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AIIMS 2008
Exam
AIIMS
Year
2008
Subject
Physics
Chapter
Electrostatics

AIIMS 2008 · Physics · Electrostatics

Electric field at the centre of a uniformly charged semicircle of...

Electric field at the centre of a uniformly charged semicircle of radius aa is Electric field at the centre of a uniformly charged semicircle of radius a is

Options

  1. A(\frac{\lambda}{2 \pi \varepsilon_0 a^2})
  2. B(\frac{\lambda}{4 \pi^2 \varepsilon_0 a^2})
  3. C(\frac{\lambda^2}{2 \pi \varepsilon_0 a})
  4. D(\frac{\lambda}{2 \pi \varepsilon_0 a})

Correct answer

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

Step-by-step solution

Electric field at the centre of a uniformly charged semicircle of radius a is
Electric field intensity at OO due to small elemental length dld l of charged ring, dE=14πε0⋅λdla2=14πε0⋅λadθa2dE=14πε0⋅λadθ\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}\thereforeNetelectricfieldatcentreNet electric field at centreOisisE=∫dEsin⁡θ=∫0π14πε0λasin⁡θdθ=λ4πε0a[−cos⁡θ]0π∴E=λ2πε0a.\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}$
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