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A point charge of 2.0μC is at the centre of a cubic Gaussian surface 9.0cm on edge. What is the net electric flux through the surface?

Answer
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Hint: Since we are given a Gaussian surface here, so we must use Gauss’ law to calculate the net electric flux through the surface.

Formula used
SEdS=qε0ϕ=qε0
Where E is the electric field due to a point charge, qis the charge enclosed by the surface S,ϕ is the electric flux flowing through the surface and ε0 is the permittivity of free space.

Complete step by step answer
Gauss’ Law states that the total electric flux over any closed surface is equal to 1ε0times the total charge enclosed by the surface, where ε0 is the permittivity of free space.
In this question, we are told that a charge of 2.0μC is enclosed by a cubic Gaussian surface. According to the law, no matter what the shape of the surface is, but if a charge is enclosed by any closed surface, then the flux flowing through the surface must be 1ε0 times the charged enclosed.
Therefore, according to gauss’ law,
SEdS=qε0ϕ=qε0
Where ϕ is the total electric flux flowing through the Gaussian surface.
We know ε0=8.85×1012N1m2C2
Substituting this value, we get,
ϕ=2×1068.85×1012Nm2C1
ϕ=2.26×105Nm2C1

Therefore, the net electric flux flowing through the surface is 2.26×105Nm2C1

Note: The equation used above is the integral form of the Gauss’ law.
If we use divergence theorem on both the sides of the equation, we will get its differential form given by the expression, E=ρε0where ρ is defined as the charge per unit volume. This is one of the four Maxwell’s laws of electromagnetic induction.
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A point charge of 2.0μC is at the centre of a cubic Gaussian surface 9.0cm on edge. What is the net electric flux through the surface?

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Electric Charges & Fields Class 12 Physics - NCERT EXERCISE 1.18 | Physics NCERT | Vishal Kumar Sir
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