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Two metal spheres of different sizes are joined by a thin wire. Charge flows until something becomes equal, and students often guess "the charge". It is the potential. This post shows why, how to split the charge, why the smaller sphere ends up with the stronger surface field, and how much energy is lost, with two worked examples.
Is this in the syllabus?
NCERT's chapter Electrostatic Potential and Capacitance gives the potential of a charged spherical shell, V = q/(4πε₀R) at and outside its surface (equation 2.19), and section 2.9 shows that a conductor in electrostatics has the same potential throughout, with a surface field E = σ/ε₀. The CBSE Physics curriculum for 2026-27 lists "Conductors and insulators, free charges and bound charges inside a conductor", and the Syllabus for JEE (Main) 2026 lists "electric potential and its calculation for a point charge" and "conductors and insulators". Check the current year's documents before your exam.
Why does the potential, not the charge, become equal?
The spheres and the wire together form one conductor. NCERT's section 2.9 shows that the electrostatic field inside a conductor is zero, so no work is done moving charge within it, and every point is at the same potential. Charge flows along the wire until that is true.
Assume the spheres are far apart compared with their radii, so each one's charge spreads uniformly and the other's field is negligible. Then:
equal potential: Q₁/R₁ = Q₂/R₂
charge conserved: Q₁ + Q₂ = Q, the total before joining
So Q₁ = Q × R₁/(R₁ + R₂) and Q₂ = Q × R₂/(R₁ + R₂). The bigger sphere takes the bigger share.
Worked example 1: a charged sphere and a neutral one
Sphere A (radius 2.0 cm) carries +12 nC. Sphere B (radius 4.0 cm) is uncharged. They are far apart and joined by a thin wire. Take k = 1/(4πε₀) ≈ 9.0 × 10⁹ N m²/C², from the NIST value of ε₀.
Before joining: V_A = kQ/R = 9.0 × 10⁹ × 12 × 10⁻⁹/0.020 = 5400 V.
After: Q_A = 12 × 2/(2 + 4) = 4.0 nC and Q_B = 12 × 4/6 = 8.0 nC.
Common potential: V = 9.0 × 10⁹ × 4.0 × 10⁻⁹/0.020 = 1800 V. Check with B: 9.0 × 10⁹ × 8.0 × 10⁻⁹/0.040 = 1800 V.
Charge that flowed through the wire: 8.0 nC.
Which sphere has the stronger field at its surface?
The surface charge density is σ = Q/(4πR²). With Q proportional to R, σ is proportional to 1/R. So the smaller sphere has the larger σ and, from E = σ/ε₀, the stronger field just outside. In example 1, σ_A/σ_B = R_B/R_A = 2. This is the usual textbook explanation for why charge crowds at sharp, highly curved parts of a conductor.
Is energy conserved?
Charge is conserved; electrostatic energy is not. Each isolated sphere acts as a capacitor with C = 4πε₀R. In example 1:
before: U = ½QV = ½ × 12 × 10⁻⁹ × 5400 = 3.24 × 10⁻⁵ J
after: U = ½ × 12 × 10⁻⁹ × 1800 = 1.08 × 10⁻⁵ J
loss: 2.16 × 10⁻⁵ J, dissipated while the charge flows, mainly as heat in the wire
The capacitor formula ΔU = ½ × C₁C₂/(C₁ + C₂) × (V₁ − V₂)² gives the same number. Our post on charge sharing between capacitors derives it.
Worked example 2: opposite charges
Sphere A (radius 3.0 cm) carries +10 nC; sphere B (radius 1.0 cm) carries −2.0 nC. They are far apart and then joined.
Total charge: +10 − 2.0 = +8.0 nC.
Q_A = 8.0 × 3/4 = 6.0 nC and Q_B = 8.0 × 1/4 = 2.0 nC. B's charge has changed sign.
Common potential: 9.0 × 10⁹ × 6.0 × 10⁻⁹/0.030 = 1800 V. Check with B: 9.0 × 10⁹ × 2.0 × 10⁻⁹/0.010 = 1800 V.
Before joining, A was at +3000 V and B at −1800 V. A net +4.0 nC moved from A to B; in the metal, it is electrons that actually move, from B to A.
Which mistakes cost marks?
Splitting the charge equally. That holds only for identical spheres, the case NCERT's chapter Electric Charges and Fields describes for Coulomb's method of halving a charge.
Ignoring signs when adding the initial charges.
Using these formulas for spheres that touch or sit close together. Each sphere's field then changes the other's charge distribution, and the simple Q ∝ R result no longer holds.
Assuming energy is conserved, which gives a wrong final potential.
Saying the larger sphere has the larger surface field.
How should you practise this?
Testing yourself beats rereading for long-term memory. Roediger and Karpicke (2006) found that students who took recall tests on a passage retained more after two days and after a week than students who restudied it, even though restudying made them more confident. So close this post and try:
Day 1: from memory, write the three relations: equal potential, charge conservation, energy before and after.
Day 2: three problems with different radius ratios, including one with opposite charges.
Day 3: one problem on surface charge density ratios and one on the heat produced.
Day 4: a mixed set with our post on placing a third charge for equilibrium, so you must choose the method.
If you want to revisit potential and conductors from the start, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
Tonight, redo example 1 with B's radius doubled to 8.0 cm, and check that the common potential falls to 1080 V.
