Answer:
Step-by-step explanation:
The charge on 10μF capacitor = 10 x 12 x 10⁻⁶ = 120 μC
when it is connected with 20μF capacitor both acquires common potential whose value is
= 120 x 10⁻⁶ /( 10 +20) x 10⁻⁶ = 4 V.
Energy stored in 20μF capacitor =1/2 x 20 x 10⁻⁶ x 4 x 4 = 160 x 10⁻⁶ J.