The electric field at the centre of a semi-circular arc of radius 'r', which is charged uniformly and has linear charge density \( \lambda \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{\lambda}{2\pi\varepsilon_0 r} \)
Divide the arc into small elements and add their fields at the centre using Coulomb's law. An element at angle \( \theta \) has length \( r\,d\theta \) and charge \( \lambda r\,d\theta \).
\( \displaystyle dE = \frac{1}{4\pi\varepsilon_0}\,\frac{\lambda r\,d\theta}{r^{2}} = \frac{\lambda\,d\theta}{4\pi\varepsilon_0\,r} \)
By symmetry, the components along the diameter cancel in pairs; only the components along the line of symmetry, \( dE\sin\theta \), add up.
\( \displaystyle E = \int_{0}^{\pi} \frac{\lambda \sin\theta\,d\theta}{4\pi\varepsilon_0\,r} = \frac{\lambda}{4\pi\varepsilon_0\,r}\,\Big[-\cos\theta\Big]_{0}^{\pi} = \frac{2\lambda}{4\pi\varepsilon_0\,r} \)
\( \displaystyle E = \frac{\lambda}{2\pi\varepsilon_0\,r} \)
Shortcut: The answer must have the units of \( \lambda/(\varepsilon_0 r) \), which is the form of a field (charge per unit length over \( \varepsilon_0 \) times a length). Option (2) has \( r^2 \) and Option (3) has \( \lambda^2 \), so both are dimensionally wrong; Option (1) leaves out the factor 2 from the integral.
Note: A similar question was also asked in CUET (UG) 2025 on 21-05-2025 (Slot 2).