AP EAMCET · PHYSICS · Electrostatics
A wire of length \(L\) has a charge \(Q\) distributed uniformly along its length. The wire is bent in the shape of a semicircle. The magnitude of the electric field at the centre of curvature of the semicircle is
- A \(\frac{1}{4 \pi \varepsilon_0} \frac{Q}{L^2}\)
- B \(\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{L}\)
- C \(\frac{\mathrm{Q}}{2 \varepsilon_{\mathrm{o}}} \frac{1}{\mathrm{~L}^2}\)
- D \(\frac{1}{2 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{L}^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{Q}}{2 \varepsilon_{\mathrm{o}}} \frac{1}{\mathrm{~L}^2}\)
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