Charge Redistribution
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
When conductors are connected, charge flows until they reach the same potential. The total charge is conserved. For spheres joined by a wire, that means the charge splits in proportion to radius, not equally.
1. Joined spheres
2. Other redistribution
Grounding connects a conductor to Earth, a huge conductor at V = 0. Charge flows until the object is at V = 0 too.
Induction: with a charged rod nearby, ground the object, cut the ground, then remove the rod. The object keeps a charge opposite to the rod’s.
Nested conductors: a charge +q at the center of a metal shell induces −q on the shell’s inner surface. The outer surface gets the shell’s net charge minus that: a shell with −3q total has −2q outside.
3. Example, then try it
Two identical metal spheres carry +8.0 µC and −2.0 µC. They touch and separate. Find each one’s charge.
Total +6.0 µC, shared equally by identical spheres: +3.0 µC each.
Metal spheres of radius 0.10 m and 0.30 m, far apart, are joined by a thin wire and carry a total of 8.0 nC. Find the charge on each, their common potential, and the field at each surface.
Split: q ∝ R, so q1 : q2 = 1 : 3. q1 = 2.0 nC, q2 = 6.0 nC.
Potential: kq1/R1 = (9.0 × 10⁹)(2.0 × 10⁻⁹) ÷ 0.10 = 180 V (same for the large one: 54 ÷ 0.30 ✓).
Fields: small: kq1/R1² = 18 ÷ 0.010 = 1800 N/C; large: 54 ÷ 0.090 = 600 N/C.
Ratio: 1800/600 = 3 = R2/R1: the smaller sphere has the stronger field.
Answer: 2.0 nC and 6.0 nC; 180 V; 1800 N/C and 600 N/C.
4. Common mistakes
Two metal spheres of different sizes are joined by a wire. Which has the larger surface charge density?
q ∝ R, so σ = q/(4πR²) ∝ 1/R. The smaller sphere has the larger σ and the stronger surface field.
Ready to practice?
5 questions on Charge Redistribution, each with a hint and an explanation for every wrong answer.