A thread carrying a uniform charge $\mathrm{\lambda }$ per unit length has the configurations shown in a) and (b). Assuming a curvature radius R to be considerably less than length of the thread, find the magnitude of the electric field strength at the point O.

a) i) ${\mathrm{E}}_1={\mathrm{E}}_1\hat{\mathrm{i}}-{\mathrm{E}}_1\hat{\mathrm{j}}$

$\Rightarrow {\mathrm{E}}_1=\frac{\mathrm{\lambda }}{4\mathrm{\pi }{\in }_0\mathrm{R}}\hat{\mathrm{i}}-\frac{\mathrm{\lambda }}{4\mathrm{\pi }{\in }_0\mathrm{R}}\hat{\mathrm{j}}$

ii) A circular arc which is making angle $\mathrm{\pi }$/ 2 at the centre of curvature ‘o’

${\mathrm{E}}_2=\frac{\mathrm{q}}{2{\mathrm{\pi ϵ}}_0{\mathrm{\varphi R}}^2}\sin \frac{0}{2}$

$\begin{array}{l}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi E}}_0\mathrm{R}}\sin \mathrm{\pi }/4(\because \mathrm{\varphi }=\mathrm{\pi }/2)\\ \overrightarrow{{\mathrm{E}}_2}={\mathrm{E}}_2\cos 45\widehat{\mathrm{i}}+{\mathrm{E}}_2\sin 45\widehat{\mathrm{j}}\\ \Rightarrow \overrightarrow{{\mathrm{E}}_2}=\frac{\mathrm{\lambda }}{4{\mathrm{\pi E}}_0\mathrm{R}}\widehat{\mathrm{i}}+\frac{\mathrm{\lambda }}{4{\mathrm{\pi E}}_0\mathrm{R}}\widehat{\mathrm{j}}\end{array}$

iii) $\overrightarrow{{\mathrm{E}}_3}=-{\mathrm{E}}_{\parallel }\hat{\mathrm{i}}+{\mathrm{E}}_{\perp }\hat{\mathrm{j}}$

$\Rightarrow \overrightarrow{{\mathrm{E}}_3}=\frac{-\mathrm{\lambda }}{4{\mathrm{\pi ϵ}}_0\mathrm{R}}\hat{\mathrm{i}}+\frac{\mathrm{\lambda }}{4{\mathrm{\pi ϵ}}_0\mathrm{R}}\hat{\mathrm{j}}$

net electric field at point ‘o’ is

$\overrightarrow{\mathrm{E}}=\overrightarrow{{\mathrm{E}}_1}+\overrightarrow{{\mathrm{E}}_2}+\overrightarrow{{\mathrm{E}}_3}\Rightarrow \overrightarrow{\mathrm{E}}=\frac{\sqrt{2}{\displaystyle \mathrm{\lambda }}}{4\mathrm{\pi }{\in }_0\mathrm{R}}$

b) i) 

Electric field at point ‘o’

${\mathrm{E}}_1={\mathrm{E}}_{\perp }\hat{\mathrm{i}}-{\mathrm{E}}_{\Vert }\hat{\mathrm{j}}=\frac{\lambda }{4\pi {ϵ}_0\mathrm{R}}\hat{\mathrm{i}}-\frac{\lambda }{4\pi {ϵ}_0\mathrm{R}}\hat{\mathrm{j}}$

ii) Electric field at circular arc

$\begin{array}{l}{\overrightarrow{\mathrm{E}}}_2=\frac{\mathrm{\lambda }}{2{\mathrm{\pi E}}_0\mathrm{R}}\sin \frac{\mathrm{\theta }}{2}\\ \Rightarrow \overrightarrow{{\mathrm{E}}_2}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }{\in }_0\mathrm{R}}\\ \Rightarrow \overrightarrow{{\mathrm{E}}_2}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }{\in }_0\mathrm{R}}\widehat{\mathrm{j}}(\because \mathrm{\theta }=\mathrm{\pi })\end{array}$

iii) $\begin{array}{r}\overrightarrow{{\mathrm{E}}_3}=\frac{-\mathrm{\lambda }}{4{\mathrm{\pi ϵ}}_0\mathrm{R}}\widehat{\mathrm{i}}-\frac{\mathrm{\lambda }}{4{\mathrm{\pi ϵ}}_0\mathrm{R}}\widehat{\mathrm{j}}\\ \overrightarrow{\mathrm{E}}=\overrightarrow{{\mathrm{E}}_1}+\overrightarrow{{\mathrm{E}}_2}+\overrightarrow{{\mathrm{E}}_3}\Rightarrow \overrightarrow{\mathrm{E}}=0\end{array}$