Gauss's law, fully solved

The problem

Step 1: Choose the Gaussian surface

Gauss's law is true for any closed surface:

It is only useful, though, when we can pull E out of the integral. That needs a surface where:

1. E has the same magnitude at every point, and
2. E is parallel to dA (perpendicular to the surface) everywhere.

The charge is spherically symmetric, so the field must point radially, and its size can depend only on r. A concentric sphere of radius r meets both conditions. On that sphere:

This left-hand side is the same in every part. The only thing that changes from part to part is Q_enc.

Part (a): Outside the sphere (r > R)

The Gaussian sphere encloses all of the charge, so Q_enc = Q:

Part (b): Inside the sphere (r < R)

Now the Gaussian sphere encloses only part of the charge. Because the charge is uniform, use the volume charge density:

The charge enclosed is ρ times the volume of the Gaussian sphere:

Put this into Gauss's law:

Check: at r = R both answers give kQ/R². The field is continuous at the surface, as it should be for a volume charge with no surface layer.

Part (c): Graph of E(r)

Part (d): Put in numbers

With Q = 4.0 × 10⁻⁹ C, R = 0.050 m and k = 8.99 × 10⁹ N·m²/C²:

Notice the pattern: halfway in (r = R/2), E is half its surface value, because it's linear. Twice as far out (r = 2R), E is a quarter of it, because it's inverse-square.

Part (e): Add a neutral conducting shell

Now surround the sphere with a neutral conducting shell, with inner radius 2R and outer radius 3R. Work through it region by region, always with a concentric Gaussian sphere.

Inside the metal (2R < r < 3R)

In electrostatic equilibrium, E = 0 inside a conductor. So a Gaussian sphere drawn inside the metal must enclose zero net charge:

The shell is neutral overall, so whatever isn't on the inner surface must be on the outer one:

The field in every region

• r < R: E = kQr/R³. The shell doesn't change Q_enc here.
• R < r < 2R (the gap): E = kQ/r².
• 2R < r < 3R (in the metal): E = 0.
• r > 3R: Q_enc = Q − Q + Q = Q, so E = kQ/r² again.

Common mistakes

• Using Q instead of Q_enc inside the sphere. Only the charge inside your Gaussian surface counts.
• Confusing an insulator with a conductor. A charged conducting sphere has E = 0 inside, because all its charge is on the surface. A uniformly charged insulating sphere has E ∝ r inside.
• Using r in place of R (or the reverse) in ρ. The density uses the sphere's radius R; the enclosed volume uses the Gaussian radius r.
• Thinking the shell "blocks" the field. Outside a neutral shell the field is unchanged. Only grounding the shell would remove it.