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A classic exam question connects a charged capacitor to another one and asks for the final voltage and the energy. The answer always shows energy missing, which puzzles students who have been told energy is conserved. This post shows how to find the final state using charge conservation, how to calculate the energy lost, and where that energy goes.
What does the syllabus and textbook say?
The Syllabus for JEE (Main) 2026 lists capacitors in series and parallel and "energy stored in a capacitor" under Electrostatics. The NCERT Class 12 chapter Electrostatic Potential and Capacitance works the simplest case in Example 2.10: a charged 900 pF capacitor connected to an identical uncharged one ends with half the original energy. NCERT explains that a transient current flows before the system settles, and "energy is lost during this time in the form of heat and electromagnetic radiation". Check the syllabus for your exam year.
How do you find the common potential?
When two capacitors are joined plate to plate, three things fix the final state.
Charge is conserved on each connected pair of plates.
The connected plates end at the same potential, so both capacitors have the same final voltage V.
Each capacitor then has Q = CV.
Joining positive to positive (like plates): V = (C₁V₁ + C₂V₂) / (C₁ + C₂).
Joining positive to negative (opposite plates): charges partly cancel, so V = |C₁V₁ − C₂V₂| / (C₁ + C₂).
How much energy is lost?
Subtract the final energy from the initial energy. For like plates, the algebra simplifies to
ΔU = C₁C₂ (V₁ − V₂)² / [2(C₁ + C₂)].
For opposite plates, replace (V₁ − V₂) with (V₁ + V₂). The loss is zero only if the capacitors start at the same voltage, in which case no charge flows.
Worked example 1: like plates
A 6 μF capacitor charged to 100 V is connected, positive to positive, to a 3 μF capacitor charged to 40 V.
Initial charges: Q₁ = 6 × 100 = 600 μC, Q₂ = 3 × 40 = 120 μC. Total 720 μC.
Common potential: V = 720 / 9 = 80 V.
Final charges: 6 × 80 = 480 μC and 3 × 80 = 240 μC. So 120 μC flowed from the first to the second.
Initial energy: ½ × 6 × 10⁻⁶ × 100² + ½ × 3 × 10⁻⁶ × 40² = 0.0300 + 0.0024 = 0.0324 J.
Final energy: ½ × 9 × 10⁻⁶ × 80² = 0.0288 J.
Loss: 0.0324 − 0.0288 = 0.0036 J.
Formula check: (6 × 3 / 9) × 10⁻⁶ × 60² / 2 = 2 × 10⁻⁶ × 3600 / 2 = 0.0036 J. Both routes agree.
Worked example 2: opposite plates
Now connect the same two capacitors positive to negative.
Net charge on each connected pair: 600 − 120 = 480 μC.
Common potential: V = 480 / 9 ≈ 53.3 V.
Final energy: 480² / (2 × 9) μJ = 12,800 μJ = 0.0128 J.
Loss: 0.0324 − 0.0128 = 0.0196 J.
Formula check: 2 × 10⁻⁶ × 140² / 2 = 0.0196 J. Reversing the connection loses more than five times as much energy.
Is the same thing happening when a battery charges a capacitor?
Yes. Charging a 10 μF capacitor to 12 V moves Q = 120 μC through the battery, which does work QV = 120 × 10⁻⁶ × 12 = 1.44 × 10⁻³ J. The capacitor stores ½QV = 0.72 × 10⁻³ J. Exactly half is lost, again independent of the resistance.
Where does the energy really go?
In an ideal circuit with zero resistance the question looks unanswerable, which is why it is called the two-capacitor paradox. Real circuits are never ideal. Powell (1979) argued that the problem should be treated more realistically by including the circuit's self-inductance, and that for practical circuit values the usual series RC model can be inappropriate. For exam purposes, NCERT's answer holds: heat in the connecting wires and electromagnetic radiation.
Which mistakes are most common?
Conserving energy instead of charge. Energy is not conserved in the capacitors; charge is.
Adding charges when plates are reversed. Opposite plates mean subtracting.
Forgetting the final voltage is common. After connection the capacitors are in parallel.
Using U = ½CV² with the initial V for the final state.
Thinking the loss depends on the wire. It does not.
For series and parallel combinations and dielectrics, read our guide to capacitors and dielectrics problems.
A practice routine
Solve three charge-sharing problems: like plates, opposite plates, and one capacitor initially uncharged.
For each, calculate the loss twice: by subtracting energies and by the formula.
Add one battery-charging problem and confirm the half-lost result.
If a step in the working does not make sense, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can see why charge, not energy, is the conserved quantity.
Today, rework NCERT Example 2.10 with unequal capacitors and check your energy loss against the formula.
