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Electrostatics with Conductors

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletCavity Lab · put a charge in the hollow, ground the shell or not, and read the field a probe finds inside, in the metal and outside

In electrostatic equilibrium the field inside the conducting material is zero, which forces excess charge to the surface, makes the whole conductor one equipotential, and sets the field just outside to $\sigma/\varepsilon_0$, perpendicular to the surface. A cavity is a separate region: a charge $+q$ inside a hollow fills the hollow with field and induces $-q$ on the cavity wall, with $+q$ left on the outer surface if the conductor was neutral. Both results come from the same move, a Gaussian surface walked through the metal where $E = 0$. On a shape without spherical symmetry the surface density is not uniform; it climbs where the curvature is sharp, so points and corners carry the strongest fields.

Three errors dominate. Extending the metal's zero field into a cavity that contains charge, which erases both the cavity field and the induced wall charge. Writing $\sigma = Q/A$ on a pointed or pear-shaped conductor, which then makes the sparking at the tip unexplainable. And treating a metal enclosure as a symmetric shield: it does keep outside fields out of an empty cavity, but a charge inside still announces itself through induced charge on the outer surface, and only grounding removes that.

three regions, three different answers +q − − − − + + + + metal cavity: E is the field of +q, NOT zero wall: induced −q, by Gauss in the metal metal: E = 0, no charge in the volume outer surface: +q, and it radiates wrong picture: E = 0 everywhere inside erases the cavity field AND the wall charge ground it: only the green layer leaves; the blue and pink layers do not move
Zero field is a property of the metal, not of everything the metal surrounds. Walking a Gaussian surface through the shaded ring is what pins the wall charge at exactly minus q.
same conductor, same potential, very different sigma + + + + + + + + + + + + blunt sparks here first small radius of curvature: sigma large, E = sigma / eps0 large NOT sigma = Q / A everywhere. That is the sphere's privilege alone
Equal potential across the whole surface is what forces the pile-up: joining two spheres gives sigma proportional to one over R, so the tip wins.

The work

3 ways in · any order
Lesson
Electrostatics with Conductors ›

Pins down the one axiom (zero field in the metal) and then draws the line the errors cross: cavities, induced surface charge, one-way shielding, and why surface density peaks at a point.

Skill check · 10 scenarios
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Diagnostic
10-item topic check ›

Ten items spanning the failure modes of this topic: erasing the field in a cavity that contains charge, forgetting the induced charge on the cavity wall and the outer surface, spreading charge uniformly over a conductor that has corners, and expecting a metal box to hide an enclosed charge from the outside. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
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Targeted Practice
Drill a single misconception ›

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions
Diagnostic first