Consider a thin, uniformly charged rod 18 cm long, that is, bent into a semicircle. The total charge on the rod is 0.36 $μ C .$ The magnitude of the electric field at the centre of the semicircle is $n π \times 10^{5} N / C .$ then $n =$

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Detailed Solution

$\begin{array}{l} E = \int 2 dEcosθ = \int \frac{2 kQdθ}{R^{2}} . cosθ \\ = \int \frac{2 k}{R^{2}} (λdl) cosθ = \frac{2 kλ}{R^{2}} \int Rdθcosθ \\ = \frac{2 kλ}{R} \int_{0}^{π / 2} cosθdθ = \frac{2 kλ}{R} \\ = \frac{2 k}{R} (\frac{Q}{πR}) = \frac{2 kQ}{{πR}^{2}} = \frac{2 kQπ}{{(πR)}^{2}} = \frac{2 kQπ}{l^{2}} \\ = \frac{2 \times 9 \times 10^{9} \times 0.36 \times 10^{- 6} \times 3.14}{{(18 \times 10^{- 2})}^{2}} \\ = 0.0628 \times 10^{9 - 6 + 4} = 0.0628 \times 10^{7} \\ = 6.28 \times 10^{5} N / C \end{array}$