Physics - Electrostatics Question with Solution | TestHub

A semi-infinite wire gives $\frac{k\lambda }{R}$ along and $\frac{k\lambda }{R}$ perpendicular to itself at C.

Horizontal wire: ${\vec{E}}_1=\frac{k\lambda }{R}(-\hat{ı}-\hat{ȷ})$; vertical wire: ${\vec{E}}_2=\frac{k\lambda }{R}(\hat{ı}+\hat{ȷ})$. They cancel.

Quarter arc (second quadrant) at its centre: ${\vec{E}}_{arc}=\frac{k\lambda }{R}(\hat{ı}-\hat{ȷ})$, pointing away from the arc along its bisector.

Total: $\vec{E}=\frac{k\lambda }{R}\hat{ı}-\frac{k\lambda }{R}\hat{ȷ}$. Answer: C