Mistake Master

Redistribution of Charge between Conductors

Two conductors joined by a wire form one conductor, so they come to one potential. For two distant spheres $V_1 = V_2$ gives $Q_1/R_1 = Q_2/R_2$, so the charge splits in proportion to radius and the common potential is $kQ/(R_1 + R_2)$. Because $E = V/R$ at each surface, the smaller sphere ends with the stronger field and the higher surface density even though it holds less charge. Grounding is the same condition with earth as the partner: it fixes $V = 0$ and lets whatever charge that demands flow in or out.

Two errors dominate. Splitting the charge evenly between unequal conductors, or equalizing surface charge density instead of potential, both of which leave the system with a potential difference across a conducting wire. And treating a ground wire as a drain that always empties a conductor: it empties an isolated sphere with nothing nearby, but next to a fixed positive charge it delivers negative charge instead, which is exactly how charging by induction works.

joined spheres: same V, unequal Q, and the small one has the strong field R 3R Q / 4 3Q / 4 E = V / R E = V / 3R V is the same on both: kQ₁/R = kQ₂/3R NOT Q/2 each. That leaves a factor of 3 in potential across a wire NOT equal sigma either. That would give Q/10 and 9Q/10
Equal potential is the equilibrium condition. Equal charge is what it happens to produce when the two conductors are twins, and nothing more general than that.
grounding fixes V = 0. the charge does whatever that requires 0 alone: V = kQ/R = 0 forces Q = 0 it really does drain, which is where the false rule comes from +Q net − near +Q: electrons flow IN to pull V down to 0 cut the wire and the sphere is left negative
Same wire, opposite outcomes. Ask what charge makes the potential zero given the neighbors, and the direction of the flow comes out of the answer.

The work

Lesson live · diagnostic and drills coming soon
Lesson
Redistribution of Charge between Conductors

Derives the connected-conductor split from equal potential, shows why the smaller sphere ends up with the stronger surface field, and replaces the folk rule about grounding with the condition that actually holds.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: halving the total between unequal conductors, equalizing surface charge density instead of potential, expecting the bigger sphere to have the bigger field, and treating a ground wire as a drain that always empties a conductor. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not yet available · 10 items
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.

Unlocks from the diagnostic, which is not published yet