Electric field question

1. Feb 5, 2010

camrylx

1. The problem statement, all variables and given/known data
A thin, semi-infinite rod with a uniform linear charge density (lambda) (in units of
C/m) lies along the positive x axis from x = 0 to x = 1; a similar rod lies
along the positive y axis from y = 0 to y = 1. Calculate the
electric field at a point in the x-y plane in the first quadrant.

2. Relevant equations
This is a calc based course. Intergration is required for this problem. If you need a diagram I do have one.

3. The attempt at a solution

2. Feb 5, 2010

fluidistic

Hi and welcome to the forum!
In this forum you have to show an attempt, so that we can help you where you're stuck.

3. Feb 5, 2010

camrylx

This is one question that I am not that sure how to get started with.

4. Feb 5, 2010

fluidistic

$$d\vec E = \frac{kdQ \vec r}{r^3}$$. What is worth dQ in the case of a straight segment of length dx of the rod?
Once you have $$d\vec E$$, you just have to integrate (choosing the appropriate limits of integration) to get $$\vec E$$.
I suggest you to start by drawing the situation. Put a point $$(x_0,y_0)$$ in the first quadrant. Tackle the problem first with the x-axis rod. Calculate the E field due a small element dx of it, then integrate to get the E field (in point $$(x_0,y_0)$$) due to the whole x-axis rod.
Do the same for the y-axis rod and sum them up. Don't forget that they are vectors.

I hope it helps. Feel free to post any difficulties you encounter.

5. Feb 8, 2010

camrylx

ok so I understand that for intergrating the 2 rods you set them up as
$$\int dE1x+dE2x$$ and the same for the y component but I am kinda stuck on what is next.

6. Feb 8, 2010

camrylx

ok so I understand that for intergrating the 2 rods you set them up as
$$\int dE1x+dE2x$$ and the same for the y component but I am kinda stuck on what is next.