For Such Question start solving from the farthest point. The circuit will keep on reducing.

Capacitors circled in the diagram are short circuited so, they can be removed from the circuit.

To solve this, we determine the combination of capacitors required to achieve the desired capacitance and voltage.

Given:

• Individual capacitor:
  $C = 8 \, \mu\text{F}$ 

  , $V_{\text{max}} = 250 \, \text{V}$ 

• Desired combination:
  $C_{\text{req}} = 16 \, \mu\text{F}$ 

  , $V_{\text{req}} = 1000 \, \text{V}$ 

Step 1: Voltage requirement

To achieve

$V_{\text{req}} = 1000 \, \text{V}$

, multiple capacitors must be connected in series because the voltage across a series combination adds up. The number of capacitors required in series is:

$$n = \frac{V_{\text{req}}}{V_{\text{max}}} = \frac{1000}{250} = 4$$

Thus, 4 capacitors in series are required to handle 1000 V.

Step 2: Capacitance in series

The effective capacitance of

$n$

capacitors in series is given by:

$$C_{\text{series}} = \frac{C}{n} = \frac{8}{4} = 2 \, \mu\text{F}$$

So, a series of 4 capacitors provides

$C_{\text{series}} = 2 \, \mu\text{F}$

.

Step 3: Capacitance requirement

To achieve

$C_{\text{req}} = 16 \, \mu\text{F}$

, multiple such series groups must be connected in parallel because capacitance in parallel adds up. The number of such series groups required is:

$$m = \frac{C_{\text{req}}}{C_{\text{series}}} = \frac{16}{2} = 8$$

Thus, 8 series groups are required.

Step 4: Total capacitors

Each series group contains 4 capacitors, and there are 8 such groups. Therefore, the total number of capacitors is:

$$\text{Total capacitors} = n \cdot m = 4 \cdot 8 = 32$$

Final Answer:

The minimum number of capacitors required is:

$$\boxed{32}$$

When Switch is open C_eq= C/4 Charge given by the Battery is CE/4.

When Switch is open C_eq= C/3 Charge given by the Battery is CE/3.

Extra Charge flowing through the circuit it = \( \frac{CE}{3}-\frac{CE}{4}= \frac{CE}{12}\)

Circircled ones are in parallel

Here’s a shorter solution:

Given:

• 
  $V = 100 \, \text{V}$ 

  (total battery voltage)

• Dielectric constant
  $K = 3$ 

  for capacitor $A$ 

• Identical capacitors
  $A$ 

  and $B$ 

Step 1: Capacitance

• 
  $C_A = 3C_0$ 

  (because of the dielectric in $A$ 

  )

• 
  $C_B = C_0$ 

  (no dielectric in $B$ 

  )

Step 2: Total capacitance in series:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_A} + \frac{1}{C_B} = \frac{1}{3C_0} + \frac{1}{C_0} = \frac{4}{3C_0}$$

$$C_{\text{eq}} = \frac{3C_0}{4}$$

Step 3: Voltage division:

The voltage is divided in proportion to the inverse of capacitances:

$$V_A = \frac{C_B}{C_A + C_B} \times 100 = \frac{C_0}{3C_0 + C_0} \times 100 = \frac{1}{4} \times 100 = 25 \, \text{V}$$

$$V_B = 100 - 25 = 75 \, \text{V}$$

Final Answer:

• 
  $V_A = 25 \, \text{V}$ 
• 
  $V_B = 75 \, \text{V}$ 

To calculate the work required to separate the plates of the capacitor, let’s analyze the situation step by step:

1. Initial Setup

• The capacitor has:
  • Plate area =
    $A$ 
  • Initial separation =
    $d$ 
  • Initial potential difference =
    $V$ 
• After charging, the battery is disconnected, so the charge
  $Q$ 

  on the plates remains constant.

The charge stored in the capacitor is given by:

$$Q = C V$$

where

$C$

is the capacitance:

$$C = \frac{\varepsilon_0 A}{d}$$

Thus, the charge is:

$$Q = \frac{\varepsilon_0 A}{d} V$$

2. Energy Stored in the Capacitor

The energy stored in the capacitor is:

$$U = \frac{1}{2} C V^2$$

Substituting

$C = \frac{\varepsilon_0 A}{d}$

:

$$U = \frac{1}{2} \cdot \frac{\varepsilon_0 A}{d} \cdot V^2$$

Initially, the energy is:

$$U_{\text{initial}} = \frac{\varepsilon_0 A V^2}{2d}$$

When the separation is increased to

$2d$

, the capacitance decreases to:

$$C_{\text{new}} = \frac{\varepsilon_0 A}{2d}$$

The energy becomes:

$$U_{\text{final}} = \frac{1}{2} C_{\text{new}} V_{\text{new}}^2$$

Since the charge

$Q$

is constant and

$Q = C_{\text{new}} V_{\text{new}}$

:

$$V_{\text{new}} = \frac{Q}{C_{\text{new}}} = \frac{\frac{\varepsilon_0 A}{d} V}{\frac{\varepsilon_0 A}{2d}} = 2V$$

Substitute

$C_{\text{new}}$

and

$V_{\text{new}}$

:

$$U_{\text{final}} = \frac{1}{2} \cdot \frac{\varepsilon_0 A}{2d} \cdot (2V)^2$$

$$U_{\text{final}} = \frac{\varepsilon_0 A V^2}{2d}$$

3. Work Done to Separate the Plates

The work done to separate the plates is equal to the increase in energy:

$$W = U_{\text{final}} - U_{\text{initial}}$$

Substitute the values:

$$W = \frac{\varepsilon_0 A V^2}{2d} - \frac{\varepsilon_0 A V^2}{2d}$$

$$W = \frac{\varepsilon_0 A V^2}{2d}$$

Final Answer:

The work required to separate the plates is:

$$W = \frac{\varepsilon_0 A V^2}{2d}$$

Let's analyze the problem step by step and derive why the potential difference is 10 V.

1. Capacitors in Parallel (Initial Condition)

When the 10 capacitors, each with capacitance

$C$

, are connected in parallel:

• The equivalent capacitance is:
  $$C_{\text{parallel}} = 10C$$ 
• The total charge stored when connected to a battery of potential
  $V$ 

  is: $$Q = C_{\text{parallel}} \cdot V = (10C) \cdot V = 10CV$$ 

2. Capacitors in Series (Reconfigured System)

After disconnecting the battery, the capacitors are joined in series:

• The equivalent capacitance for 10 capacitors in series is:
  $$C_{\text{series}} = \frac{C}{10}$$ 
• The charge
  $Q$ 

  stored on the series combination remains the same (as charge is conserved): $$Q_{\text{series}} = Q = 10CV$$ 

3. Potential Across the Series Combination

The potential difference across a capacitor or combination of capacitors is related to the charge

$Q$

and capacitance

$C$

:

$$V_{\text{series}} = \frac{Q}{C_{\text{series}}}$$

Substitute

$Q = 10CV$

and

$C_{\text{series}} = \frac{C}{10}$

:

$$V_{\text{series}} = \frac{10CV}{\frac{C}{10}} = V$$

Thus, the potential across the series combination is

$V = 10V$

,.

Final Answer:

The potential of the series combination is:

$$\boxed{10V}$$

We need to check each option separately. We 5 capacitors are connected  in parallel, 2 capacitors are connected in series.

C_eq= (5C×C/2)/ (5C+C/2)= 5C/11 = 5×2/11 = 10/11 μF.