A 5 μF capacitor is fully charged to a potential difference of . . .

Question : A 5 μF capacitor is fully charged to a potential difference of 20 V. After removing the source voltage, it is connected to an uncharged capacitor of 10 μF in parallel. The charge on the second capacitor is 

a) 50 μC

b) 100/3 μC

c) 200/3 μC

d) 40 μC

Doubt by Sonali

Solution :

C₁ = 5 μF

V₁ = 20 V

C₂ = 10 μF

V₂ = 0 V

Common Potential (V_C)
= (C₁V₁+C₂V₂) / (C₁+C₂)
V_C = (5×20+10×0) / (5+10)
V_C = 100/15
V_C = 20/3 V

Charge on the second Capacitor
q=CV
q = 10 μF×(20/3) V
q = (200/3) µC

Hence, c) would be the correct answer.