To solve this, we will use the concept of common potential and energy conservation. Let's derive it step-by-step:

1. Initial Energy Stored in Capacitor 1
   Let the initial capacitance of the first capacitor be $C$ 

   , and the battery's voltage be $V$ 

   .
   The energy stored in the capacitor is: 

   $$U = \frac{1}{2} C V^2$$ 

2. When the second capacitor is connected
   After disconnecting the battery, an identical capacitor (with capacitance $C$ 

   ) is connected across the first one. The total charge remains conserved because the battery is removed. Let the final voltage be $V_f$ 

   .Total charge initially:

    

   $$Q_{\text{initial}} = CV$$After connecting the second capacitor, the total capacitance becomes:

    

   $$C_{\text{total}} = C + C = 2C$$Common potential

   $V_f$:

    

   $$V_f = \frac{\text{Total charge}}{\text{Total capacitance}} = \frac{CV}{2C} = \frac{V}{2}$$ 

3. Final Energy Stored in Both Capacitors
   The final energy stored in the system is the sum of the energy in both capacitors: 

   $$U_{\text{final}} = \frac{1}{2} C V_f^2 + \frac{1}{2} C V_f^2 = C V_f^2$$Substituting

   $V_f = \frac{V}{2}$:

    

   $$U_{\text{final}} = C \left( \frac{V}{2} \right)^2 = C \frac{V^2}{4}$$Simplifying:

    

   $$U_{\text{final}} = \frac{1}{2} \cdot C V^2 \cdot \frac{1}{2} = \frac{U}{2}$$ 

Thus, the final energy stored in the system is

$\frac{U}{2}$

.

When the distance between the plates of a charged capacitor is decreased after disconnecting the battery, the following happens:

1. Charge remains constant: Since the battery is disconnected, the charge
   $Q$ 

   on the plates does not change.

2. Capacitance increases: The capacitance of a parallel plate capacitor is given by: 

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

   $d$is the distance between the plates. As

   $d$decreases,

   $C$increases.

3. Potential difference decreases: The voltage
   $V$ 

   across the capacitor is related by: 

   $$V = \frac{Q}{C}$$Since

   $Q$is constant and

   $C$increases,

   $V$decreases.

4. Potential energy decreases: The energy stored in a capacitor is: 

   $$U = \frac{1}{2} \frac{Q^2}{C}$$As

   $C$increases,

   $U$decreases because

   $Q^2$is constant.

Thus, the potential energy decreases when the distance between the plates is reduced.

To solve this problem, let's analyze the situation step by step:

1. Initial Charges on Capacitors

The charge on each capacitor is given by

$Q = CV$

:

• For the first capacitor (
  $C_1 = 2 \, \mu\text{F}, V_1 = 10 \, \text{V}$ 

  ): 

  $$Q_1 = C_1 V_1 = 2 \times 10^{-6} \cdot 10 = 20 \, \mu\text{C}$$ 

• For the second capacitor (
  $C_2 = 3 \, \mu\text{F}, V_2 = 20 \, \text{V}$ 

  ): 

  $$Q_2 = C_2 V_2 = 3 \times 10^{-6} \cdot 20 = 60 \, \mu\text{C}$$ 

2. Connecting Capacitors with Opposite Polarity

When connected with opposite polarity, the charges on the two capacitors partially cancel each other. The net charge is:

$$Q_{\text{net}} = Q_2 - Q_1 = 60 \, \mu\text{C} - 20 \, \mu\text{C} = 40 \, \mu\text{C}$$

The equivalent capacitance of the two capacitors in parallel is:

$$C_{\text{eq}} = C_1 + C_2 = 2 \, \mu\text{F} + 3 \, \mu\text{F} = 5 \, \mu\text{F}$$

3. Final Energy Stored in the System

The energy stored in a capacitor is given by:

$$U = \frac{1}{2} \frac{Q_{\text{net}}^2}{C_{\text{eq}}}$$

Substitute the values:

$$U = \frac{1}{2} \cdot \frac{(40 \times 10^{-6})^2}{5 \times 10^{-6}}$$

$$U = \frac{1}{2} \cdot \frac{1600 \times 10^{-12}}{5 \times 10^{-6}}$$

$$U = \frac{1}{2} \cdot 320 \times 10^{-6} = 160 \times 10^{-6} \, \text{J}$$

$$U = 540 \, \mu\text{J}$$

Final Answer:

The final energy stored in the system is 540 μJ.

