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:

$$\boxed{32}$$

Let's verify and calculate the correct answer step-by-step:

Given Data:

• Capacitances:
  $C_1 = 2 \, \mu\text{F}, C_2 = 3 \, \mu\text{F}, C_3 = 5 \, \mu\text{F}$ 
• Maximum voltages:
  $V_1 = 3 \, \text{V}, V_2 = 2 \, \text{V}, V_3 = 1 \, \text{V}$ 

Step 1: Maximum charge each capacitor can store:

$$Q_1 = C_1 \cdot V_1 = 2 \cdot 3 = 6 \, \mu\text{C}$$

$$Q_2 = C_2 \cdot V_2 = 3 \cdot 2 = 6 \, \mu\text{C}$$

$$Q_3 = C_3 \cdot V_3 = 5 \cdot 1 = 5 \, \mu\text{C}$$

The capacitor with the minimum charge capacity limits the system. Here,

$Q_{\text{max}} = 5 \, \mu\text{C}$

, dictated by

$C_3$

.

Step 2: Voltage distribution across each capacitor:

In series, charge

$Q$

is the same on all capacitors, and the voltage across each capacitor is:

$$V_1 = \frac{Q}{C_1}, \quad V_2 = \frac{Q}{C_2}, \quad V_3 = \frac{Q}{C_3}$$

Total voltage across the series combination:

$$V_{\text{total}} = V_1 + V_2 + V_3$$

Substitute

$Q = 5 \, \mu\text{C}$

:

$$V_1 = \frac{5}{2} = 2.5 \, \text{V}, \quad V_2 = \frac{5}{3} \approx 1.67 \, \text{V}, \quad V_3 = \frac{5}{5} = 1 \, \text{V}$$

Step 3: Total voltage:

$$V_{\text{total}} = V_1 + V_2 + V_3 = 2.5 + 1.67 + 1 = \frac{15}{6} + \frac{10}{6} + \frac{6}{6} = \frac{31}{6} \, \text{V}.$$

Final Answer:

The maximum voltage the series combination can withstand is:

$$\boxed{\frac{31}{6} \, \text{V}} \approx 5.17 \, \text{V}.$$

The problem involves two spherical conductors

$A_1$

and

$A_2$

connected by a copper wire. Let’s analyze and compute the equivalent capacitance of the system.

Given:

• 
  $A_1$ 

  and $A_2$ 

  are concentric spherical conductors.

• Radii of the spheres:
  $r_1$ 

  (inner) and $r_2$ 

  (outer).

• The medium is air, so the permittivity is
  $\varepsilon_0$ 

  .

Key Concepts:

1. Potential Difference Between the Spheres: The two conductors are connected by a wire, meaning they are at the same potential. As a result, the electric field exists only between the two spheres.
2. Capacitance of a Single Isolated Sphere: If only
   $A_2$ 

   existed as a spherical conductor, its capacitance would be: 

   $$C_{\text{single}} = 4 \pi \varepsilon_0 r_2.$$ 

3. Why the System is Equivalent to an Isolated Sphere: Since
   $A_1$ 

   is connected to $A_2$ 

   via a conducting wire, any charge added to $A_1$ 

   immediately flows to $A_2$ 

   , making the system behave as if there is only one conductor of radius $r_2$ 

   .

Equivalent Capacitance:

Thus, the capacitance of the system is:

$$C_{\text{equivalent}} = 4 \pi \varepsilon_0 r_2.$$

Final Answer:

The equivalent capacitance of the system is:

$$\boxed{4 \pi \varepsilon_0 r_2}.$$

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

First Branch has a capacitance of 1F , for second branch it is 1/2F , for third branch it is 1/4 F and so, on . As all these branches are in parallel

\[ C_{eq}=C_{1}+C_{2}+C_{3}+....\]

\[ C_{eq}= 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8}+....=\frac{1}{1-\frac{1}{2}}=2\mu F\]

To find the equivalent capacitance for this cubical capacitor network, where each edge of the cube has a capacitance

$C$

, here’s the shortest solution:

Step-by-Step:

1. Symmetry analysis:
   • By symmetry, all corners of the cube can be grouped into equivalent potential nodes.
   • The cube's symmetry allows reduction to a simpler circuit.
2. Key nodes:
   • Node
     $A$ 

     is connected to one corner of the cube.

   • Node
     $B$ 

     is connected to the diagonally opposite corner.

3. Effective connections:
   • Due to symmetry, three capacitors are effectively in parallel between
     $A$ 

     and an intermediate point.

   • Similarly, three capacitors are effectively in parallel between
     $B$ 

     and the same intermediate point.

   • Two capacitors remain directly between
     $A$ 

     and $B$ 

     .

4. Simplification:
   • The three parallel capacitors at each node result in:
     $$C_{\text{parallel}} = 3C$$ 
   • The equivalent circuit becomes two
     $3C$ 

     capacitors in series with a $2C$ 

     capacitor: $$\text{Series combination:} \quad \frac{1}{C_{\text{eq}}} = \frac{1}{3C} + \frac{1}{3C} + \frac{1}{2C}$$ 

5. Calculation:
   • Combine series:
     $$\frac{1}{C_{\text{eq}}} = \frac{2}{3C} + \frac{1}{2C} = \frac{4}{6C} + \frac{3}{6C} = \frac{7}{6C}$$ 
   • Invert to find
     $C_{\text{eq}}$ 

     : $$C_{\text{eq}} = \frac{6C}{7} \times 2 = \frac{12C}{7}$$ 

Thus, the equivalent capacitance is:

$$C_{\text{eq}} = \frac{12C}{7}$$

Let's break down the situation step by step:

Given:

• A capacitor is charged using a battery and then disconnected (so no current can flow after disconnection).
• A dielectric slab is inserted between the plates of the capacitor after disconnecting the battery.

Key concepts:

• Capacitance with Dielectric: When a dielectric slab is inserted, the capacitance of the capacitor increases. The new capacitance
  $C'$ 

  is related to the original capacitance $C$ 

  by the dielectric constant $K$ 

  : 

  $$C' = K \cdot C$$where

  $K$is the dielectric constant of the material.

• Charge on the Plates: Since the capacitor is disconnected from the battery, no additional charge can flow onto the plates. Thus, the charge
  $Q$ 

  remains the same, given by: 

  $$Q = C \cdot V$$where

  $V$is the potential difference across the plates. Since the charge remains constant, the equation becomes:

   

  $$Q = C' \cdot V'$$where

  $V'$is the new potential difference across the plates.

• Potential Difference: Since the capacitance increases and the charge stays constant, the potential difference
  $V'$ 

  must decrease (because $Q = C' \cdot V'$ 

  and $C' > C$ 

  ).

• Stored Energy: The energy stored in a capacitor is given by: 

  $$U = \frac{Q^2}{2C}$$Since the capacitance increases and the charge is constant, the stored energy

  $U$decreases, as it is inversely proportional to the capacitance.

Conclusion:

• Decrease in potential difference: The potential difference across the plates decreases because the capacitance increases while the charge remains constant.
• Reduction in stored energy: The energy stored in the capacitor decreases because the capacitance increases.
• No change in charge: The charge on the plates remains the same since the capacitor is disconnected from the battery.

Thus, the correct answer is: "Decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates."