Redistribute charge by equal potential

Choose the isolated system

Consider two widely separated conducting spheres joined by an ideal connector. The approximation neglects mutual potentials and charge on the connector. At final equilibrium their potentials agree, but the pair’s total charge remains fixed. For radii R1 and R2, kQ1/R1 = kQ2/R2 and Q1 + Q2 = Qt. Thus Qi = Qt Ri/(R1 + R2). Equal charge follows only for equal radii in this model.

Compare density rather than only inventory

The surface density is Qi/(4pi Ri²). At common potential it scales as 1/Ri, so the smaller sphere carries less total charge but greater surface density and stronger surface field. Sharp points and edges of a conductor commonly concentrate surface charge relative to flatter areas. The two-sphere result illustrates this effect; it is not a universal formula replacing a full geometry calculation at any curved surface.

Account for energy during redistribution

If all charge starts on sphere 1, initial energy is kQt²/(2R1). Final energy is k(Q1²/R1 + Q2²/R2)/2. The reduction leaves the electrostatic account through the actual relaxation process, which this endpoint model does not simulate. An ideal final connector is not permission to claim energy disappeared.

Change to an open reservoir boundary

Ground is an ideal reference reservoir whose potential remains fixed while charge is exchanged. A conductor near a positive external source can receive electrons from that reservoir. In a conceptual virtual sequence, isolating the conductor before removing the source can retain excess electrons; allowing reservoir contact while the source is removed can instead allow that charge to return. Predict the net-charge difference from the order of boundary changes, not from attraction alone. This is a diagram account, not a physical grounding procedure.

Worked original case

With R1 = 1 m, R2 = 2 m and Qt = 3 nC, final charges are 1 and 2 nC. Both potentials are 8.98755 V. The smaller sphere’s surface field and density are twice those of the larger. Initial energy is 40.4440 nJ and final energy 13.4813 nJ.

Investigation

Compare radius ratios 0.5, 1 and 2 while holding total charge fixed. Predict charge, potential, density and field ratios separately. Retain the missing electrostatic energy account. Contrast the closed pair with the virtual ground-reservoir sequence and record which total is no longer conserved.

Changed-case practice

Sources and boundaries

Original cases checked against the official framework on 30 October 2026, with OpenStax verification on 31 October 2026 against the current official framework, Unit 10 and OpenStax capacitance. These are virtual supplied models, not instructions to charge, ground, wire or handle real equipment. Public answers are practice; retained plans and self-reports do not establish independent mastery, actual laboratory participation or AP authorization.