Mistake Master
Student view — seeing the site as a student does
CED objectives

Gauss's Law

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletGauss Lab · swap shell for solid sphere and watch the interior field change its mind

Gauss's law states that the net flux through any closed surface equals the enclosed net charge over $\varepsilon_0$, where the field being integrated is the total field, outside charges included. It yields a value of $E$ only when symmetry makes $E$ uniform and normal on the flux-carrying part of the surface: a concentric sphere for spherical charge, a coaxial cylinder for a line, a pillbox for a sheet. Inside a uniform shell $E = 0$; inside a uniform solid sphere $E = kQr/R^3$; an infinite line gives $2k\lambda/r$; an infinite sheet gives $\sigma/2\varepsilon_0$ at any distance, and a conductor's surface gives $\sigma/\varepsilon_0$.

Five errors recur. Erasing an outside charge from the field on the surface because it does not count in the flux. Factoring $E$ out of the integral over a cube face or around a dipole, where $E$ is neither uniform nor normal. Swapping the shell's zero interior for the solid sphere's linear rise, or using total $Q$ inside the solid sphere. Treating the imagined surface as a physical object, so that a bigger sphere seems to change the field. And making the infinite sheet's field fade with distance, or handing a free-standing sheet the conductor's $\sigma/\varepsilon_0$.

E versus r, same total charge Q, radius R E r R 2R kQ / R² kQ / 4R² shell: E = 0 inside solid: E = kQr / R³ outside R the two curves are identical and match a point charge Q at the center inside, the difference is what a sphere of radius r encloses: nothing, or (r/R)³ of Q
Same charge, same outside field. Inside, the shell encloses nothing until r reaches R while the solid sphere encloses the fraction (r/R) cubed, so one curve sits at zero and the other rises in a straight line to the same surface value.
same pillbox, same enclosed charge, different number of exits sheet, σ flux leaves BOTH caps: 2EA E = σ / 2ε₀, at any distance conductor E = 0 inside surface, σ flux leaves ONE cap: EA E = σ / ε₀ just outside
Both pillboxes enclose the same charge, sigma times the cap area. The free-standing sheet sends field out through two caps and the conductor through one, and that single fact is the factor of two between the two results.

The work

3 ways in · any order
Lesson
Gauss's Law

Separates what Gauss's law always gives (the net flux) from what it gives only under symmetry (the field), then works the shell, solid sphere, line and sheet with the enclosed-charge question asked out loud each time.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: erasing outside charges from the field on the surface, factoring E out of the integral without symmetry, swapping the shell and solid-sphere interiors, treating the Gaussian surface as a physical object, and making the infinite sheet's field fade with distance. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
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