To derive the potential difference between the plates when one plate is given a charge

$Q$

and the other plate is uncharged, let's go step by step:

1. Induced Charge on the Opposite Plate

When a charge

$Q$

is placed on one plate, the other uncharged plate will develop an induced charge of

$-Q$

(due to electrostatic induction). This creates an electric field between the plates.

2. Net Capacitance of the System

For a parallel plate capacitor, the capacitance is given by:

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

However, we are not dealing with the conventional capacitor configuration here. Instead, only one plate is directly charged while the other has an induced charge.

The effective charge separation across the plates is the same, but the field contributions are halved because the uncharged plate contributes only via induction. This effectively makes the potential difference behave as if the charge on the capacitor were shared equally across its plates.

3. Potential Difference

The potential difference

$V$

across the plates is related to the charge

$Q$

by:

$$V = \frac{Q_{\text{effective}}}{C}$$

Here, the effective charge separation contributes as if the charge on each plate were effectively halved due to induction:

$$Q_{\text{effective}} = \frac{Q}{2}$$

Substitute

$Q_{\text{effective}}$

into the equation for

$V$

:

$$V = \frac{\frac{Q}{2}}{C}$$

$$V = \frac{Q}{2C}$$

Final Answer:

The potential difference between the plates is:

$$V = \frac{Q}{2C}$$

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:

$$\boxed{10V}$$

To solve this, let's calculate the percentage of energy dissipated when the switch

$S$

is turned to position 2.

Given:

• 
  $C_1 = 2 \, \mu\text{F}$ 
• 
  $C_2 = 8 \, \mu\text{F}$ 
• Initial voltage across
  $C_1 = V$ 

Key Concept:

When two capacitors are connected, the redistribution of charge causes energy loss. The energy dissipated can be calculated using the direct formula:

$$\text{Energy dissipated} = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$$

Here:

• 
  $V_1$ 

  is the initial voltage of $C_1$ 

  .

• 
  $V_2 = 0$ 

  (initially, $C_2$ 

  is uncharged).

Step-by-step Calculation:

1. Common Voltage after Connection: The final voltage across both capacitors is: 

   $$V_{\text{common}} = \frac{C_1 V}{C_1 + C_2} = \frac{2V}{2 + 8} = \frac{V}{5}.$$ 

2. Initial Energy Stored in
   $C_1$ 

   : 

   $$E_{\text{initial}} = \frac{1}{2} C_1 V^2 = \frac{1}{2} (2 \times 10^{-6}) V^2 = 10^{-6} V^2 \, \text{J}.$$ 

3. Final Energy Stored: The total energy stored after redistribution is: 

   $$E_{\text{final}} = \frac{1}{2} (C_1 + C_2) V_{\text{common}}^2 = \frac{1}{2} (2 + 8) \left(\frac{V}{5}\right)^2.$$Substituting values:

    

   $$E_{\text{final}} = \frac{1}{2} (10) \left(\frac{V^2}{25}\right) = \frac{10 V^2}{50} = \frac{V^2}{5} \times 10^{-6} \, \text{J}.$$ 

4. Energy Dissipated: 

   $$E_{\text{dissipated}} = E_{\text{initial}} - E_{\text{final}} = 10^{-6} V^2 - \frac{10^{-6} V^2}{5} = \frac{4}{5} \times 10^{-6} V^2.$$ 

5. Percentage of Energy Dissipated: 

   $$\% \text{Dissipated} = \frac{E_{\text{dissipated}}}{E_{\text{initial}}} \times 100 = \frac{\frac{4}{5} \times 10^{-6} V^2}{10^{-6} V^2} \times 100 = 80\%.$$ 

Final Answer:

The percentage of energy dissipated is 80%.

The capacitance of a parallel-plate capacitor is given by the general formula:

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

where:

• 
  $C$ 

  = capacitance,

• 
  $\varepsilon_0$ 

  = permittivity of free space,

• 
  $A$ 

  = effective overlapping area of the two plates,

• 
  $d$ 

  = separation between the plates.

For Plates of Unequal Areas ( $A_1 < A_2$

):

When two plates of areas

$A_1$

and

$A_2$

are used to form a capacitor, only the smaller area (

$A_1$

) determines the effective overlapping area for charge storage. This is because the excess area of the larger plate (

$A_2 - A_1$

) does not contribute to the capacitance.

Thus, the capacitance is:

$$C = \frac{\varepsilon_0 A_1}{d}$$

Final Answer:

The capacitance of the capacitor is:

$$\boxed{C = \frac{\varepsilon_0 A_1}{d}}$$