Chapter 21 Problem 49

Problem

A thin rod bent into the shape of an arc of a circle of radius $R$ carries a uniform charge per unit length $\lambda$ The arc subtends a total angle $2\theta _{0}$ symmetric about the $x$ axis. Determine the electric field ${\vec {E}}$ at the origin $0$

Solution

Select a differential element of the arc which makes an angle of $\theta$ with the x axis. The length of this element is $Rd\theta$, and the charge on that element is $dq=\lambda Rd\theta$. This element generates an electric field

$dE={\frac {1}{4\pi \epsilon _{0}}}{\frac {\lambda Rd\theta }{R^{2}}}$

The arc is symmetric about the x axis, therefore the total field will only have x component (the vertical components will cancel out)

$dE_{\text{hor.}}={\frac {1}{4\pi \epsilon _{0}}}{\frac {\lambda Rd\theta }{R^{2}}}\cos \theta$

$E_{\text{hor.}}=\int _{-\theta _{0}}^{\theta _{0}}{\frac {1}{4\pi \epsilon _{0}}}\cos \theta {\frac {\lambda Rd\theta }{R^{2}}}\cos \theta$ $={\frac {\lambda }{4\pi \epsilon _{0}R}}\int _{-\theta _{0}}^{\theta _{0}}\cos \theta d\theta$ $={\frac {\lambda }{4\pi \epsilon _{0}R}}\left[\sin \theta _{0}-\sin(-\theta _{0})\right]$ $={\frac {2\lambda \sin \theta _{0}}{4\pi \epsilon _{0}R}}$

The field points in the negative x direction

${\vec {E}}=-{\frac {2\lambda \sin \theta _{0}}{4\pi \epsilon _{0}R}}{\hat {i}}$