Animated Solution for Physics - Electrostatics: Find the potential difference between the points A and B and between the points B and C in the steady state.

Imagine a bustling city where traffic suddenly comes to a complete halt. That is exactly what happens in a DC circuit containing capacitors once the steady state is reached. The capacitors act as roadblocks, completely stopping the flow of direct current.

When you first look at this circuit, the combination of resistors and capacitors might seem intimidating. However, the key to unlocking this problem lies in the phrase 'steady state'. In this state, the capacitors are fully charged, and the current $I$ in the circuit drops to absolute zero.

According to Ohm's Law, the voltage drop across any resistor is given by $V=IR$. Since $I=0$, the voltage drop across both the $20\Omega$ and $10\Omega$ resistors is exactly zero! This is a beautiful simplification. It means we can treat these resistors as ideal wires for the purpose of finding potentials. The entire $100\text{V}$ from the battery is applied directly across the main nodes of the capacitor network, which we will call Node A and Node C.

Now, let's look at the capacitor network. By tracing the continuous wires, we can identify the true nodes. The left vertical wire connects points F, D, and A together, forming a single super-node (Node A). Similarly, the right vertical wire connects points G, E, and C together, forming another super-node (Node C).

With this perspective, the circuit folds neatly: 1. The two $3\mu \text{F}$ capacitors are both connected directly between Node A and Node B. They are in parallel, giving an equivalent capacitance of $C_{AB}=3+3=6\mu \text{F}$. 2. The two $1\mu \text{F}$ capacitors are connected between Node B and Node C. They are also in parallel, giving $C_{BC}=1+1=2\mu \text{F}$. 3. The remaining $1\mu \text{F}$ capacitor is connected directly between Node A and Node C.

We are asked to find the potential differences $V_{AB}$ and $V_{BC}$. The $1\mu \text{F}$ capacitor connected across A and C is in parallel with the A-B-C branch, so it doesn't affect the voltage distribution along the A-B-C path. We only need to focus on the $6\mu \text{F}$ and $2\mu \text{F}$ capacitors, which are in series across the $100\text{V}$ source.

In a series circuit, the charge $Q$ on each equivalent capacitor is the same. Since $V=\frac{Q}{C}$, the voltage across a capacitor is inversely proportional to its capacitance. The formula for voltage division between two series capacitors is:

Applying this to our circuit:

The remaining voltage must drop across the second capacitor:

And there we have it! By understanding the physical reality of the steady state and carefully tracing our nodes, a complex web of components simplifies into a straightforward series circuit.