# How to calculate the electric field due to a thin arc?

I have been given to calculate the electric field at the centre of a thin arc with linear charge density as a function of $\cos\theta$ as $\lambda(\theta)=\lambda_0 \cos\theta$.

How I approached: The angle subtended by the ends of the arc at the centre is $\theta$. Now I considered a very thin segment at an angle $\alpha$ with the vertical with a small angle $d\alpha$ such that $d\alpha = \Delta \theta$. The sines of the electric field due to all points is $0$. All we are left with are the cosines of electric field. Thus I got an equation to integrate involving both $\cos\alpha$ and $\cos\theta$.

My problem: I am not sure whether the above mentioned approach to calculate the field is correct and if it is correct,then how should I proceed, what limits should I use ? $-\dfrac{\theta}{2} \to +\dfrac{\theta}{2}$?

• Because you are using a cosine function the values of your limits work. (The function varies between 0 and 1 between those degrees) – user97261 Jun 14 '16 at 19:12

It's a good idea to approach this as you did, and you're certainly correct to choose a thin segment at an angle $\alpha$. This small segment subtends an angle $d\theta$. Remember that $\alpha$ is some arbitrary value of the variable $\theta$.

The charge carried by this segment is then $dq=\lambda d\theta=(\lambda_0\cos\theta) d\theta$. This segment will produce an electric field

$d\vec{E}=\dfrac{k\cdot dq}{r^2}\hat{r}\quad\quad\quad$(you'll also see $d\vec{E}=\dfrac{k\cdot dq}{r^3}\vec{r}$, but these are the same),

where $k=\dfrac{1}{4\pi\epsilon_0}$. As you've said, symmetry arguments guarantee that electric fields perpendicular to the arc's central axis cancel ($\vec{E}_y=0$).

The $x$-components, on the other hand, add. The $x$-component of the electric field from your single element is, like you said,

$d\vec{E}_x=d\vec{E}\cdot\cos\alpha=d\vec{E}\cdot\cos\theta$

Again, remember that your $\alpha$ is just some particular value of the variable $\theta$. This leaves your $x$-component as

$dE_x=\dfrac{k\lambda \cdot d\theta\cos\theta}{r^2}\\ =\dfrac{k(\lambda_0\cos\theta)\cos\theta\cdot d\theta}{R^2}\quad\quad\text{(r is constant)}\\ =\dfrac{k\lambda_0}{R^2}\cos^2\theta\cdot d\theta.$

And yes, your integration limits are spot on.