Animated Solution for Physics - Electrostatics: In the given circuit, a charge of $+80\mu \text{C}$ is given to the upper plate of the $4\mu \text{F}$ capacitor. Then in the steady state, the charge on the upper plate of the $3\mu \text{F}$ capacitor is

The behavior of capacitors in a circuit can sometimes feel like a complex puzzle, but the key to unlocking it often lies in identifying isolated systems. Let's embark on a journey to understand exactly how charge distributes itself in this fascinating circuit!

We are presented with a circuit featuring a $4\mu \text{F}$ capacitor connected in series with a parallel combination of a $2\mu \text{F}$ and a $3\mu \text{F}$ capacitor.

The problem states that a charge of $+80\mu \text{C}$ is deposited onto the upper plate of the $4\mu \text{F}$ capacitor. Our goal is to track this charge and determine how much of it ends up on the upper plate of the $3\mu \text{F}$ capacitor once the system reaches a steady state.

When $+80\mu \text{C}$ of charge is placed on the top plate of the $4\mu \text{F}$ capacitor, it creates a strong electric field. This field attracts electrons from the lower parts of the circuit, inducing an equal and opposite charge of $-80\mu \text{C}$ on its bottom plate.

But where do these electrons come from?

If we look closely at the circuit diagram, the bottom plate of the $4\mu \text{F}$ capacitor is connected directly to the top plates of the $2\mu \text{F}$ and $3\mu \text{F}$ capacitors. This entire H-shaped conducting section is physically separated from the rest of the circuit by the dielectric gaps of the three capacitors. It forms an isolated system.

Before the charge was applied, this isolated system was electrically neutral. By the law of conservation of charge, its net charge must remain zero.

Since the bottom plate of the first capacitor holds $-80\mu \text{C}$, the top plates of the other two capacitors must share a total charge of $+80\mu \text{C}$ to balance it out!

Now that we know the $2\mu \text{F}$ and $3\mu \text{F}$ capacitors share a total of $+80\mu \text{C}$, how exactly do they divide it?

Because these two capacitors are connected in parallel, the potential difference ($V$) across them must be identical. We know the fundamental relationship for a capacitor is:

Since $V$ is constant for both, the charge $q$ is directly proportional to the capacitance $C$. This means the $+80\mu \text{C}$ charge will divide itself in the exact ratio of their capacitances, which is $2:3$.

We are specifically interested in the charge on the $3\mu \text{F}$ capacitor ($q_3$).

Using the ratio we just discovered, the $3\mu \text{F}$ capacitor will take 3 parts out of the total 5 parts ($2+3=5$) of the available charge.

Substituting our values:

And there we have it! The upper plate of the $3\mu \text{F}$ capacitor will hold exactly $+48\mu \text{C}$ in the steady state. By simply following the charge and respecting the boundaries of isolated systems, even the most daunting capacitor circuits become beautifully simple.