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Question
A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.
Solution
Consider an element of angular width dθ, as shown in the figure.
\[dq = \frac{Q}{L}Rd\theta\]
The net electric field has a vertical component only.
\[\text{So}, E_{net} = \int dE \sin\theta\]
\[ = \frac{1}{4\pi \epsilon_0}\frac{Q}{L} \int\limits_0^\pi \frac{Rd\theta}{R^2}\sin\theta\]
\[= \frac{- 1}{4\pi \epsilon_0}\frac{Q}{LR} \left[ \cos\theta \right]_0^\pi \]
\[ = \frac{2Q}{4\pi \epsilon_0 LR}\]
\[ = \frac{Q}{2 \epsilon_0 L^2} \left( \because \pi R = L \right)\]
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