Consider a square frame made of uniformly charged non conducting wires of length $\mathcal{l}$ and linear charge density $\lambda C/m.$

P is a point at a perpendicular distance of $\mathcal{l}$ from centre of a square. Choose the CORRECT statement(s) 

 $r=\frac{\sqrt{5}\mathcal{l}}{2},{\theta }_1={\theta }_2$
 
$\sin {\theta }_1=\sin {\theta }_2=\left(\frac{1}{\sqrt{6}}\right)$ 
$\tan \alpha =\frac{1}{2},\cos \alpha =\left(\frac{2}{\sqrt{5}}\right)$ 
$E_{\perp }=\frac{K\lambda }{r}(\sin {\theta }_1+\sin {\theta }_2)$ 
$E_{\left|\right|}=\frac{K\lambda }{r}(\cos {\theta }_1-\cos {\theta }_2)$ 
$E_{\left|\right|}=0$ 
$dV=\frac{Kdq}{\sqrt{r^2+x^2}}$ 
$dV=\frac{K\lambda dx}{\sqrt{r^2}+x^2}$ 
 
$\tan \theta =\left(\frac{x}{r}\right)$ 
$x=r\tan \theta$ 
$\frac{dx}{d\theta }-r{\sec }^2\theta$ 
$dV=K\lambda \underset{-{\theta }_2}{\overset{{\theta }_1}{\int }}\frac{r{\sec }^2\theta d\theta }{r\sec \theta }=K\lambda \underset{{\theta }_2}{\overset{{\theta }_1}{\int }}\sec \theta d\theta$ 
$=K\lambda {\log }_e(\sec \theta +\tan \theta ){|}_{-{\theta }_2}^{{\theta }_1}$ 
$V=K\lambda {\log }_e\left(\frac{\sec \theta +\tan \theta }{\sec \theta -\tan \theta }\right)$ 
$E_{\perp }=\frac{K}{\frac{\sqrt{5}\mathcal{l}}{2}}\left(2.\frac{1}{\sqrt{6}}\right)=\frac{4K\lambda }{\sqrt{30}\mathcal{l}}$ 
$E_0=4E_{\left|\right|}\cos \alpha =4\left(\frac{4K\lambda }{\sqrt{30}\mathcal{l}}\right)\left(\frac{2}{\sqrt{5}}\right)=\left[\frac{32K\lambda }{\sqrt{150}\mathcal{l}}\right]$