Question

All capacitors are identical in the figure. Capacitor C₁ is charged to 48 µC by the emf source when the switch is in position 1. The switch is then moved to position 2 and the charge redistributes among all the capacitors. After this redistribution, the charge on C₁ is:

Answer 1

When the switch is moved to position 2, charge on the positive plate moves to the other capacitors until the voltages balance. The equivalent capacitance of the 3 capacitors in series is C/3. Let Q₀ = 48 µC be the initial charge on C₁ and ΔQ be the charge transferred to the three uncharged capacitors. The equal potential condition then yields (Q₀− ΔQ)/ C = 3ΔQ / C or ΔQ = 12µC . Thus the remaining charge on C₁ is 48 – 12 = 36 µC.