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 ? $-\dfrac{\theta}{2} \to +\dfrac{\theta}{2}$?

  • $\begingroup$ Because you are using a cosine function the values of your limits work. (The function varies between 0 and 1 between those degrees) $\endgroup$ – 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,


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.