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.
Step 1: Find the initial charge on $C_1$. Capacitor $C_1 = 10\,\mu\text{F}$ is charged by a $9\,\text{V}$ battery. The charge stored on it is: \[ Q_i = C_1 V = 10 \times 10^{-6} \times 9 = 90 \times 10^{-6}\,\text{C} = 9.0 \times 10^{-5}\,\text{C} \] Step 2: Understand charge conservation after connection. When the battery is removed and $C_1$ is connected to the uncharged $C_2 = 20\,\mu\text{F}$, the total charge is conserved (it simply redistributes between the two capacitors). The total charge remains $Q_i = 9.0 \times 10^{-5}\,\text{C}$. Step 3: Apply the equilibrium condition. At equilibrium, both capacitors reach the same potential difference $V_f$ (because they are connected in parallel). So: \[ Q_i = (C_1 + C_2) V_f \] \[ 9.0 \times 10^{-5} = (10 + 20) \times 10^{-6} \times V_f \] Step 4: Calculate the final voltage. \[ V_f = \frac{9.0 \times 10^{-5}}{30 \times 10^{-6}} = \frac{9.0}{3.0} = 3\,\text{V} \] Step 5: Find the charge on $C_2$. The charge on $C_2$ at equilibrium is: \[ Q_2 = C_2 V_f = 20 \times 10^{-6} \times 3 = 60 \times 10^{-6}\,\text{C} = 6.0 \times 10^{-5}\,\text{C} \] Step 6: State the final answer. The charge on $C_2$ after equilibrium is: \[ \boxed{6.0 \times 10^{-5}\,\text{C}} \]
