# Electric Field of a circular arc at a point

Tags:
1. Nov 7, 2014

### Hijaz Aslam

1. The problem statement, all variables and given/known data

Given that the circular arc wire with radius 'r' has a linear charge density $\lambda$. What is the Electric field at the origin?

2. Relevant equations
$\vec{E}=\frac{kq}{r^2}$ where $k=9\times10^9$ is a constant.

3. The attempt at a solution

I took a small segment dy $\theta$ above the x-axis with charge $dq=\lambda dy$. Therefore $d\vec{E}=\frac{k\lambda cos\theta dy}{r^2}$ as all other charges along the y-axis cancel out each other.

Now $cos\theta=\frac{x}{r}$. And $x^2+y^2=r^2$ is the equation of the arc.
Therefore $cos\theta=\frac{\sqrt{r^2-y^2}}{r}$. And then proceeding to integrate $d\vec{E}=\frac{k\lambda \sqrt{r^2-y^2} dy}{r^3}$ and arrive at an answer.

But my text tackles the question the same way until, at a point it takes $dy=rd\theta$ and then substitutes and integrates $d\vec{E}=\frac{k\lambda cos\theta d\theta}{r}$ and arriving at an answer. But my answer differs from the one arrived by my textbook. Am I wrong somewhere?

2. Nov 7, 2014

### Staff: Mentor

$dq=\lambda dy$ would mean every "step" dy gives the same charge, and equivalently the same length of the wire. This is not true.

3. Nov 7, 2014

### Hijaz Aslam

mfb : That is the condition given in the question. Moreover my text starts the solution in the same way. (The wire is uniformly charged)

4. Nov 8, 2014

### rude man

Use symmetry to conclude what about the y component of E at the origin?
For the x component, dEx = k cosθ dq/R2 should be apparent.
What is dq in terms of arc length ds?
Go that way & you will wind up agreeing with your textbook.

As mfb implied, dy = R dθ is incorrect. R dθ is an element of arc length ds, not an element of distance along the y axis.