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AP® Physics C: Electricity and Magnetism · Unit 10 · Topic 10.2

Redistribution of Charge Between Conductors: Revision Notes (Physics C: E&M 10.2)

One-page recap of charge sharing between conductors: equal potential on contact, spheres joined by a wire, nested conductors, ground and the induced charge on a grounded sphere.

Short on time? This page is the recap. For the derivations, Figure 1 and the worked examples, use the full study guide. This topic is part of the calculus-based course.

Recap

  • Conductors in electrical contact become one conductor, so they end at one potential. The flow is effectively instant.
  • Total charge is conserved. Electrons move towards higher potential (positive charge “flows” from high V to low V).
  • Equal charges result only for identical conductors.
  • A charged conductor joined to a hollow conductor that surrounds it loses all its charge to the outer surface.
  • Ground is an ideal reference at V = 0 that can absorb or supply any amount of charge without changing its potential.
  • Grounding an isolated conductor makes it neutral. Grounding a conductor near a charge leaves an induced charge of opposite sign.
  • To keep an induced charge: break the ground first, then remove the nearby charge.

Key relationships

Situation Result
Any conductors in contact V₁ = V₂; q₁ + q₂ = Q_total
Distant spheres joined by a thin wire q₁/R₁ = q₂/R₂; q₁ = QR₁/(R₁ + R₂)
Same pair: densities and fields σ₁/σ₂ = E₁/E₂ = R₂/R₁; E = V/R at each surface
Grounded sphere (radius R), point charge q at distance d from centre Q′ = −qR/d
Grounded shell around a charge q outer surface charge → 0; no field outside

Assumptions

  • “Far apart”: each sphere’s potential comes only from its own charge.
  • “Thin wire”: the wire holds negligible charge.
  • Ground is ideal: its potential never changes.

Mistakes to avoid

  1. Sharing equally between different-sized conductors. Use equal V.
  2. Bigger sphere → bigger surface field. It is the smaller sphere.
  3. Grounding always gives zero charge. Not with a charge nearby.
  4. Wrong order of steps. Remove the ground before the inducing charge.
  5. Using the uneven spread of Q′ to block the calculation. At the centre, all of Q′ is at distance R, so its potential is kQ′/R.

Quick self-check

  1. Distant spheres of radii 1.0 cm and 4.0 cm share 10 nC through a wire. Charges? (2.0 nC and 8.0 nC)
  2. Which of those spheres has the stronger surface field, and by what factor? (The 1.0 cm sphere, 4 times stronger)
  3. A grounded sphere of radius 0.10 m is 0.50 m (centre to charge) from +2.0 nC. Induced charge? (−0.40 nC)

Next: practice questions.

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