Three capacitors having capacitances of `8.4 muF, 8.4 muF` and `4.2 muF` are connected in series across a 36 potential difference.
(a) What is the charge on `4.2muF` capacitor?
(b) What is the total energy stored in all three capacitors?
(c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination?
(d) What is the total energy now stored in the capacitors?

Explanation

To solve the problem step by step, we will address each part of the question systematically.

### Given:
- Capacitors: \( C_1 = 8.4 \, \mu F \), \( C_2 = 8.4 \, \mu F \), \( C_3 = 4.2 \, \mu F \)
- Voltage across the capacitors: \( V = 36 \, V \)

### (a) Charge on the \( 4.2 \, \mu F \) capacitor

1. **Calculate the equivalent capacitance for capacitors in series:**
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
\[ \frac{1}{C_{eq}} = \frac{1}{8.4} + \frac{1}{8.4} + \frac{1}{4.2} \]
\[ \frac{1}{C_{eq}} = \frac{1}{8.4} + \frac{1}{8.4} + \frac{2}{8.4} = \frac{4}{8.4} = \frac{2}{4.2} \]
\[ C_{eq} = 2.1 \, \mu F \]

2. **Calculate the total charge \( Q \) on the equivalent capacitor:**
\[ Q = C_{eq} \times V = 2.1 \, \mu F \times 36 \, V = 75.6 \, \mu C \]

3. **Since the charge is the same on all capacitors in series, the charge on the \( 4.2 \, \mu F \) capacitor is:**
\[ Q_{4.2} = 75.6 \, \mu C \]

### (b) Total energy stored in all three capacitors

1. **Use the formula for energy stored in a capacitor:**
\[ U = \frac{1}{2} C_{eq} V^2 \]
\[ U = \frac{1}{2} \times 2.1 \times 10^{-6} \, F \times (36 \, V)^2 \]
\[ U = \frac{1}{2} \times 2.1 \times 10^{-6} \times 1296 \]
\[ U = 1.36 \, mJ \]

### (c) Voltage across each capacitor in the parallel combination

1. **When the capacitors are disconnected and reconnected in parallel, the charge on each capacitor remains the same:**
- Charge on \( C_1 \) and \( C_2 \): \( Q_1 = Q_2 = 75.6 \, \mu C \)
- Charge on \( C_3 \): \( Q_3 = 75.6 \, \mu C \)

2. **Calculate the voltage across each capacitor using \( V = \frac{Q}{C} \):**
- For \( C_1 \) and \( C_2 \):
\[ V_1 = V_2 = \frac{Q_1}{C_1} = \frac{75.6 \, \mu C}{8.4 \, \mu F} = 9 \, V \]
- For \( C_3 \):
\[ V_3 = \frac{Q_3}{C_3} = \frac{75.6 \, \mu C}{4.2 \, \mu F} = 18 \, V \]

### (d) Total energy now stored in the capacitors

1. **Calculate the total energy stored in the parallel combination:**
\[ U_{total} = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 + \frac{1}{2} C_3 V_3^2 \]
\[ U_{total} = \frac{1}{2} \times 8.4 \times 10^{-6} \times (9)^2 + \frac{1}{2} \times 8.4 \times 10^{-6} \times (9)^2 + \frac{1}{2} \times 4.2 \times 10^{-6} \times (18)^2 \]
\[ U_{total} = 2 \times \frac{1}{2} \times 8.4 \times 10^{-6} \times 81 + \frac{1}{2} \times 4.2 \times 10^{-6} \times 324 \]
\[ U_{total} = 0.0003402 + 0.0006804 = 1.02 \, mJ \]

### Summary of Answers:
(a) Charge on \( 4.2 \, \mu F \) capacitor: \( 75.6 \, \mu C \)
(b) Total energy stored: \( 1.36 \, mJ \)
(c) Voltage across each capacitor in parallel: \( 9 \, V \) for \( C_1 \) and \( C_2 \), \( 18 \, V \) for \( C_3 \)
(d) Total energy in parallel: \( 1.02 \, mJ \)