Animated Solution for Physics - Electrostatics: In the following circuit $C_1=12\mu \text{F}$, $C_2=C_3=4\mu \text{F}$ and $C_4=C_5=2\mu \text{F}$. The Charge stored in $C_3$ is _______ $\mu \text{C}$.

Imagine you are sitting in the exam hall, and you are presented with a massive, tangled web of wires and capacitors. Your first instinct might be to panic, to start writing down massive nodal equations, and to brute-force your way through the math. But wait! Take a breath. The most elegant solutions in physics often come not from heavy computation, but from sharp observation.

When we first look at this circuit, it seems incredibly complex. We have five different capacitors—$C_1,C_2,C_3,C_4,$ and $C_5$—arranged in a bridge-like structure with diagonal connections. We also have two voltage sources: a $6\text{V}$ battery on the left and a $2\text{V}$ battery on the right.

A common pitfall for students is to immediately assign an unknown variable, say $V_M$, to the central node and start writing a massive Kirchhoff's Current Law (KCL) equation. While mathematically sound, this approach is a trap designed to consume your most precious resource: time.

Let's take a step back and establish our reference points. The entire bottom wire is continuous, meaning it acts as a single node. We can safely assign this as our ground, setting its potential to $0\text{V}$.

Because the negative terminals of both batteries are connected to this ground, the potentials at the top nodes are immediately known. The node directly above the left battery is at $6\text{V}$, and the node directly above the right battery is at $2\text{V}$.

Identifying the Short

Now, let's focus on the central top node, which we will call Node $M$. This is the node where the top plate of our target capacitor, $C_3$, is connected.

Trace the wire extending to the right from Node $M$. Does it pass through any resistor or capacitor before reaching the $2\text{V}$ node? Look closely at the diagram. The line is perfectly solid! There is absolutely no component between Node $M$ and the positive terminal of the $2\text{V}$ battery.

In circuit theory, an ideal wire connecting two points means those two points are at the exact same electrical potential. Therefore, Node $M$ is directly shorted to the $2\text{V}$ source. Without writing a single equation, we have deduced that:

The question asks for the charge stored in capacitor $C_3$. To find the charge, we only need two things: the capacitance and the potential difference across it.

We know the capacitance is given as $C_3=4\mu \text{F}$.

What about the potential difference? The top plate of $C_3$ is connected to Node $M$ (which is at $2\text{V}$), and the bottom plate is connected to the ground wire (which is at $0\text{V}$). Therefore, the potential difference $\Delta V_3$ is simply:

Now, we just apply the fundamental capacitor equation, $Q=C\Delta V$:

And just like that, we have our answer! The values of $C_1,C_2,C_4,$ and $C_5$ were nothing more than clever distractors placed there to test your conceptual clarity. Always remember: before you calculate, observe!