Animated Solution for Physics - Electrostatics: In the circuit shown in the figure, the total charge is $750\mu \text{C}$ and the voltage across capacitor $C_2$ is $20\text{ V}$. Then, the charge on capacitor $C_2$ is

When tackling mixed capacitor circuits, the first and most crucial step is to clearly identify which components are in series and which are in parallel. Looking at our circuit diagram, we can see that the total charge $Q$ supplied by the battery flows entirely through capacitor $C_1$.

After passing through $C_1$, the circuit splits into two branches at a junction. One branch contains $C_2$ and the other contains $C_3$. Because these two capacitors are connected across the same two nodes, they are in a parallel combination. This entire parallel block is, in turn, connected in series with $C_1$.

A fundamental property of parallel circuits is that the potential difference (voltage) across all parallel branches is identical. The problem states that the voltage across $C_2$ is $20\text{ V}$.

Because $C_3$ is in parallel with $C_2$, it must share the exact same voltage. Therefore, we immediately know that $V_3=20\text{ V}$. This is a massive stepping stone because we now have both the capacitance ($C_3=8\mu \text{F}$) and the voltage ($V_3=20\text{ V}$) for the third capacitor.

Using the core electrostatic formula $Q=CV$, we can calculate the charge stored on $C_3$:

Now, let's trace the flow of charge. The total charge $Q=750\mu \text{C}$ arrives at the junction and must split between the two parallel branches. According to the principle of conservation of charge, the total incoming charge must equal the sum of the outgoing charges in the branches.

Mathematically, this is expressed as:

We know the total charge is $750\mu \text{C}$ and we just found that $160\mu \text{C}$ goes to $C_3$. Substituting these values into our conservation equation gives us a simple linear equation to solve for $Q_2$:

Subtracting $160\mu \text{C}$ from both sides, we find the charge on $C_2$:

And there we have it! The charge on capacitor $C_2$ is $590\mu \text{C}$.