A hollow conducting shell carries total charge Q. What is the field in the region outside the shell (r > R)?

A hollow conducting shell carries total charge Q. What is the field in the region outside the shell (r > R)?

• A. E outside is that of a point charge Q located at the center: E = (1/(4π ε0)) Q / r^2.
• B. E outside is zero.
• C. E outside is a uniform field.
• D. E outside depends on the internal charge distribution and is not simply described.

Answer: (A)

When a conductor carries charge, all the excess charge resides on its outer surface, and the field inside the conductor is zero. For the region outside a hollow conducting shell, the shell looks like a single point charge Q located at the center because of spherical symmetry. To find the field, use a spherical Gaussian surface of radius r larger than the shell’s outer radius. The electric field on that surface is purely radial and has the same magnitude at all points, so the flux is E(r) times the surface area 4πr^2. Gauss’s law then gives E(r) · 4πr^2 = Q/ε0, so E(r) = (1/4π ε0) Q / r^2, directed radially outward for Q > 0.

When a conductor carries charge, all the excess charge resides on its outer surface, and the field inside the conductor is zero. For the region outside a hollow conducting shell, the shell looks like a single point charge Q located at the center because of spherical symmetry. To find the field, use a spherical Gaussian surface of radius r larger than the shell’s outer radius. The electric field on that surface is purely radial and has the same magnitude at all points, so the flux is E(r) times the surface area 4πr^2. Gauss’s law then gives E(r) · 4πr^2 = Q/ε0, so E(r) = (1/4π ε0) Q / r^2, directed radially outward for Q > 0.