Let's solve this using the direct formula.

Formula:

The final common voltage

$V$

across the capacitors after they are connected is given by:

$$V = \frac{C_1 V_1 - C_2 V_2}{C_1 + C_2}$$

Step 1: Given Data

• 
  $C_1 = 1 \, \mu\text{F}, \, Q_1 = 1 \, \mu\text{C} \Rightarrow V_1 = \frac{Q_1}{C_1} = \frac{1}{1} = 1 \, \text{V}$ 
• 
  $C_2 = 2 \, \mu\text{F}, \, Q_2 = 4 \, \mu\text{C} \Rightarrow V_2 = \frac{Q_2}{C_2} = \frac{4}{2} = 2 \, \text{V}$ 

Step 2: Final Voltage

Substitute into the formula:

$$V = \frac{C_1 V_1 - C_2 V_2}{C_1 + C_2} = \frac{1 \cdot 1 - 2 \cdot 2}{1 + 2} = \frac{1 - 4}{3} = \frac{-3}{3} = -1 \, \text{V}.$$

(Note: The negative sign indicates the direction of potential difference.)

Step 3: Final Charges

The final charges on the capacitors are calculated using

$Q = C \cdot V$

:

• For
  $C_1$ 

  : $Q_1' = C_1 \cdot V = 1 \cdot (-1) = -1 \, \mu\text{C}$ 

  , but in magnitude, $Q_1' = 1 \, \mu\text{C}$ 

  .

• For
  $C_2$ 

  : $Q_2' = C_2 \cdot V = 2 \cdot (-1) = -2 \, \mu\text{C}$ 

  , but in magnitude, $Q_2' = 2 \, \mu\text{C}$ 

  .

Final Answer:

The final charges are:

$$\boxed{1 \, \mu\text{C} \, \text{and} \, 2 \, \mu\text{C}}.$$

Initially, the uncharged ball is attracted to the positive plate by induction. Upon touching, it acquires positive charge and is repelled by the positive plate, moving towards the negative plate. Upon touching the negative plate, it acquires negative charge and is repelled by the negative plate, moving back to the positive plate. This continuous charge transfer and repulsion causes the ball to oscillate between the plates, touching each in turn.

To handle \( 1000\text{V} \), \( N_s = \frac{1000}{250} = 4 \) capacitors must be connected in series. The capacitance of one row is \( C_s = \frac{8}{4} = 2\ \mu\text{F} \). To get a total capacitance of \( 16\ \mu\text{F} \), we need \( N_p = \frac{16}{2} = 8 \) parallel rows. Total number \( = N_s \times N_p = 4 \times 8 = 32 \).

Initially, \( C_1(120) = C_2(80) \Rightarrow C_2 = 1.5 C_1 \). Finally, with parallel connection: \( (C_1 + 2)(40) = C_2(160) \). Substituting \( C_2 = 1.5 C_1 \) gives \( 40 C_1 + 80 = 240 C_1 \Rightarrow C_1 = 0.4\ \mu\text{F} \) and \( C_2 = 0.6\ \mu\text{F} \).

When connected in series, the charges do not change, so the total energy stored remains the same: \(U = N \times \left(\frac{1}{2} C V^2\right)\). However, the individual potential differences add up, so the total potential difference becomes \(NV\).

Charge on A is \(15 \mu\text{F} \times 100\text{ V} = 1500 \mu\text{C}\) and on B is \(1 \mu\text{F} \times 100\text{ V} = 100 \mu\text{C}\). When dielectric of A is removed, its capacitance becomes \(1 \mu\text{F}\). The common potential is \(V_c = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{1500 + 100}{1 + 1} = 800\text{ V}\).