Electric field in two dimensions

AI Thread Summary
To find the electric field at point P due to a positively charged rod, it is essential to break the rod into differential segments and integrate the contributions from each segment. The electric field components can be expressed as dEx = dEcos(theta) and dEy = dEsin(theta), where dE is derived from the charge density and the distance to point P. The distance r from the charge element to point P is given by r = sqrt(L^2 + d^2), with d being the vertical distance from the line to point P. Integrating these components along the length of the rod allows for the calculation of the total electric field at point P. This approach effectively addresses the complexities of the geometry involved.
stunner5000pt
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I have this problem which I am stuck on: Find an expression for the electric field at point P as in this figure
P
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+++++++++++++++++++++++++++++++++++++++++++++++++++++++
<----------------------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 something like this -
Ey = (integrate)-L to +L Q/(4pi Epsilon(o) d^2)dq
However i am not sure about the X components and how taht is calculated, please help.
 
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Why don't u apply Gauss Th
 
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
 
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