A point charge of $2.0 \, \mu ext{C}$ is at the center of a cubic Gaussian surface $9.0 \, ext{cm}$ on edge

Question

A point charge of $2.0 \, \mu	ext{C}$ is at the center of a cubic Gaussian surface $9.0 \, 	ext{cm}$ on edge. What is the net electric flux through the surface?

Accepted Answer

According to Gauss's Law, the net electric flux $\Phi_E$ through a closed surface depends only on the net charge enclosed ($q$) and the permittivity of free space ($\varepsilon_0$), independent of the size or shape of the surface. 
Formula: $\Phi_E = \frac{q}{\varepsilon_0}$ 
Given: 
$q = 2.0 \, \mu	ext{C} = 2.0 	imes 10^{-6} \, 	ext{C}$ 
$\varepsilon_0 \approx 8.854 	imes 10^{-12} \, 	ext{C}^2 	ext{N}^{-1} 	ext{m}^{-2}$ 
Calculation: 
$\Phi_E = \frac{2.0 	imes 10^{-6}}{8.854 	imes 10^{-12}} \approx 0.2258 	imes 10^6 = 2.26 	imes 10^5 \, 	ext{N m}^2 	ext{C}^{-1}$. 
(See NCERT Physics Class 12, Exercise 1.18).

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