AP EAPCET 06-Jul-2022 Shift 1 Paper

The given situation is shown below.On straight wire $\lambda =$ charge per unit length $=\frac{Q}{L}$ small charge on length element $d l$ is$dq=\lambda \times dl$ .....(i) Electric field at centre due to a point charge$=K \times \frac{dq}{r^2}$ When each electric line is broken into 2 components, horizontal component of electric field balance each other, only vertical components remain.So, net electric field along vertical $dE_{\text{vertical}} = 2 dE \cos\theta = 2 \frac{K dq}{r^2} \cos\theta$$= \frac{2 K \cos\theta}{r^2} \times \lambda dl$( But $d\theta = \frac{dl}{r} \Rightarrow dl = r d\theta$)Integrating the electric field,$E = \int\limits_{0}^{\frac{\pi}{2}} \frac{2 K \lambda \cos\theta d\theta}{r}$but length of semi-circular path is$L = \pi r$$\Rightarrow r = \frac{L}{\pi}$$\therefore E = \frac{2 K Q}{L \times \frac{L}{\pi}} = \frac{2 K \pi Q}{L^2}$$\therefore E = \frac{2}{4 \pi \varepsilon_0} \times \frac{\pi Q}{L^2} = \frac{Q}{2 \varepsilon_0 L^2} .$