If I have a semi-circular ring of charges (charges uniformly distributed), centred at the origin of the $x-y$ plane, with radius $r$, I can easily find that the electric field at the origin is

$$-\frac{2kQ}{\pi r^2} $$ in the $y$-direction (not sure how to illustrate y-hat using TeX).

However, I am having a lot of difficulty working out the electric field on the $x$-axis for any point inside the ring, i.e. for $-r<x<r$.

Is this a complicated generalisation? How do I tackle it? (NB: I prefer a vector-based approach, rather than a magnitude based approach.)


1 Answer 1


Yes it is a complicated generalization.

The electric field at a point $\mathbf r$ is

$$ \mathbf E(\mathbf r) = k\int \frac{\mathbf r - \mathbf r'}{|\mathbf r - \mathbf r'|^3} dq'. $$

For the problem you're attempting to solve, let $R$ be the radius of the ring to avoid notational confusion with other "r" variables, then

$$ \mathbf r = (x,0,0), \qquad \mathbf r'= (R\cos\theta', R\sin\theta', 0). $$

It follows that

\begin{align} \frac{\mathbf r - \mathbf r'}{|\mathbf r - \mathbf r'|^3} = \frac{(x - R\cos\theta', -R\sin\theta', 0)}{((x-R\cos\theta')^2 + R^2\sin^2\theta')^{3/2}}. \end{align}

For the problem at hand, the charge measure $dq'$ is

$$ dq' = \frac{Q}{\pi R}(Rd\theta') = \frac{Q}{\pi}d\theta'. $$

Plugging these in reveals that to compute the field at a given $x\neq 0$, we'd need to compute the following integral:

$$ k\int_{-\pi}^0 \frac{(x - R\cos\theta', -R\sin\theta', 0)}{((x-R\cos\theta')^2 + R^2\sin^2\theta')^{3/2}} \frac{Q}{\pi}d\theta'. $$

This is a hard integral compared to the case in which $x=0$ because in that case it collapses to

$$ -\frac{kQ}{\pi R^2}\int_{-\pi}^0 (\cos\theta', \sin\theta', 0)d\theta' = \left(0, -\frac{2kQ}{\pi R^2}, 0\right). $$


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