Physics - Electrostatics Question with Solution | TestHub

Inradius $\rho =\frac{\ell }{2\sqrt{3}}$; height $h=\frac{\ell }{2\sqrt{6}}$

The triangular sheet subtends solid angle

$\Omega =6{\int }_0^{\pi /3}\left(1-\frac{h\cos \varphi }{\sqrt{{\rho }^2+h^2{\cos }^2\varphi }}\right)d\varphi$

${\int }_0^{\pi /3}\frac{h\cos \varphi d\varphi }{\sqrt{{\rho }^2+h^2{\cos }^2\varphi }}={\sin }^{-1}\left(\frac{h\sin (\pi /3)}{\sqrt{{\rho }^2+h^2}}\right)=\frac{\pi }{6}$

$\Omega =6\left(\frac{\pi }{3}-\frac{\pi }{6}\right)=\pi$

$E=\frac{\sigma \Omega }{4\pi {\epsilon }_0}=\frac{\sigma }{4{\epsilon }_0}$. Therefore $\alpha =4$.