Question:
A capacitor of capacitance \(C_1=10\,\mu\text{F}\) is charged using \(9\,\text{V}\) battery. It is then removed from the battery and connected to another capacitor \(C_2=20\,\mu\text{F}\) as shown in the figure. The charge on \(C_2\) after equilibrium has reached is
A capacitor of capacitance \(C_1=10\,\mu\text{F}\) is charged using \(9\,\text{V}\) battery. It is then removed from the battery and connected to another capacitor \(C_2=20\,\mu\text{F}\) as shown in the figure. The charge on \(C_2\) after equilibrium has reached is
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When a charged capacitor is disconnected from a battery and connected to another capacitor, total charge is conserved and the final potential becomes common.
Updated On: Jun 26, 2026
- \(6.0\times 10^{-5}\,\text{C}\)
- \(6.0\times 10^{-6}\,\text{C}\)
- \(3.0\times 10^{-5}\,\text{C}\)
- \(3.0\times 10^{-6}\,\text{C}\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the initial charge on \(C_1\).
Initially, \[ C_1=10\,\mu\text{F}=10\times 10^{-6}\,\text{F} \] and \[ V=9\,\text{V} \] The initial charge on \(C_1\) is \[ Q=C_1V \] \[ Q=(10\times 10^{-6})(9) \] \[ Q=90\times 10^{-6}\,\text{C} \] \[ Q=9.0\times 10^{-5}\,\text{C} \]
Step 2: Use charge sharing after connection.
After removing the battery, \(C_1\) is connected to \(C_2\).
The total charge remains conserved.
At equilibrium, both capacitors have the same potential difference \(V_f\).
So, \[ Q_{\text{total}}=(C_1+C_2)V_f \] Given, \[ C_2=20\,\mu\text{F}=20\times 10^{-6}\,\text{F} \] Thus, \[ 9.0\times 10^{-5}=(10\times 10^{-6}+20\times 10^{-6})V_f \] \[ 9.0\times 10^{-5}=30\times 10^{-6}V_f \] \[ V_f=3\,\text{V} \]
Step 3: Find the charge on \(C_2\).
The charge on \(C_2\) is \[ Q_2=C_2V_f \] \[ Q_2=(20\times 10^{-6})(3) \] \[ Q_2=60\times 10^{-6}\,\text{C} \] \[ Q_2=6.0\times 10^{-5}\,\text{C} \]
Step 4: Final conclusion.
Hence, the charge on \(C_2\) after equilibrium is \[ \boxed{6.0\times 10^{-5}\,\text{C}} \]
Initially, \[ C_1=10\,\mu\text{F}=10\times 10^{-6}\,\text{F} \] and \[ V=9\,\text{V} \] The initial charge on \(C_1\) is \[ Q=C_1V \] \[ Q=(10\times 10^{-6})(9) \] \[ Q=90\times 10^{-6}\,\text{C} \] \[ Q=9.0\times 10^{-5}\,\text{C} \]
Step 2: Use charge sharing after connection.
After removing the battery, \(C_1\) is connected to \(C_2\).
The total charge remains conserved.
At equilibrium, both capacitors have the same potential difference \(V_f\).
So, \[ Q_{\text{total}}=(C_1+C_2)V_f \] Given, \[ C_2=20\,\mu\text{F}=20\times 10^{-6}\,\text{F} \] Thus, \[ 9.0\times 10^{-5}=(10\times 10^{-6}+20\times 10^{-6})V_f \] \[ 9.0\times 10^{-5}=30\times 10^{-6}V_f \] \[ V_f=3\,\text{V} \]
Step 3: Find the charge on \(C_2\).
The charge on \(C_2\) is \[ Q_2=C_2V_f \] \[ Q_2=(20\times 10^{-6})(3) \] \[ Q_2=60\times 10^{-6}\,\text{C} \] \[ Q_2=6.0\times 10^{-5}\,\text{C} \]
Step 4: Final conclusion.
Hence, the charge on \(C_2\) after equilibrium is \[ \boxed{6.0\times 10^{-5}\,\text{C}} \]
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