MHT CET · Physics · Electrostatics
A uniformly charged semicircular arc of radius ' \(r\) ' has linear charge density \((\lambda)\). What is the electric field at its centre?
( \(\epsilon_0=\) permittivity of free space)
- A \(\frac{\lambda}{2 \pi \in_0 r}\)
- B \(\frac{2 \pi \epsilon_0}{\lambda}\)
- C \(\frac{\lambda}{4 \epsilon_0}\)
- D \(\frac{2 \epsilon_0}{\lambda}\)
Answer & Solution
Correct Answer
(A) \(\frac{\lambda}{2 \pi \in_0 r}\)
Step-by-step Solution
Detailed explanation
Electric field due to a circular arc at its center,
\(\mathrm{E}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}} \sin \alpha\), where \(\alpha\) is the angle which arc subtends at its center
For semicircle, \(\alpha=90^{\circ}\)
Hence, \(E=\frac{\lambda}{2 \pi \epsilon_0 r}\)
\(\mathrm{E}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}} \sin \alpha\), where \(\alpha\) is the angle which arc subtends at its center
For semicircle, \(\alpha=90^{\circ}\)
Hence, \(E=\frac{\lambda}{2 \pi \epsilon_0 r}\)
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