# Semi-infinite non-conducting rod

1. Mar 7, 2013

### kinof

1. The problem statement, all variables and given/known data
A "semi-infinite" nonconducting rod has a uniform linear charge density λ. Show that the electric field E at point P a distance R above one end of the rod makes an angle of 45° with the rod and that this result is independent of the distance R.

2. Relevant equations
$$\vec{E}=\int \frac{k\lambda }{r^2}dx$$

3. The attempt at a solution
I have found that $$\vec{E_x}=\frac{-k\lambda}{\sqrt{R^2+x^2}}$$ and $$\vec{E_y}=\frac{k\lambda x}{R\sqrt{R^2+x^2}}$$

I have tried to show that, if Ex and Ey make a 45° angle, then they must be equal, but that leads nowhere. I have tried simply taking the arctan of Ey/Ex, but that leads to -x/R, which doesn't make sense. I have also tried evaluating the integrals at infinity and then using L'Hospital's rule, but it doesn't work for Ey.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 7, 2013

### TSny

After integrating, did you evaluate at the limits of integration? $x$ should not appear in the results.

3. Mar 7, 2013

### kinof

Oh, woops. I got $$\frac{k\lambda}{R}$$ for both. Of course the ratio of these is 1, so the angle is 45 degrees. Thanks.

4. Mar 7, 2013

### TSny

Looks good!

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