Consider a spherically symmetric charge distribution with total charge Q enclosed by a Gaussian surface of radius r > R. What are the electric field E(r) and the electric flux Φ_E through the surface?

Consider a spherically symmetric charge distribution with total charge Q enclosed by a Gaussian surface of radius r > R. What are the electric field E(r) and the electric flux Φ_E through the surface?

• A. E(r) = (1/(4π ε0)) Q / r^2; Φ_E = Q/ε0.
• B. E(r) = (1/(4π ε0)) Q / r; Φ_E = Q.
• C. E(r) = (1/(4π ε0)) Q / r^2; Φ_E = Q/ε0.
• D. E(r) = (1/(4π ε0)) Q / r^3; Φ_E = Q/ε0.

Answer: (C)

The key idea is Gauss’s law together with symmetry. Outside a spherically symmetric charge distribution, the field behaves as if all the charge Q were a point at the center, so the electric field is radial and falls off as 1/r^2: E(r) = (1/(4π ε0)) Q / r^2. To find the flux through a spherical surface of radius r, use Φ_E = ∮ E · dA. On a sphere, E is perpendicular to the surface and has the same magnitude everywhere, so Φ_E = E(r) × (surface area) = (1/(4π ε0)) Q / r^2 × 4π r^2 = Q/ε0. Thus, for r > R, the field is E(r) = (1/(4π ε0)) Q / r^2 and the electric flux through the surface is Φ_E = Q/ε0. The flux is independent of r as long as the surface encloses the charge.

The key idea is Gauss’s law together with symmetry. Outside a spherically symmetric charge distribution, the field behaves as if all the charge Q were a point at the center, so the electric field is radial and falls off as 1/r^2: E(r) = (1/(4π ε0)) Q / r^2.

To find the flux through a spherical surface of radius r, use Φ_E = ∮ E · dA. On a sphere, E is perpendicular to the surface and has the same magnitude everywhere, so Φ_E = E(r) × (surface area) = (1/(4π ε0)) Q / r^2 × 4π r^2 = Q/ε0.

Thus, for r > R, the field is E(r) = (1/(4π ε0)) Q / r^2 and the electric flux through the surface is Φ_E = Q/ε0. The flux is independent of r as long as the surface encloses the charge.