The Electric Field Due to a Line of Charge — Question 7

Question 7

A fixed-charge arc has an angular cancellation factor

Positive charge Q is uniformly spread over a circular arc of radius R, spanning $-\alpha\le\phi\le\alpha$ around the positive x axis, with $0<\alpha\le\pi$.

1. Find the center field in terms of Q, R and $\alpha$.

2. Examine the limits $\alpha\to0$ and $\alpha\to\pi$, and explain why holding Q fixed matters.

Question 7 - Solution

A fixed-charge arc has an angular cancellation factor

(a) Method and calculation

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

(b) Interpretation and further calculation

$$\alpha\to0:\quad E_x\to-kQ/R^2;\qquad\alpha\to\pi:\quad E_x\to0.$$$$\sin\alpha/\alpha\to1\quad\text{as }\alpha\to0.$$With fixed total Q, a narrow arc approaches a point charge at distance R. Holding density fixed instead would shrink its total charge to zero.

Physical check

The angular parameter is measured in radians in the line-density formula.