Q · 45

Two switches — charge finally on C₂

Charge sharing

Question

In the circuit shown here \(C_1 = 6\,\mu F\), \(C_2 = 3\,\mu F\) and battery \(B = 20\ V\). The switch \(S_1\) is first closed. It is then opened and afterwards \(S_2\) is closed. What is the charge finally on \(C_2\)?

The Circuit

C₁ C₂ B = 20 V S₂ S₁

Closing \(S_1\) completes the battery–\(C_1\) loop (right edge); \(C_2\) stays cut off behind the open \(S_2\). Later, opening \(S_1\) and closing \(S_2\) drops the battery and links \(C_2\) across the charged \(C_1\).

Two phases

Phase 1 — S₁ closed: charge C₁

C₁ 20 V S₁ ✓ 120 µC

Phase 2 — S₂ closed: C₁ ∥ C₂

C₁ C₂ S₂ ✓ share

Concept

Charge the bigger \(C_1\) fully, isolate it, then let it share with \(C_2\). Once the battery is gone the total charge is fixed, so the common voltage is \(Q/(C_1+C_2)\) and \(C_2\) takes \(C_2\times\) that.

Method & Steps

  1. \(S_1\): \(Q_1 = C_1V = 6\times20 = 120\,\mu C\)
  2. \(S_1\) open: \(C_1\) keeps 120 µC
  3. \(S_2\): \(V'=\dfrac{120}{6+3}=13.\overline{3}\,V\)
  4. \(Q_2 = C_2V' = 3\times13.\overline{3}=40\,\mu C\)

Easy Trick

In sharing, charge splits in the ratio of the capacitances. \(C_1:C_2 = 6:3 = 2:1\), so of the 120 µC the \(C_2\) gets the \(\tfrac{1}{3}\) share → 40 µC (and \(C_1\) keeps \(\tfrac{2}{3}=80\) µC). The 120 µC option is the trap for forgetting the sharing step.

c
\(Q_2 = 40\ \mu C\)
\(C_1\) charged to 120 µC first, then shares one-third of it with \(C_2\).