For a uniformly charged solid sphere with total charge Q and radius R, what is the electric field inside (r < R) and outside (r > R)?

For a uniformly charged solid sphere with total charge Q and radius R, what is the electric field inside (r < R) and outside (r > R)?

• A. Inside: E = k Q / R^3; Outside: E = k Q / r^2
• B. Inside: E = k Q r / R^2; Outside: E = k Q / r^2
• C. Inside: E = k Q r / R^3; Outside: E = k Q / r^2
• D. Inside: E = 0; Outside: E = k Q / r^2

Answer: (C)

This uses Gauss’s law with a uniformly charged solid sphere. The key is that the charge is spread throughout the volume, so the field inside isn’t zero the way it would be for a conductor, and the amount of charge enclosed by a Gaussian surface of radius r (< R) grows with r. First find the volume charge density: ρ = Q / (4/3 π R^3) = 3Q / (4π R^3). For a Gaussian surface of radius r inside the sphere, the enclosed charge is q_enclosed = ρ × (4/3)π r^3 = Q (r^3 / R^3). Gauss’s law says E (4π r^2) = q_enclosed / ε0, so E = (1 / (4π ε0)) × [Q (r^3 / R^3)] / r^2 = k Q r / R^3, where k = 1/(4π ε0). Outside the sphere (r > R), all the charge is enclosed, so E = k Q / r^2. Thus inside: E = k Q r / R^3; outside: E = k Q / r^2. The inside field grows linearly with r and matches the boundary value at r = R, ensuring continuity.

This uses Gauss’s law with a uniformly charged solid sphere. The key is that the charge is spread throughout the volume, so the field inside isn’t zero the way it would be for a conductor, and the amount of charge enclosed by a Gaussian surface of radius r (< R) grows with r.

First find the volume charge density: ρ = Q / (4/3 π R^3) = 3Q / (4π R^3).

For a Gaussian surface of radius r inside the sphere, the enclosed charge is q_enclosed = ρ × (4/3)π r^3 = Q (r^3 / R^3).

Gauss’s law says E (4π r^2) = q_enclosed / ε0, so E = (1 / (4π ε0)) × [Q (r^3 / R^3)] / r^2 = k Q r / R^3, where k = 1/(4π ε0).

Outside the sphere (r > R), all the charge is enclosed, so E = k Q / r^2.

Thus inside: E = k Q r / R^3; outside: E = k Q / r^2. The inside field grows linearly with r and matches the boundary value at r = R, ensuring continuity.