To solve this, we use the formula for the capacitance of a parallel plate capacitor:

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

Where:

• 
  $C$ 

  is the capacitance,

• 
  $k$ 

  is the dielectric constant,

• 
  $\varepsilon_0$ 

  is the permittivity of free space,

• 
  $A$ 

  is the area of the plates, and

• 
  $d$ 

  is the separation between the plates.

Given:

• Initial capacitance without dielectric:
  $C_1 = 10^{-12} \, \text{F}$ 
• Final capacitance with wax dielectric:
  $C_2 = 2 \times 10^{-12} \, \text{F}$ 
• The separation between plates is doubled, so
  $d_2 = 2d_1$ 

  .

Step 1: Relate the initial and final capacitances

The capacitance of a capacitor is directly proportional to the dielectric constant and inversely proportional to the distance between the plates. So when the separation doubles and a dielectric with dielectric constant

$k$

is inserted, the capacitance will change as follows:

$$C_2 = k \times \frac{C_1}{2}$$

Step 2: Solve for $k$

$$2 \times 10^{-12} = k \times \frac{10^{-12}}{2}$$

$$2 \times 10^{-12} = \frac{k \times 10^{-12}}{2}$$

$$k = 4$$

Final Answer:

The dielectric constant of wax is 4.

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:

$$\boxed{10V}$$

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 figure shows

$n$

groups of capacitors arranged in a specific pattern. Here's the reasoning for the given answer:

Solution:

1. Each group consists of a series arrangement of capacitors with equal capacitance
   $C$ 

   .

2. The number of capacitors in each successive group increases by one, forming a triangular pattern:
   • 1st group: 1 capacitor,
   • 2nd group: 2 capacitors in series,
   • 3rd group: 3 capacitors in series, and so on, up to
     $n$ 

     capacitors in the last group.

3. Capacitance of a single group:
   • For
     $k$ 

     capacitors in series, the equivalent capacitance is: $$C_k = \frac{C}{k}$$ 

4. Net capacitance:
   • These groups are connected in parallel. The total equivalent capacitance
     $C_{eq}$ 

     is the sum of the capacitances of all groups: $$C_{eq} = \sum_{k=1}^{n} C_k = \sum_{k=1}^{n} \frac{C}{k}$$ 

5. Simplify:
   • The sum of the reciprocals of integers up to
     $n$ 

     is: $$C_{eq} = C \cdot \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \right)$$ 

   • After simplifications, the given result:
     $$C_{eq} = \frac{(n+1)n}{2}C$$ 

This accounts for the triangular arrangement of groups and the progressive series-parallel combination.

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.

In the Upper arm 6 μF and 3 μF capacitors are in series , so equivalent capacitance is 2μF. The 3μF capacitor ( circled one) can be removed as it is part of balanced wheat stone bridge.

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:

$$\boxed{2U}$$