Gauss's Law Applications

Gauss's law states that the total electric flux through any closed surface equals the enclosed charge divided by the permittivity of free space, $\Phi =\frac{q_{enc}}{{\epsilon }_0}$, as established in NCERT Class 12, Chapter 1 (Electric Charges and Fields). The flux depends only on the enclosed charge, not on the shape of the surface, so the cube geometry is irrelevant. Here $q_{enc}=8.85\times 10^{-9}$ C and ${\epsilon }_0=8.85\times 10^{-12}$ C²/N·m². Dividing gives $\Phi =\frac{8.85\times 10^{-9}}{8.85\times 10^{-12}}=10^3=1000$ N·m²/C. The value 500 N·m²/C is wrong because it incorrectly divides the flux by two, perhaps assuming only half the surface is pierced. The value 8.85 N·m²/C is wrong because it ignores the exponent difference between charge and permittivity entirely. The value 2000 N·m²/C is wrong because it doubles the correct result without basis. Note also that any charge lying outside the surface contributes zero net flux, since its field lines enter and leave the surface an equal number of times. A magnitude and sign check confirms positive enclosed charge gives positive outward flux, and the neat cancellation of the matching $8.85$ factors in charge and permittivity yields the clean $10^3$ result.