Capacitor P (10 × 10⁻⁶ F) is charged to 6.0 V.
Initial charge: QP=10×10−6×6=60×10−6 C.
When connected to uncharged capacitor Q (20 × 10⁻⁶ F), charge redistributes until both reach same potential.
Final voltage: Vf=CtotalQtotal=30×10−660×10−6=2 V.
Charge on Q: QQ=20×10−6×2=40×10−6 C = 4×10−5 C.
Thus α=4.