Animated Solution for Physics - Electrostatics: In the circuit shown, charge on the $5\mu \text{F}$ capacitor is

This problem presents a classic multi-loop capacitor circuit. While it can certainly be solved using Kirchhoff's Voltage Law (KVL) by assigning loop charges, that method often leads to a system of simultaneous equations which can be tedious and prone to algebraic errors. Instead, we will use a much more elegant and powerful technique: Nodal Analysis.

The first step in nodal analysis is to choose a reference node, or 'ground', and assign it a potential of $0\text{V}$. Looking at the circuit, the junction between the two $6\text{V}$ batteries at the bottom is an excellent choice because it connects to multiple branches.

Once we set the bottom-middle node to $0\text{V}$, we can easily determine the potentials of the other nodes connected directly to the batteries.

Let's examine the left battery. The longer vertical line represents the positive terminal, which is facing left. This means the potential increases by $6\text{V}$ as we move from the $0\text{V}$ reference node to the left wire. Therefore, the entire left vertical wire is at a potential of $+6\text{V}$.

Similarly, the right battery has its positive terminal facing right. Moving from the $0\text{V}$ reference node to the right wire, the potential again increases by $6\text{V}$. Thus, the entire right vertical wire is also at a potential of $+6\text{V}$.

Now, we have only one node with an unknown potential: the top-middle junction where the $2\mu \text{F}$, $4\mu \text{F}$, and $5\mu \text{F}$ capacitors meet. Let's call the potential at this node $V$.

According to the principle of conservation of charge (which forms the basis of Kirchhoff's Current Law for isolated nodes), the net charge leaving this isolated junction must be zero. We can express the charge leaving through each capacitor as the product of its capacitance and the potential difference across it (from $V$ to the adjacent node).

For the $2\mu \text{F}$ capacitor, the charge leaving is $2(V-6)$. For the $4\mu \text{F}$ capacitor, the charge leaving is $4(V-6)$. For the $5\mu \text{F}$ capacitor, the charge leaving is $5(V-0)$.

Summing these up gives us our master nodal equation:

Now, we simply solve this linear equation for $V$:

Combining the $V$ terms and the constant terms:

With the potential $V$ known, finding the charge on the $5\mu \text{F}$ capacitor is straightforward. The potential difference across it is simply $V-0=V$.

This matches option (c). By choosing nodal analysis, we bypassed complex simultaneous equations and arrived at the solution swiftly and accurately.