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 plot a graph for the current

$i$

versus time

$t$

, let's break the problem into steps:

1. Capacitor with Dielectric Slab

• Initial Configuration: The capacitor has plates of total area
  $2A$ 

  and plate separation $d$ 

  . It is connected to a battery with emf $E$ 

  .

• Dielectric Slab: A dielectric slab of area
  $A$ 

  is inserted at a constant speed $v$ 

  .

2. Capacitance with Partial Dielectric

When a dielectric is partially inserted into the capacitor, the total capacitance is the sum of two capacitors:

• One part with the dielectric slab (
  $C_1$ 

  ).

• One part without the dielectric slab (
  $C_2$ 

  ).

Capacitance of the two regions:

1. With Dielectric Slab: 

   $$C_1 = \frac{\kappa \varepsilon_0 A}{d}$$where

   $\kappa$is the dielectric constant.

2. Without Dielectric Slab: 

   $$C_2 = \frac{\varepsilon_0 (2A - A)}{d} = \frac{\varepsilon_0 A}{d}$$ 

Thus, the total capacitance

$C_{\text{total}}$

is:

$$C_{\text{total}} = C_1 + C_2 = \frac{\kappa \varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{d}$$

$$C_{\text{total}} = \frac{\varepsilon_0 A}{d} (\kappa + 1)$$

3. Change in Capacitance with Time

As the slab moves with speed

$v$

, the area covered by the slab changes with time:

$$\text{Covered area } A_{\text{covered}} = v \cdot t$$

The effective capacitance changes as:

$$C(t) = \frac{\kappa \varepsilon_0 (v \cdot t)}{d} + \frac{\varepsilon_0 (2A - v \cdot t)}{d}$$

$$C(t) = \frac{\varepsilon_0}{d} \left[ \kappa (v \cdot t) + (2A - v \cdot t) \right]$$

Simplify:

$$C(t) = \frac{\varepsilon_0}{d} \left[ 2A + (\kappa - 1)(v \cdot t) \right]$$

4. Current in the Circuit

The current in the circuit is related to the rate of change of capacitance:

$$i(t) = E \cdot \frac{dC}{dt}$$

Differentiate

$C(t)$

with respect to

$t$

:

$$\frac{dC}{dt} = \frac{\varepsilon_0}{d} (\kappa - 1) v$$

Thus:

$$i(t) = E \cdot \frac{\varepsilon_0}{d} (\kappa - 1) v$$

5. Graph of $i$

vs. $t$

The current

$i(t)$

is constant because it does not depend on

$t$

(the rate of capacitance change is constant). Hence, the graph of

$i$

vs.

$t$

will be a horizontal line at:

$$i = E \cdot \frac{\varepsilon_0}{d} (\kappa - 1) v$$

The arrangement appears to be a system of parallel plates connected alternately to terminals

$a$

and

$b$

. Let’s determine the net capacitance.

Key Observations:

1. The plates form a series-parallel combination.
2. The area of each plate is
   $A$ 

   , and the separation between adjacent plates is $d$ 

   .

3. The effective configuration can be reduced to find the equivalent capacitance.

Equivalent Capacitance Derivation:

1. Pairing of Plates:
   • Adjacent plates (connected alternately) act as capacitors.
   • Each capacitor has a capacitance
     $C = \frac{\varepsilon_0 A}{d}$ 

     .

2. Parallel and Series Combination:
   • There are three capacitors in the arrangement, effectively forming a single network.
   • The middle plate shares equal charge with both sides, simplifying to an equivalent capacitance of
     $\frac{\varepsilon_0 A}{d}$ 

     .

Thus, the net capacitance is:

$$C_{\text{net}} = \frac{\varepsilon_0 A}{d}.$$

Let’s solve step by step why the energy stored becomes

$2U$

:

1. Initial Setup

• Capacitance of the capacitor: 

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

  • 
    $A$ 

    = area of the plates,

  • 
    $d$ 

    = distance between the plates,

  • 
    $\varepsilon_0$ 

    = permittivity of free space.

• When the capacitor is charged by a battery to a voltage
  $V$ 

  , the charge stored is: 

  $$Q = CV$$The energy stored in the capacitor is:

   

  $$U = \frac{1}{2} C V^2$$ 

2. When the Distance is Doubled

After the battery is removed, the charge

$Q$

on the capacitor remains constant because there’s no external connection. However, the capacitance changes due to the increased plate separation.

• New capacitance: 

  $$C' = \frac{\varepsilon_0 A}{2d}$$ 

• Energy stored in a capacitor is given by: 

  $$U' = \frac{Q^2}{2C'}$$Substitute

  $C' = \frac{\varepsilon_0 A}{2d}$:

   

  $$U' = \frac{Q^2}{2 \cdot \frac{\varepsilon_0 A}{2d}} = \frac{Q^2 \cdot 2d}{2 \varepsilon_0 A}$$Simplify:

   

  $$U' = \frac{Q^2 d}{\varepsilon_0 A}$$ 

3. Relating $U'$

and $U$

From the initial setup:

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

Substitute

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

and

$Q = CV$

:

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

From the doubling of plate separation:

$$U' = 2 \cdot \frac{Q^2}{2C} = 2U$$

Final Answer:

The energy stored in the capacitor after doubling the separation is:

$$\boxed{2U}$$

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:

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

On inserting dielectric Capacitance of the system increases so more charge is given by the battery.

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:

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