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Question: A point charge \[4\,\mu{\text{ C}}\] is at the centre of a cubic Gaussian surface \[10\,{\text{cm}}\...

A point charge 4 μ C4\,\mu{\text{ C}} is at the centre of a cubic Gaussian surface 10 cm10\,{\text{cm}} on edge. Net electric flux through the surface is
A. 2.5×105 N⋅m2⋅C - 12.5 \times {10^5}\,{\text{N}} \cdot {{\text{m}}^2} \cdot {{\text{C}}^{{\text{ - 1}}}}
B. 4.5×105 N⋅m2⋅C - 14.5 \times {10^{5\,}}{\text{N}} \cdot {{\text{m}}^2} \cdot {{\text{C}}^{{\text{ - 1}}}}
C. 4.5×106 N⋅m2⋅C - 14.5 \times {10^6}\,{\text{N}} \cdot {{\text{m}}^2} \cdot {{\text{C}}^{{\text{ - 1}}}}
D. 2.5×106 N⋅m2⋅C - 12.5 \times {10^6}\,{\text{N}} \cdot {{\text{m}}^2} \cdot {{\text{C}}^{{\text{ - 1}}}}

Explanation

Solution

Use the formula for the net electric flux of the Gaussian surface. This equation gives the relation between the net electric flux of the Gaussian surface, net electric charge enclosed in the surface and the permittivity of the medium.

Formula used:
The net flux ϕnet{\phi _{net}} through a Gaussian surface is
ϕnet=qencε0{\phi _{net}} = \dfrac{{{q_{enc}}}}{{{\varepsilon _0}}} …… (1)
Here, qq is the net charge enclosed in the Gaussian surface and ε0{\varepsilon _0} is the permittivity of the medium.

Complete step by step answer:
A point charge 4 μ C4\,\mu{\text{ C}} is at the centre of a cubic Gaussian surface 10 cm10\,{\text{cm}} on edge.
The value of the permittivity ε0{\varepsilon _0} of the medium is 8.85×10−12 C2/N⋅m28.85 \times {10^{ - 12}}\,{{\text{C}}^{\text{2}}}{\text{/N}} \cdot {{\text{m}}^{\text{2}}}.
ε0=8.85×10−12 C2/N⋅m2{\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{{\text{C}}^{\text{2}}}{\text{/N}} \cdot {{\text{m}}^{\text{2}}}
Convert the unit of the charge qenc{q_{enc}} in the Gaussian surface from microcoulomb to coulomb.
qenc=(4 μ C)(10−6 C1 μ C){q_{enc}} = \left( {4\,\mu{\text{ C}}} \right)\left( {\dfrac{{{{10}^{ - 6}}\,{\text{C}}}}{{1\,\mu{\text{ C}}}}} \right)
⇒qenc=4×10−6 C\Rightarrow {q_{enc}} = 4 \times {10^{ - 6}}\,{\text{C}}
Hence, the charge enclosed in the Gaussian surface is 4×10−6 C4 \times {10^{ - 6}}\,{\text{C}}.
Calculate the net electric flux through the Gaussian surface.
Substitute 4×10−6 C4 \times {10^{ - 6}}\,{\text{C}} for qenc{q_{enc}} and 8.85×10−12 C2/N⋅m28.85 \times {10^{ - 12}}\,{{\text{C}}^{\text{2}}}{\text{/N}} \cdot {{\text{m}}^{\text{2}}} for ε0{\varepsilon _0} in equation (1).
ϕnet=4×10−6 C8.85×10−12 C2/N⋅m2{\phi _{net}} = \dfrac{{4 \times {{10}^{ - 6}}\,{\text{C}}}}{{8.85 \times {{10}^{ - 12}}\,{{\text{C}}^{\text{2}}}{\text{/N}} \cdot {{\text{m}}^{\text{2}}}}}
⇒ϕnet=4.5×105 N⋅m2⋅C - 1\Rightarrow {\phi _{net}} = 4.5 \times {10^{5\,}}{\text{N}} \cdot {{\text{m}}^2} \cdot {{\text{C}}^{{\text{ - 1}}}}
Therefore, the net electric flux through the Gaussian surface is 4.5×105 N⋅m2⋅C - 14.5 \times {10^{5\,}}{\text{N}} \cdot {{\text{m}}^2} \cdot {{\text{C}}^{{\text{ - 1}}}}.

So, the correct answer is “Option B”.

Additional Information:
The net electric flux of an enclosed surface is given by Gauss's law.
According to Gauss's law, the net electric flux of any Gaussian surface (closed surface) is the ratio of the net electric charge enclosed in that surface and the permittivity of the medium.

Note:
Convert the unit of the net electric charge enclosed in the Gaussian surface in SI system of units.
If the charge is not uniformly distributed in the Gaussian surface and is discrete then the net electric charge is taken as the sum of all the electric charges.