Electrostatics Practice Test 2026 – Comprehensive Exam Preparation

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For a solid sphere with total charge Q, what is the electric field outside the sphere for r > R?

E(r) = Q r / (4π ε0 r^3)

E(r) = Q / (4π ε0 r^2)

Outside the sphere, the electric field depends only on the total charge and on the distance from the center, falling off with 1/r^2. Using a spherical Gaussian surface of radius r > R, the flux is E(r)·4πr^2. Gauss's law relates this to the enclosed charge: E(r)·4πr^2 = Q/ε0, giving E(r) = Q/(4π ε0 r^2). This is the familiar inverse-square form for a point charge located at the center, which holds here because the sphere is spherically symmetric and all the charge acts as if it’s concentrated at the center from outside.

The expression E(r) = Q r /(4π ε0 r^3) is algebraically the same as E(r) = Q/(4π ε0 r^2) for r ≠ 0, so it describes the same field. The other forms would imply a constant field with respect to r or zero field, which does not match the outside-field behavior.

E(r) = Q / (4π ε0 R^2)

E(r) = 0