Question

A thin nonconducting ring of radius R has linear charge density $\lambda ={\lambda }_{0}cos\theta$, where ${\lambda }_0$ is a constant, $\theta$ is azimuthal angle. The electric field at the centre of the ring is $\frac{{\lambda }_0}{x{\epsilon }_{0}R}$. Find $x$ :

Answer 1

Consider two small elements of length $dl=Rd\theta$ symmetrically at angle $\theta$ on both sides as shown in the figure. Here, $d{E}_1=d{E}_2$ and sin component cancel out each other. Charge on the element , $dq=\lambda dl=\left({\lambda }_0\cos \theta \right)Rd\theta$
Net field at center due to both elements , $dE=2d{E}_1\cos \theta =2\left(\frac{dq}{4\pi {ϵ}_0{R}^2}\right)\cos \theta =\frac{{\lambda }_0}{4\pi {ϵ}_{0}R}2{\cos }^2\theta d\theta$
or $E=\frac{{\lambda }_0}{4\pi {ϵ}_{0}R}{\int }_0^{\pi }(1+\cos 2\theta )d\theta =\frac{{\lambda }_0}{4\pi {ϵ}_{0}R}\left[\right(\pi -0)+\frac{1}{2}(\sin 2\pi -\sin 0\left)\right]=\frac{{\lambda }_0}{4{ϵ}_{0}R}$
thus, $x=4$