A `4.00 muF` capacitor and a `6.00muF` capacitor are connected in parallel across a `660 V` supply line
(a) Find the charge on each capacitor and the voltage across each.
(b) The charged capacitors are disconnected from the line and from each other, and reconnected to each other with terminals of unlike sign together. Find the final charge on each and the voltage across each.

Explanation

### Solution:

**(a)** To find the charge on each capacitor and the voltage across each:

1. **Identify the capacitances and voltage:**
- Capacitance of capacitor 1, \( C_1 = 4.00 \, \mu F = 4.00 \times 10^{-6} \, F \)
- Capacitance of capacitor 2, \( C_2 = 6.00 \, \mu F = 6.00 \times 10^{-6} \, F \)
- Voltage across both capacitors, \( V = 660 \, V \)

2. **Calculate the charge on capacitor 1:**
\[ Q_1 = C_1 \times V = 4.00 \times 10^{-6} \, F \times 660 \, V = 2.64 \times 10^{-3} \, C = 2640 \, \mu C \]

3. **Calculate the charge on capacitor 2:**
\[ Q_2 = C_2 \times V = 6.00 \times 10^{-6} \, F \times 660 \, V = 3.96 \times 10^{-3} \, C = 3960 \, \mu C \]

4. **Voltage across each capacitor:**
Since the capacitors are connected in parallel, the voltage across each capacitor is the same:
\[ V_{C1} = V_{C2} = 660 \, V \]

**Final Results for (a):**
- Charge on capacitor 1, \( Q_1 = 2640 \, \mu C \)
- Charge on capacitor 2, \( Q_2 = 3960 \, \mu C \)
- Voltage across each capacitor, \( V = 660 \, V \)

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**(b)** To find the final charge on each capacitor and the voltage across each after they are disconnected and reconnected with unlike terminals together:

1. **Initial charges:**
- Charge on capacitor 1, \( Q_1 = 2640 \, \mu C \) (positive)
- Charge on capacitor 2, \( Q_2 = -3960 \, \mu C \) (negative)

2. **Calculate the total charge when they are connected:**
\[ Q_{\text{total}} = Q_1 + Q_2 = 2640 \, \mu C - 3960 \, \mu C = -1320 \, \mu C \]

3. **Let the final voltage across both capacitors be \( V_f \).**
- Charge on capacitor 1 after reconnection:
\[ Q_{1f} = C_1 \times V_f = 4.00 \times 10^{-6} \, F \times V_f \]
- Charge on capacitor 2 after reconnection:
\[ Q_{2f} = C_2 \times V_f = 6.00 \times 10^{-6} \, F \times V_f \]

4. **Apply conservation of charge:**
\[ Q_{\text{total}} = Q_{1f} + Q_{2f} \]
\[ -1320 \, \mu C = 4.00 \times 10^{-6} \, F \times V_f + 6.00 \times 10^{-6} \, F \times V_f \]
\[ -1320 \, \mu C = (4 + 6) \times 10^{-6} \, F \times V_f = 10 \times 10^{-6} \, F \times V_f \]

5. **Solve for \( V_f \):**
\[ V_f = \frac{-1320 \, \mu C}{10 \times 10^{-6} \, F} = -132 \, V \]

6. **Calculate final charges:**
- Charge on capacitor 1:
\[ Q_{1f} = 4.00 \times 10^{-6} \, F \times (-132 \, V) = -528 \, \mu C \]
- Charge on capacitor 2:
\[ Q_{2f} = 6.00 \times 10^{-6} \, F \times (-132 \, V) = -792 \, \mu C \]

**Final Results for (b):**
- Final charge on capacitor 1, \( Q_{1f} = -528 \, \mu C \)
- Final charge on capacitor 2, \( Q_{2f} = -792 \, \mu C \)
- Final voltage across each capacitor, \( V_f = -132 \, V \)

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