Q.Three capacitors of capacitances $2\ \text{pF}$, $3\ \text{pF}$ and $4\ \text{pF}$ are connected in parallel.

Concept: Capacitor Network Analysis — In a parallel combination, the total capacitance is the sum of individual capacitances, and each capacitor sees the same voltage.

Step 1 — Total capacitance in parallel

$C_{\text{eq}} = C_1 + C_2 + C_3 = 2 + 3 + 4 = 9\ \text{pF}$.

Step 2 — Charge on each capacitor

For a parallel connection, voltage across each is the same: $V = 100\ \text{V}$.

Using $Q = CV$:

• $Q_1 = 2 \times 10^{-12} \times 100 = 2 \times 10^{-10}\ \text{C}$
• $Q_2 = 3 \times 10^{-12} \times 100 = 3 \times 10^{-10}\ \text{C}$ …

When capacitors are connected in parallel, the key idea is that the voltage across each capacitor is the same. Think of it like water pipes branching off from a single main pipe — each branch sees the same water pressure. Here, the battery supplies a fixed voltage, and every parallel branch gets that full voltage.

Why does this matter for capacitance? Capacitance tells us how much charge a capacitor can store per volt. If you put capacitors side by side, you're effectively increasing the plate area available for storing charge. The total charge stored by the combination is just the sum of charges on each capacitor. Since $Q = CV$ and $V$ is common, the total charge is $(C_1 + C_2 + C_3)V$, so the equivalent capacitance is simply the sum.

Now let's apply this to the given numbers.

1. Find the total capacitance. The three capacitances are $2\ \text{pF}$, $3\ \text{pF}$, and $4\ \text{pF}$. In parallel, we add them directly:

That's part (a) done.

2. Understand the voltage condition.

   The combination is connected to a $100\ \text{V}$ supply. In a parallel connection, every capacitor sees the same voltage. So each capacitor has $V = 100\ \text{V}$ across its plates.

3. Calculate the charge on each capacitor.

   Use $Q = CV$ for each one individually.

   • For the $2\ \text{pF}$ capacitor: $Q_1 = (2 \times 10^{-12}) \times 100 = 2 \times 10^{-10}\ \text{C} = 200\ \text{pC}$
   • For the $3\ \text{pF}$ capacitor: $Q_2 = (3 \times 10^{-12}) \times 100 = 3 \times 10^{-10}\ \text{C} = 300\ \text{pC}$
   • For the $4\ \text{pF}$ capacitor: $Q_3 = (4 \times 10^{-12}) \times 100 = 4 \times 10^{-10}\ \text{C} = 400\ \text{pC}$ …

Method: Solving a Parallel Capacitor Network

This method finds the equivalent capacitance and the charge on each capacitor for capacitors connected side-by-side across the same two nodes (parallel).

Steps

Step 1: Recognise the parallel connection and its defining property

In parallel, every capacitor's plates are tied to the same two nodes, so each one sees the full supply voltage $V$. It is the charge, not the voltage, that differs from capacitor to capacitor.

Step 2: Combine by simple addition

$C_{\text{eq}} = C_1 + C_2 + C_3 + \dots$ …