Physics - Electrostatics Question with Solution | TestHub

Charge per quarter arc is $Q$ over length $\frac{\pi R}{2}$, so $\lambda =\frac{2Q}{\pi R}$

Field at centre due to a quarter arc $=\frac{2k\lambda }{R}\sin 45°$ along its bisector

Horizontal components of the two halves cancel; vertical components add

$E=2\times \frac{2k\lambda }{R}{\sin }^{2}45°=\frac{2k\lambda }{R}$

Substituting: $E=\frac{2}{4\pi {\epsilon }_0}\times \frac{2Q}{\pi R}\times \frac{1}{R}=\frac{Q}{{\pi }^2{\epsilon }_0{R}^2}$

Answer: (D)