1.29  A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is ($\frac{\mathit{\sigma }}{2{\mathit{\epsilon }}_0}$  )  , where $\widehat{\mathit{n}}$ is the unit vector in the outward normal direction, and $\mathit{\sigma }\mathit{}$ is the surface charge density near the hole.

The following are the topics covered in this chapter: Electric Field and Field Lines, Gauss's Law, Electric Dipole, Conductors and Insulators, and Electric Flux. 

Indeed, it is an easy chapter of Class 12 Physics. In this chapter, you will learn about the foundational concepts like Gauss's law and electric fields.

Flux through the 6 sides of square (i.e. cube)

${?}_T=\frac{q_{in}}{{\epsilon }_0}=\frac{12\times 10^{-6}C}{{\epsilon }_0}$   

Flux through a square

$?=\frac{{?}_T}{6}=\frac{12\times 10^{-6}}{{\epsilon }_0\times 6}=\frac{2\times 10^{-6}}{{\epsilon }_0}$

$\Rightarrow ?=225.98\times 10^{3}N{m}^2/C?226N{m}^2/C$     

In the medical entrance test NEET, this chapter has a moderate weightage. You can expect around 2-3 questions of this chapter that contributes to the 3-5% of the total marks in the Physics section.

From Guass’s Law

${\displaystyle ?\overrightarrow{E}.\overrightarrow{dA}=\frac{q_{in}}{{\in }_0}}$            

$E.4\pi r^2=\frac{{\displaystyle \underset{0}{\overset{r}{\int }}\left(4\pi r^2.dr\right)\left(\alpha r\right)}}{{\in }_0}$            

$E=\left(\frac{\alpha r^2}{4{\in }_0}\right)$