# Homework Help: Trouble calculating the electric field at a point

1. Jan 5, 2010

### fluidistic

1. The problem statement, all variables and given/known data
The situation is the one of a hairpin such as in this website : http://www.physics.brocku.ca/Courses/1P22_Crandles/problems/hairpin.jpg.
I must prove that the electric field in "a" is worth 0. There's a tip. It says to calculate the electric field due to a differential part on the semi circle and to compare it with a differential part on the infinite length wire. With both the same angle $$\theta$$, where theta is the angle between the point over the semicircle considered, point "a" and the vertical radius (bottom one) of the semicircle.

2. Relevant equations

Linear charge density of the wire : $$\lambda$$.
3. The attempt at a solution
$$d\vec E=\frac{dQ \hat r}{r^2}$$. For the differential portion over the semicircle, I get that $$dE=\frac{\lambda \cos (\theta ) d\theta}{b}$$. Obviously I should find the same result for the dE of the differential portion of the straight wire, but I find $$\frac{\lambda \left [ \tan (\theta +d\theta) - \tan \theta \right ] }{b \cos ^2 (\theta + d \theta)}$$. I don't think they are the same. At least I've tried to equate them, but with no success.
Did I do everything wrong?

2. Jan 5, 2010

### rl.bhat

The current carrying conductor does not produce electric field, because it does not have any net charge. In every differential part of the conductor electrons are drifting in a particular direction but the net charge is zero at any instant.
In the attached link they have asked the magnetic field at a and b.
Have you changed it to electric field?

3. Jan 6, 2010

### fluidistic

Dear rl.bhat, I should have mentioned that my problem is different from the picture at http://www.physics.brocku.ca/Courses/1P22_Crandles/problems/hairpin.jpg. In my problem there's no current, but a uniformly charged wire whose linear charge density is $$\lambda$$. I'm sorry about the confusion, I realize it's entirely my fault. The website I provided is good to understand the shape of the charged wire and the position where I want to calculate the electric field.

4. Jan 7, 2010

### rl.bhat

If you consider the electric field due to two symmetric differential elements on the semicircle, you get dE= 2*k*λ*cosθ*dθ/b. If you take the integration between the limits from 0 to π/2, you get E = 2*k*λ/b.
Similarly if you consider two symmetric differential elements of length dx at a distance x from the starting point of the straight conductors, the electric field at a is given by
dE = 2*k*λ*x*dx/(x^2 + b^2)^3/2. If you take the integration between the limits from 0 to infinity, you get the same result as above. These two fields are in the opposite direction. So they cancel each other making net field at a zero.

5. Jan 7, 2010

### fluidistic

Thanks for helping!
Ok, I almost got this for dE, except for the k (that I simply forgot) and the factor 2. I follow you until here.

Wow... I don't know why your dE differs SO much from mine. I'll try to reach your dE. If I have any problem, I will ask for further help. Thank you very much for your time, effort and will to help.

6. Jan 8, 2010

### fluidistic

All these differentials, I'm having a very hard time with them and I'm spending a too big amount of time playing with them.

For the semi circular path, here is my work:
$$d\vec E=\frac{kdQ\vec r}{r^3}$$ where $$\vec r=r\cos \theta \hat i +r\sin \theta \hat j$$.
$$dQ=\lambda bd\theta$$.
Thus $$d\vec E=\frac{k\lambda bd\theta (b\cos \theta \hat i + b\sin \theta \hat j)}{b^3}=\frac{k\lambda d\theta (\cos \theta \hat i+ \sin \theta \hat j)}{b}$$. That was for the 0 to pi/2 part. Now I have to multiply by 2 the vertical component and throw off the horizontal one. Then I integrate from 0 to $$\frac{\pi}{2}$$ to get the desired result:
$$\vec E=\frac{2k\lambda}{b}\int _0 ^{\frac{\pi}{2}} \sin \theta d\theta =\frac{2k\lambda}{b} \hat j$$. I don't understand why I don't reach a negative sign...

I've tried several ways to solve for the 2 semi-infinite wires, but I can't reach your result "dE = 2*k*λ*x*dx/(x^2 + b^2)^3/2". Can someone guide me a bit further please?

7. Jan 8, 2010

### rl.bhat

Your expression for E due to semicircle is correct. Its direction is towards the right side from the semicircle. No need to put negative sign.
Now for field due to straight wire, consider a small element dx at a distance x from the
left end of the bottom wire. Field at a due to this element is
dE = k*λ*dx/r^2 = k*λ*dx/(x^2 + b^2). Similarly find the field due to the identical element on the top wire. dE*sinθ components of these fields get canceled each other.
So net field due to straight wire is
E = Intg(2*dE*cosθ) from limit x = 0 to x = infinity. Here cosθ = x /(x^2 + b^2)^1/2.
Its direction is towards left of a.

8. Jan 8, 2010

### vela

Staff Emeritus
Probably because you're thinking of $$\vec{r}$$ the wrong way. It goes from the charge to the point, not the other way around.

9. Jan 9, 2010

### fluidistic

Thanks to both!