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

  $\frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{L}$
• Correct answer
  C

  $\frac{\mathrm{Q}}{2 \varepsilon_{\mathrm{o}}} \frac{1}{\mathrm{~L}^2}$
• D

  $\frac{1}{2 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{L}^2}$

Solution

We have $$ \begin{aligned} & \mathrm{dQ}=\lambda \mathrm{dl} \\ & =\lambda \mathrm{Rd} \theta \\ & \text { So, } \mathrm{dE}=\frac{1}{4 \pi \varepsilon_0} \frac{\lambda \mathrm{Rd} \theta}{\mathrm{R}^2} \\ & =\frac{1}{4 \pi \varepsilon_0} \frac{\lambda \mathrm{d} \theta}{\mathrm{R}} \end{aligned} $$ So, $\mathrm{E}=\int \mathrm{dE} \cos \theta=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda}{\mathrm{R}} \int_{-\pi / 2}^{+\pi / 2} \cos \theta \mathrm{d} \theta$ $$ =\frac{2 \lambda}{4 \pi \varepsilon_0 \mathrm{R}}=\frac{2}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{Q} / \mathrm{L}}{\mathrm{L} / \pi}=\frac{\mathrm{Q}}{2 \varepsilon_0} \frac{1}{\mathrm{~L}_2} $$ $$ \left[\because \mathrm{L}=\pi \mathrm{R} \text { and } \lambda=\frac{\mathrm{Q}}{\mathrm{L}}\right] $$

Hence, the correct answer is option: 'C'.