Given:

• Initial charge on the capacitor:
  $Q_1 = 60 \, \mu C$ 
• After inserting the dielectric, the total charge from the battery:
  $Q_2 = 120 \, \mu C$ 

  (additional charge drawn is $120 \, \mu C$ 

  , so the total charge on the capacitor is $120 \, \mu C + 60 \, \mu C = 180 \, \mu C$ 

  ).

Key concept:

• The charge on a capacitor is given by: 

  $$Q = C \cdot V$$where

  $C$is the capacitance and

  $V$is the potential difference across the plates.

• The dielectric increases the capacitance of the capacitor. If the dielectric constant is
  $K$ 

  , the capacitance becomes $K \times C$ 

  . Since the battery is still connected, the potential difference $V$ 

  remains constant, and the charge increases proportionally with the increase in capacitance.

Step 1: Relationship between charge and capacitance

Before the dielectric, the charge was

$Q_1 = 60 \, \mu C$

, and after inserting the dielectric, the charge is

$Q_2 = 180 \, \mu C$

.

The ratio of the final charge to the initial charge is proportional to the dielectric constant

$K$

:

$$\frac{Q_2}{Q_1} = K$$

Step 2: Solve for $K$

Substitute the given values:

$$K = \frac{180 \, \mu C}{60 \, \mu C} = 3$$

Final Answer:

The dielectric constant of the material inserted is 3.

In a parallel plate capacitor made by stacking

$n$

equally spaced plates connected alternatively:

Key Points:

1. The plates connected alternatively form a series of capacitors.
2. If there are
   $n$ 

   plates, the number of gaps (capacitors in series) between them is $n - 1$ 

   .

For capacitors in series:

The total capacitance

$C_{\text{total}}$

is given by:

$$\frac{1}{C_{\text{total}}} = \frac{1}{C} + \frac{1}{C} + \dots + \frac{1}{C} \, \, (\text{n - 1 times})$$

$$\frac{1}{C_{\text{total}}} = \frac{n - 1}{C}$$

$$C_{\text{total}} = \frac{C}{n - 1}$$

However, because the plates are connected alternatively, these effectively act as

$n - 1$

capacitors in parallel.

For capacitors in parallel:

The equivalent capacitance is:

$$C_{\text{eq}} = (n - 1)C$$

Thus, the resultant capacitance is

$(n - 1)C$

.

To find the ratio of the energy stored in the capacitor to the work done by the battery, let's break it down step by step:

1. Energy Stored in the Capacitor

The energy

$U$

stored in a capacitor is given by:

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

where

$C$

is the capacitance, and

$V$

is the voltage across the capacitor (equal to the EMF of the battery once fully charged).

2. Work Done by the Battery

The work done by the battery

$W$

is equal to the total charge delivered multiplied by the voltage:

$$W = Q \cdot V$$

The charge

$Q$

stored in the capacitor is:

$$Q = C V$$

So, the work done becomes:

$$W = C V \cdot V = C V^2$$

3. Ratio of Energy Stored to Work Done

Now, the ratio of the energy stored in the capacitor to the work done by the battery is:

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

Final Answer:

The ratio is

$\frac{1}{2}$

.

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

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

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

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."

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.

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.

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:

$$\boxed{2U}$$

Reason for the Short Circuit:

The middle capacitor is bypassed by a conducting wire (short circuit). Hence, no voltage difference exists across the middle capacitor, and it can be ignored in the calculation.

Simplified Circuit:

1. The circuit reduces to two capacitors
   $C$ 

   at the top and bottom in parallel.

2. For capacitors in parallel, the equivalent capacitance is simply the sum of their capacitances:
   $$C_{\text{eq}} = C + C = 2C$$ 

Final Answer:

The equivalent capacitance of the given combination is 2C.

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