Question

A capacitor of capacitance \(C_1=1\,\mu\text{F}\) is charged using a \(9\text{ V}\) battery. \(C_1\) is then removed from the battery and connected to capacitors \(C_2\) and \(C_3\) of \(2\,\mu\text{F}\) and \(3\,\mu\text{F}\) respectively as shown in the figure. Find the charge on \(C_3\) after equilibrium has reached.

• A. \(4.5\times10^{-6}\,\text{C}\)
• B. \(3.5\times10^{-6}\,\text{C}\)
• C. \(2.5\times10^{-6}\,\text{C}\)
• D. \(1.5\times10^{-5}\,\text{C}\)

Hint:

When a charged capacitor is disconnected from the battery and then connected to other capacitors, total charge is conserved and all parallel capacitors acquire the same final potential.

Answer 1

The Correct Option is A

Step 1: Charge the first capacitor.
$C_1 = 1\,\mu\text{F}$ is charged by a $9\ \text{V}$ battery, storing \[ Q = C_1 V = (1\,\mu\text{F})(9\ \text{V}) = 9\,\mu\text{C}. \]
Step 2: See what happens after reconnection.
The battery is removed, so this $9\,\mu\text{C}$ is all the charge in the system. Now $C_1$, $C_2$ and $C_3$ share it in parallel (same voltage across each).
Step 3: Find the combined capacitance.
For capacitors in parallel, capacitances add: \[ C_{\text{eq}} = C_1 + C_2 + C_3 = 1 + 2 + 3 = 6\,\mu\text{F}. \]
Step 4: Use charge conservation for the common voltage.
The total charge is unchanged, so the shared potential is \[ V' = \frac{Q}{C_{\text{eq}}} = \frac{9\,\mu\text{C}}{6\,\mu\text{F}} = 1.5\ \text{V}. \]
Step 5: Charge on $C_3$.
\[ Q_3 = C_3 V' = (3\,\mu\text{F})(1.5\ \text{V}) = 4.5\,\mu\text{C}. \]
Step 6: Conclude.
So $C_3$ holds $4.5\times10^{-6}\ \text{C}$ at equilibrium.
\[ \boxed{4.5\times10^{-6}\ \text{C}} \]