A semicircular arc of radius R is charged uniformly and its linear charge density is \( \lambda \). What will the electric field be at its centre?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{\lambda}{2\pi\varepsilon_0 R} \)
Divide the arc into small elements and add their fields by superposition. An element at angle \( \theta \), of length \( R\,d\theta \), carries charge \( dq = \lambda R\,d\theta \) and is at distance \( R \) from the centre.
\( \displaystyle dE = \frac{1}{4\pi\varepsilon_0}\,\frac{\lambda R\,d\theta}{R^{2}} = \frac{\lambda\,d\theta}{4\pi\varepsilon_0 R} \)
By symmetry the components along the diameter cancel in pairs, and only the components along the symmetry axis (\( dE\sin\theta \)) add up.
\( \displaystyle E = \int_{0}^{\pi}\frac{\lambda}{4\pi\varepsilon_0 R}\sin\theta\,d\theta = \frac{\lambda}{4\pi\varepsilon_0 R}\Big[-\cos\theta\Big]_{0}^{\pi} = \frac{2\lambda}{4\pi\varepsilon_0 R} \)
\( \displaystyle E = \frac{\lambda}{2\pi\varepsilon_0 R} \)
Note: A similar question was also asked in CUET (UG) 2023 on 23-06-2023 (Slot 1).