The Electric Field Due to a Line of Charge — Question 6

Question 6

A charged semicircle gives a finite center field despite equal source distances

A thin uniformly charged upper semicircle has radius R and total charge $Q>0$. Parameterize its points by $(R\cos\phi,R\sin\phi)$ with $0\le\phi\le\pi$.

1. Integrate the center field using $dq=\lambda R\,d\phi$.

2. Express the result in terms of Q and compare with the field of a complete uniformly charged circle carrying the same Q.

Question 6 - Solution

A charged semicircle gives a finite center field despite equal source distances

(a) Method and calculation

$$\lambda=Q/(\pi R),\quad d\mathbf E=-\frac{k\lambda}{R}(\cos\phi,\sin\phi)\,d\phi.$$$$E_x=0,\quad E_y=-\frac{k\lambda}{R}\int_0^\pi\sin\phi\,d\phi=-\frac{2kQ}{\pi R^2}.$$

(b) Interpretation and further calculation

$$\text{full circle}:\quad\int_0^{2\pi}\cos\phi\,d\phi=\int_0^{2\pi}\sin\phi\,d\phi=0.$$$$\mathbf E_{\mathrm{full}}(0)=\mathbf0.$$Equal distances set equal weighting per charge element, but vector cancellation depends on the angular distribution.

Physical check

The semicircle field points downward, away from the positively charged upper arc.