Electric Field of a circular arc at a point

[Hijaz Aslam]

Homework Statement

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

Homework 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?

 

[mfb]

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

 

[Hijaz Aslam]

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

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

 

[rude man]

Use symmetry to conclude what about the y component of E at the origin?
For the x component, dE_x = 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.

 

[HalfLostAndSearching]

Your approach seems correct, but it's possible that there may be a mistake in your integration or substitution. It's always a good idea to double check your work and make sure all the steps are correct. Additionally, it's helpful to compare your answer to the one given in the textbook to see where the discrepancy may be coming from. If you're still unsure, you can always ask a classmate or your teacher for help.