Consider the circuit shown in the figure. The capacitors C1 and C2 have capacitances \(2\ \mu\text{F}\) and \(8\ \mu\text{F}\), respectively. The switch can connect point X to either Y or Z. Initially XY is connected until the capacitor is fully charged by the battery. Then the switch connects X and Z, and the final charges on C1 and C2 are \(Q_1\) and \(Q_2\), respectively. What is the value of the ratio \(\frac{Q_2}{Q_1 + Q_2}\)?
Show Hint
When capacitors are connected in parallel, the charge distributes in direct proportion to their capacitances (\(Q \propto C\)).
Thus, the fraction of charge on \(C_2\) is simply \(\frac{C_2}{C_1 + C_2} = \frac{8}{2+8} = \frac{4}{5}\).
You do not need to calculate the actual voltages or intermediate charges.
Step 1: Understanding the Question:
This is a standard charge-sharing problem involving capacitors.
Initially, C1 is charged by a battery. It is then connected in parallel with C2.
We need to find the fraction of the total charge that ends up on capacitor C2. Step 2: Key Formula or Approach:
• Charge on a capacitor:
\[ Q = C V \]
• When connected in parallel, the capacitors share a common potential \(V_f\).:
\[ Q_1 = C_1 V_f \]
\[ Q_2 = C_2 V_f \]
• The ratio of the charge on one capacitor to the total charge is:
\[ \frac{Q_2}{Q_1 + Q_2} = \frac{C_2 V_f}{C_1 V_f + C_2 V_f} = \frac{C_2}{C_1 + C_2} \]
Step 3: Detailed Explanation:
• Let the electromotive force (EMF) of the battery be \(V\).
• When X is connected to Y, C1 is fully charged to the potential \(V\).:
\[ Q_{\text{initial}} = C_1 V \]
• When the switch is moved to connect X to Z, the battery is disconnected, and C1 is placed in parallel with the uncharged capacitor C2.
• The total charge is conserved:
\[ Q_{\text{total}} = Q_1 + Q_2 = Q_{\text{initial}} \]
• Let the final common potential across both capacitors be \(V_f\).:
\[ V_f = \frac{Q_{\text{total}}}{C_1 + C_2} \]
• The charge on C2 is:
\[ Q_2 = C_2 V_f \]
• The desired ratio is:
\[ \frac{Q_2}{Q_1 + Q_2} = \frac{C_2}{C_1 + C_2} \]
• Substituting the values \(C_1 = 2\ \mu\text{F}\) and \(C_2 = 8\ \mu\text{F}\).:
\[ \frac{Q_2}{Q_1 + Q_2} = \frac{8}{2 + 8} = \frac{8}{10} = \frac{4}{5} \]
Step 4: Final Answer:
The ratio \(\frac{Q_2}{Q_1 + Q_2}\) is \(\frac{4}{5}\).
