Charge Distribution on Capacitors in Steady State Circuit

To find the charge on the upper plate of the $3\ \mu\text{F}$ capacitor in the steady state, we analyze the charge distribution on the isolated portion of the circuit.

Step 1: Identify the Isolated Node

The node connected to:

1. The bottom plate of the $4\ \mu\text{F}$ capacitor,
2. The top plate of the $2\ \mu\text{F}$ capacitor, and
3. The top plate of the $3\ \mu\text{F}$ capacitor

forms an isolated conductor system (an isolated "island") that is not connected to any external voltage source directly. By the principle of conservation of charge, the total net charge on this isolated island must be zero (assuming it was initially uncharged):

$q_{\text{bottom}, 4\mu\text{F}} + q_{\text{top}, 2\mu\text{F}} + q_{\text{top}, 3\mu\text{F}} = 0$

Step 2: Induced Charge on the $4\ \mu\text{F}$ Capacitor

A charge of $+80\ \mu\text{C}$ is supplied to the top plate of the $4\ \mu\text{F}$ capacitor. By electrostatic induction, an equal and opposite charge of $-80\ \mu\text{C}$ is induced on the inner surface of its bottom plate:

$q_{\text{bottom}, 4\mu\text{F}} = -80\ \mu\text{C}$

Step 3: Charge Division Between the Parallel Capacitors

Substituting $q_{\text{bottom}, 4\mu\text{F}} = -80\ \mu\text{C}$ into the charge conservation equation gives:

$-80\ \mu\text{C} + q_{\text{top}, 2\mu\text{F}} + q_{\text{top}, 3\mu\text{F}} = 0$

$q_{\text{top}, 2\mu\text{F}} + q_{\text{top}, 3\mu\text{F}} = +80\ \mu\text{C}$

The $2\ \mu\text{F}$ and $3\ \mu\text{F}$ capacitors are connected in parallel, so they share the same potential difference $V$ across them:

$V = \frac{q_{\text{top}, 2\mu\text{F}}}{2\ \mu\text{F}} = \frac{q_{\text{top}, 3\mu\text{F}}}{3\ \mu\text{F}}$

The charge distributes across parallel capacitors in direct proportion to their capacitances:

$q_{\text{top}, 3\mu\text{F}} = \left( \frac{C_2}{C_1 + C_2} \right) \cdot Q_{\text{total}}$

$q_{\text{top}, 3\mu\text{F}} = \left( \frac{3\ \mu\text{F}}{2\ \mu\text{F} + 3\ \mu\text{F}} \right) \times 80\ \mu\text{C}$

$q_{\text{top}, 3\mu\text{F}} = \frac{3}{5} \times 80\ \mu\text{C} = +48\ \mu\text{C}$

Thus, the charge on the upper plate of the $3\ \mu\text{F}$ capacitor in the steady state is $+48\ \mu\text{C}$.

Correct Answer: (C) $+48\ \mu\text{C}$