Three capacitors of capacitances $2 \;pF$,$3 \;pF$,and $4 \;pF$ are connected in parallel.
$(a)$ What is the total capacitance of the combination?
$(b)$ Determine the charge on each capacitor if the combination is connected to a $100 \;V$ supply.

(A) Capacitances of the given capacitors are $C_{1} = 2 \;pF$,$C_{2} = 3 \;pF$,and $C_{3} = 4 \;pF$.
For a parallel combination of capacitors,the equivalent capacitance $C_{eq}$ is the algebraic sum of individual capacitances:
$C_{eq} = C_{1} + C_{2} + C_{3} = 2 + 3 + 4 = 9 \;pF$.
Therefore,the total capacitance of the combination is $9 \;pF$.
$(b)$ The supply voltage is $V = 100 \;V$. In a parallel combination,the potential difference across each capacitor is the same and equal to the supply voltage $V = 100 \;V$.
The charge $q$ on a capacitor with capacitance $C$ is given by $q = CV$.
For $C_{1} = 2 \;pF$,charge $q_{1} = C_{1}V = 2 \;pF \times 100 \;V = 200 \;pC = 2 \times 10^{-10} \;C$.
For $C_{2} = 3 \;pF$,charge $q_{2} = C_{2}V = 3 \;pF \times 100 \;V = 300 \;pC = 3 \times 10^{-10} \;C$.
For $C_{3} = 4 \;pF$,charge $q_{3} = C_{3}V = 4 \;pF \times 100 \;V = 400 \;pC = 4 \times 10^{-10} \;C$.