line of charge?
If that is the case then integrating along the length from 0 to L might be easier...
anyway
if you look at the picture, as you move from the left to the right, and angle is formed with the line extending from P along L. as you move down L this angle changes.
As the angle becomes larger the radius changes as well.
dEx=dEcos(theta)
dEy=dEsin(theta)
Now dE=Ko*dq/(r^2)
substituting the charge density lambda*dx for dq we get
dE=Ko*Lambda*dx/(r^2)
if you replace r with the varibles we do know you get
r=sqrt(L^2+d^2)
where d is the vertical distance from the line to the point.
now making the necessary trigonometric changes you can find the the integral for each component.

Hope that helps