Question

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. $\frac{\lambda^{2}}{2\pi\varepsilon_{0}a}$
• D. $\frac{\lambda}{2\pi\varepsilon_{0}a}$

Answer 1

Answer: (D)

Electric field intensity at $O$ due to small elemental length $d l$ of charged ring,
$d E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\lambda d l}{a^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\lambda a d \theta}{a^{2}}$
$d E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\lambda}{a} d \theta$
$\therefore $ Net electric field at centre $O$ is
$E=\int d E \sin \theta =\int_\limits{0}^{\pi} \frac{1}{4 \pi \varepsilon_{0}} \frac{\lambda}{a} \sin \theta d \theta$
$=\frac{\lambda}{4 \pi \varepsilon_{0} a}[-\cos \theta]_{0}^{\pi}$
$\therefore E=\frac{\lambda}{2 \pi \varepsilon_{0} a}$