Question

Four capacitors $C_1, C_2, C_3$ and $C_4$ of capacitance $2 \mu F, 3 \mu F, 4 \mu F$ and $8 \mu F$ respectively are connected in a circuit as shown. The capacitor $C_2$ is initially charged to a potential difference 20V. Now the switch is closed. A long time after, the charge (in $\mu C$) on the capacitor $C_4$ is _____.

Answer 1

\frac{480}{17}

Answer 2

The problem describes a circuit with four capacitors $C_1, C_2, C_3,$ and $C_4$. Initially, only $C_2$ is charged, and then a switch is closed, connecting all capacitors in parallel. We need to find the charge on $C_4$ after a long time, when the circuit reaches a steady state.

1. Identify Initial Conditions:

• Capacitances: $C_1 = 2 \mu F$, $C_2 = 3 \mu F$, $C_3 = 4 \mu F$, $C_4 = 8 \mu F$.
• Initial potential difference across $C_2$: $V_2 = 20 V$.
• Initial charge on $C_2$: $Q_{2,initial} = C_2 V_2 = (3 \mu F)(20 V) = 60 \mu C$. The problem states the top plate is positive and the bottom plate is negative.
• Capacitors $C_1, C_3, C_4$ are initially uncharged, so $Q_{1,initial} = Q_{3,initial} = Q_{4,initial} = 0$.

2. Analyze the Circuit After Closing the Switch: When the switch S is closed, all four capacitors become connected in parallel. This means their top plates are connected to a common node (let's call it A) and their bottom plates are connected to another common node (let's call it B). In a steady state, the potential difference across each capacitor will be the same, let's call it $V_f$.

3. Apply Principle of Charge Conservation: Consider the isolated system of the top plates of all four capacitors. Before the switch is closed, only the top plate of $C_2$ has a positive charge ($+60 \mu C$). The top plates of $C_1, C_3, C_4$ are uncharged. Initial total charge on the top plates: $Q_{top,initial} = Q_{1,initial,top} + Q_{2,initial,top} + Q_{3,initial,top} + Q_{4,initial,top}$ $Q_{top,initial} = 0 + (+60 \mu C) + 0 + 0 = +60 \mu C$.

After the switch is closed, charge redistributes, and the final charge on the top plate of each capacitor $C_i$ will be $Q_{i,final,top} = C_i V_f$. Final total charge on the top plates: $Q_{top,final} = C_1 V_f + C_2 V_f + C_3 V_f + C_4 V_f = (C_1 + C_2 + C_3 + C_4) V_f$.

By the principle of charge conservation, the total charge on the isolated system of top plates must remain constant: $Q_{top,initial} = Q_{top,final}$ $60 \mu C = (C_1 + C_2 + C_3 + C_4) V_f$

4. Calculate the Final Potential Difference ($V_f$): First, calculate the total equivalent capacitance of the parallel combination: $C_{eq} = C_1 + C_2 + C_3 + C_4 = 2 \mu F + 3 \mu F + 4 \mu F + 8 \mu F = 17 \mu F$.

Now, substitute this into the charge conservation equation: $60 \mu C = (17 \mu F) V_f$ $V_f = \frac{60}{17} V$.

5. Calculate the Charge on Capacitor $C_4$: The charge on $C_4$ in the final state is given by $Q_{C4} = C_4 V_f$. $Q_{C4} = (8 \mu F) \left( \frac{60}{17} V \right)$ $Q_{C4} = \frac{8 \times 60}{17} \mu C$ $Q_{C4} = \frac{480}{17} \mu C$.

This value can be approximated as $28.235 \mu C$.

The final answer is $\boxed{\frac{480}{17}}$.

Explanation of the solution:

1. Calculate initial charge on $C_2$: $Q_{initial} = C_2 V_2 = 3 \mu F \times 20 V = 60 \mu C$. Other capacitors are uncharged.
2. When the switch is closed, all capacitors are connected in parallel. The total initial charge from $C_2$ redistributes among all capacitors.
3. The equivalent capacitance of the parallel combination is $C_{eq} = C_1 + C_2 + C_3 + C_4 = 2+3+4+8 = 17 \mu F$.
4. The final common potential difference across all capacitors is $V_f = \frac{Q_{initial}}{C_{eq}} = \frac{60 \mu C}{17 \mu F} = \frac{60}{17} V$.
5. The charge on $C_4$ is $Q_4 = C_4 V_f = 8 \mu F \times \frac{60}{17} V = \frac{480}{17} \mu C$.