Gauss's law in integral form for electricity is which of the following?

Gauss's law in integral form for electricity is which of the following?

• A. ∮ E · dA = Q_enclosed
• B. ∮ E · dA = ε0 Q_enclosed
• C. ∮ E · dA = Q_enclosed / ε0
• D. ∇ × E = - ∂B/∂t

Answer: (C)

The main idea is that the electric flux through any closed surface is determined by how much charge is inside that surface. Gauss's law in integral form says the outward flux of E through a closed surface, ∮ E · dA, equals the enclosed charge divided by ε0. This can be seen from ∇·E = ρ/ε0 and the divergence theorem: ∮ E · dA = ∫∫∫ (∇·E) dV = ∫∫∫ (ρ/ε0) dV = Q_enclosed/ε0. The division by ε0 makes the units consistent and makes the result depend only on the total enclosed charge, not on the shape of the surface. The option that multiplies Q_enclosed by ε0 would have incorrect dimensions, and omitting ε0 would also be incorrect. The other expression, ∇×E = -∂B/∂t, is Faraday’s law about induction, not Gauss’s law.

The main idea is that the electric flux through any closed surface is determined by how much charge is inside that surface. Gauss's law in integral form says the outward flux of E through a closed surface, ∮ E · dA, equals the enclosed charge divided by ε0. This can be seen from ∇·E = ρ/ε0 and the divergence theorem: ∮ E · dA = ∫∫∫ (∇·E) dV = ∫∫∫ (ρ/ε0) dV = Q_enclosed/ε0. The division by ε0 makes the units consistent and makes the result depend only on the total enclosed charge, not on the shape of the surface. The option that multiplies Q_enclosed by ε0 would have incorrect dimensions, and omitting ε0 would also be incorrect. The other expression, ∇×E = -∂B/∂t, is Faraday’s law about induction, not Gauss’s law.