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.
The work
Lesson live · diagnostic and drills coming soon
Lesson
Redistribution of Charge between Conductors
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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.
Diagnostic
10-item topic check
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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.
Targeted Practice
Drill a single misconception
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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.