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

Hint: As three of the capacitors are connected parallelly to each other then sum of all the three capacitors gives us the total capacitance of the combination. As the capacitance is connected parallelly then the voltage across each capacitance will remain the same, now putting the voltage capacitance and charge relation we can find out the charge on each capacitance.

Complete step by step answer:
(a) As per the problem there are three capacitors of capacitance $2pF,3pF$ and $4pF$are connected parallel. When capacitors are connected parallel to each other then their total resistance is calculated by adding all the capacitance.Now we can get,
Total capacitance = capacitance one + capacitance two + capacitance three
Mathematically,
$C = {C_1} + {C_2} + {C_3} \ldots \ldots \left( 1 \right)$
Given,
${C_1} = 2pF$
$\Rightarrow {C_2} = 3pF$
$\Rightarrow {C_3} = 4pF$
Here $pF$ is the unit of capacitance.
PicoFarad = pF
Hence total capacitance,
Putting the given values in equation $\left( 1 \right)$ we get,
$C = 2pF + 3pF + 4pF$
On further solving we get,
$\therefore C = 9\,pF$

Therefore the total capacitance of the combination is $9\,pF$.

(b) As we know the supply voltage to the capacitance is $100V$. Three capacitance is connected parallelly hence voltage will remain same for all the capacitor.We know,
$q = VC \ldots \ldots \left( 2 \right)$
Where, Charge of the capacitor = $q$, Supply Voltage or voltage across capacitor = $V$ and Capacitance = $C$.
Using equation $\left( 2 \right)$ in all the three cases:
For capacitor one we get,
${q_1} = V{C_1}$
Now putting the given values in the equation we get,
${q_1} = 100V \times 2pF$
We know,
$1pF = {10^{ - 12}}F$

Now the change become,
${q_1} = 100V \times 2 \times {10^{ - 12}}F$
$\Rightarrow {q_1} = 2 \times {10^{ - 10}}C$
Unit is coulomb.
Similarly, for charge two we get,
${q_2} = V{C_2}$
Now putting the given values in the equation we get,
${q_2} = 100V \times 3pF$
$ \Rightarrow {q_2} = 100V \times 3 \times {10^{ - 12}}F$
$ \Rightarrow {q_2} = 3 \times {10^{ - 10}}C$
For charge three we get,
${q_3} = V{C_3}$
Now putting the given values in the equation we get,
${q_3} = 100V \times 4pF$
$ \Rightarrow {q_3} = 100V \times 4 \times {10^{ - 12}}F$
$ \Rightarrow {q_3} = 4 \times {10^{ - 10}}C$

Hence, upon keeping the supply voltage to the all three capacitors at $100V$, we can observe only change in their charges as $2 \times {10^{ - 10}}C$, $3 \times {10^{ - 10}}C$ and $4 \times {10^{ - 10}}C$.

Note: Always remember when capacitors are connected parallelly the the voltage across each capacitor will remain constant. Before calculating change, first change the capacitor and voltage to its SI unit to get the charge in coulomb because coulomb is voltage multiplied by farad.