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: Find the initial charge on \(C_1\).
Initially, capacitor \(C_1\) is charged by a \(9\text{ V}\) battery.
\[ C_1=1\,\mu\text{F}=1\times10^{-6}\text{ F} \]
Initial charge is
\[ Q=C_1V \]
\[ Q=(1\times10^{-6})(9) \]
\[ Q=9\times10^{-6}\text{ C} \]

Step 2: Find the equivalent capacitance after connection.
After removing the battery, \(C_1,C_2,C_3\) are connected in parallel.
So, total capacitance is
\[ C_{\text{eq}}=C_1+C_2+C_3 \]
\[ C_{\text{eq}}=1+2+3 \]
\[ C_{\text{eq}}=6\,\mu\text{F} \]

Step 3: Find the common potential difference.
Since the system is isolated, total charge remains conserved.
\[ V=\frac{Q}{C_{\text{eq}}} \]
\[ V=\frac{9\,\mu\text{C}}{6\,\mu\text{F}} \]
\[ V=1.5\text{ V} \]

Step 4: Find the charge on \(C_3\).
\[ C_3=3\,\mu\text{F} \]
\[ Q_3=C_3V \]
\[ Q_3=(3\,\mu\text{F})(1.5\text{ V}) \]
\[ Q_3=4.5\,\mu\text{C} \]
\[ Q_3=4.5\times10^{-6}\text{ C} \]

Step 5: Final conclusion.
Hence, the charge on \(C_3\) is
\[ \boxed{4.5\times10^{-6}\text{ C}} \]