Electric field at the centre of a uniformly charged semicircle of radius \(a\) is

• A

  \(\frac{\lambda}{2 \pi \varepsilon_0 a^2}\)
• B

  \(\frac{\lambda}{4 \pi^2 \varepsilon_0 a^2}\)
• C

  \(\frac{\lambda^2}{2 \pi \varepsilon_0 a}\)
• Correct answer
  D

  \(\frac{\lambda}{2 \pi \varepsilon_0 a}\)

Solution

Electric field intensity at $O$ due to small elemental length $d l$ of charged ring, $\begin{aligned} & d E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda d l}{a^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda a d \theta}{a^2} \\ & d E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda}{a} d \theta\end{aligned}$ $\therefore$ Net electric field at centre $O$ is $\begin{aligned} E=\int d E \sin \theta & =\int_0^\pi \frac{1}{4 \pi \varepsilon_0} \frac{\lambda}{a} \sin \theta d \theta \\ & =\frac{\lambda}{4 \pi \varepsilon_0 a}[-\cos \theta]_0^\pi \\ \therefore \quad E & =\frac{\lambda}{2 \pi \varepsilon_0 a} .\end{aligned}$

Hence, the correct answer is option: 'D'.