# Homework Help: Find the electric field inside a circle

1. Dec 13, 2009

### fluidistic

1. The problem statement, all variables and given/known data
I set up myself to find the electric field inside a charged circle, but not in the center of the circle. Instead, in a point on a diameter, at a distance R/2 from the center, where R is the radius.
2. The attempt at a solution
I've tried to set up an integral summing the contribution of all the circle but without success. Is it a good idea? How would you proceed?
I know that the electric field must be 0, but as I'm not really convinced, I wanted to verify it.
Any help is appreciated.
dE=dQ/r².
$$Q=2 \pi R \lambda$$.
$$dl=xd\theta$$ where dl is a differential arc on the circle, $$\theta$$ is the angle measured from the horizontal diameter to the dl element and x is the distance between the point where I want to find the electric field and the dl element.

2. Dec 13, 2009

### AEM

I would compute the field dE at your point P due to a bit of charge $dq = \lambda rd\theta$ and integrate around the circle. I would also look to see if there is any symmetry condition that makes solving the problem easier. (Note: this is my first take on how to approach it. There might be a better way.)

3. Dec 13, 2009

### fluidistic

Ok perfect! That was the kind of reply I was hoping for. Thanks, I'll try to work something out.

4. Dec 13, 2009

### fluidistic

I get $$d\vec E=\frac{Qd\theta}{2\pi Rr}\hat r$$ where R is the radius and r is the distance between P and dl.
I'm sure there's a $$cos \theta$$ that will appear from here, and integrating from 0 to $$2 \pi$$ I would eventually reach 0... but I don't know how to make the $$cos \theta$$ to appear. This is not new to me. I always struggle with this.

5. Dec 13, 2009

### AEM

Okay, so you got me hooked enough that I took another look at it. I drew a circle and drew a horizontal line I'll call the x-axis and a vertical y-axis. For each bit of charge $\lambda r d\theta$ there will be one symmetrically placed on the opposite side of the x axis. That means that the y components of the vector dE will cancel. That takes care of your $cos(\theta)$. You need that to get the x component of dE. What concerns me is that th expression for r is going to be messy, but I'm going to have to sign off soon. I'm curious as to what you're going to come up with...

6. Dec 13, 2009

### fluidistic

You drew the same sketch than I. I also realized that the symmetry will automatically tell us that the E field in the y direction is null, but after all I wanted to prove it mathematically.
I'm also stuck as to write r... I'd have an "x" variable, but it doesn't seem easy to continue.
For example $$r=\sqrt{x^2+y^2}$$ or I could choose $$r=\frac{x}{\cos \theta}$$, though I don't know how to go any further. What limit to take in the integral for example.
$$\theta$$, 0 to $$2\pi$$. But x...
I've no idea.

7. Dec 13, 2009

### rl.bhat

If you start measuring x from the point R/2, using the properties of circle, you can write
y^2 = (R/2-x)*(3R/2+x)
You can take the integration form x = 0 to x = R/2 for smaller part and x=0 to x = 3R/2 for the larger part.

8. Dec 13, 2009

### fluidistic

I thank you for your help and time. I don't understand how you reached this. Is it be simple Pythagorean theorem? I only see y²=r²-x² or y=r sin (theta).
Time for me to go to sleep, it's past 2 am and if I don't sleep now I won't sleep the night before the exam, ahah.

9. Dec 13, 2009

### rl.bhat

The relation is true for any intersecting diameter and a cord which is perpendicular to the diameter. Cord bisects the diameter.
Let AB be the diameter and PQ is the cord. Let O be the point of intersection.
Triangle OAP and OBQ are similar. So the sides are proportional.
Hence OP/OB = OA/OQ Or
OA*OB = OP*OQ. But OP = OQ. So
OA*OB = OP^2.