**Hint:** When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. The voltage across each capacitor in a parallel connection is the same as the voltage of the supply.

**Step by Step Solution:**

**Step 1:** Calculate the total capacitance of the parallel combination.

When capacitors are connected in parallel, the total capacitance $C_t$ is given by:

$C_t = C_1 + C_2 + C_3$

Given:

$C_1 = 2 \text{ pF}$

$C_2 = 3 \text{ pF}$

$C_3 = 4 \text{ pF}$

Substitute the given values:

$C_t = 2 + 3 + 4$

$C_t = 9 \text{ pF}$

**(a) The total capacitance of the combination is 9 pF.**

**Step 2:** Calculate the charge on each capacitor.

The charge Q stored on a capacitor is given by: $Q = C \times V$

Where: C is the capacitance of the capacitor.

V is the voltage across the capacitor.

Given that the voltage $V = 100 \text{ V}$ for all capacitors (since they are in parallel):

For $C_1$:

$Q_1 = C_1 \times V$

$Q_1 = 2 \text{ pF} \times 100 \text{ V}$

$Q_1 = 200 \text{ pC}$

For $C_2$:

$Q_2 = C_2 \times V$

$Q_2 = 3 \text{ pF} \times 100 \text{ V}$

$Q_2 = 300 \text{ pC}$

For $C_3$:

$Q_3 = C_3 \times V$

$Q_3 = 4 \text{ pF} \times 100 \text{ V}$

$Q_3 = 400 \text{ pC}$

**(b) The charge on the capacitors are:**

For the 2 pF capacitor: 200 pC

For the 3 pF capacitor: 300 pC

For the 4 pF capacitor: 400 pC

**Note:** The advantage of a parallel connection of capacitors is that the total capacitance increases, providing a higher storage capacity for electric charge. Also, all capacitors in a parallel connection have the same voltage across them, which is equal to the supply voltage. This characteristic allows each capacitor to store charge based on its individual capacitance value when exposed to the same voltage.