Given:

• Initial capacitance with air as the dielectric:
  $C = 9 \, \text{pF}$ 
• Dielectric constants:
  $k_1 = 3$ 

  , $k_2 = 6$ 

• Thicknesses of the dielectrics:
  $d_1 = \frac{d}{3}$ 

  and $d_2 = \frac{2d}{3}$ 

Step-by-step solution:

1. Capacitance formula:

The capacitance of a parallel plate capacitor is:

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

Where:

• 
  $k$ 

  is the dielectric constant,

• 
  $\varepsilon_0$ 

  is the permittivity of free space,

• 
  $A$ 

  is the area of the plates,

• 
  $d$ 

  is the separation between the plates.

When we insert dielectrics in series, we treat the system as two capacitors in series with different dielectric constants.

2. Capacitance of each section:

• For the dielectric with
  $k_1 = 3$ 

  and thickness $d_1 = \frac{d}{3}$ 

  , the capacitance $C_1$ 

  is:

$$C_1 = \frac{k_1 \varepsilon_0 A}{d_1} = \frac{3 \varepsilon_0 A}{d/3} = \frac{9 \varepsilon_0 A}{d}$$

• For the dielectric with
  $k_2 = 6$ 

  and thickness $d_2 = \frac{2d}{3}$ 

  , the capacitance $C_2$ 

  is:

$$C_2 = \frac{k_2 \varepsilon_0 A}{d_2} = \frac{6 \varepsilon_0 A}{2d/3} = \frac{9 \varepsilon_0 A}{d}$$

3. Total capacitance:

The total capacitance of the system is found by treating the two capacitances in series. For capacitors in series, the total capacitance

$C_{\text{total}}$

is given by:

$$\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}$$

Substitute

$C_1 = C_2 = \frac{9 \varepsilon_0 A}{d}$

:

$$\frac{1}{C_{\text{total}}} = \frac{1}{\frac{9 \varepsilon_0 A}{d}} + \frac{1}{\frac{9 \varepsilon_0 A}{d}} = \frac{2}{\frac{9 \varepsilon_0 A}{d}}$$

Thus,

$$C_{\text{total}} = \frac{9 \varepsilon_0 A}{2d}$$

4. Relating to the original capacitance:

The original capacitance with air as the dielectric is:

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

Therefore, the total capacitance becomes:

$$C_{\text{total}} = \frac{9}{2} C = 4.5 C = 4.5 \times 9 \, \text{pF} = 40.5 \, \text{pF}$$

Final Answer:

The total capacitance with the two dielectrics is 40.5 pF.

Thus, 40.5 pF is the correct answer.

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 question involves calculating the capacitance of a parallel-plate capacitor when a dielectric slab of thickness

$t$

and dielectric constant

$K = 2$

is partially inserted between the plates.

The short solution is based on treating the system as a combination of two capacitors in series:

1. Capacitor 1 (region with dielectric): The thickness of this region is
   $t$ 

   , and the capacitance is given by: 

   $$C_1 = \frac{\varepsilon_0 A K}{t} = \frac{2\varepsilon_0 A}{t}$$ 

2. Capacitor 2 (region without dielectric): The thickness of this region is
   $d - t$ 

   , and the capacitance is: 

   $$C_2 = \frac{\varepsilon_0 A}{d - t}$$ 

Since the two regions are in series, the equivalent capacitance is given by:

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$$

Substituting

$C_1$

and

$C_2$

:

$$\frac{1}{C} = \frac{t}{2\varepsilon_0 A} + \frac{d - t}{\varepsilon_0 A}$$

Simplifying:

$$\frac{1}{C} = \frac{t + 2(d - t)}{2\varepsilon_0 A} = \frac{2d - t}{2\varepsilon_0 A}$$

Therefore:

$$C = \frac{2\varepsilon_0 A}{2d - t}$$

When the dielectric slab thickness is

$t = \frac{d}{2}$

, the capacitance simplifies to:

$$C = \frac{\varepsilon_0 A}{d - \frac{t}{2}}$$

This matches the given answer. Let me know if further clarification is needed!

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:

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

The charging of a capacitor through a resistor is described by the following equation:

$$Q(t) = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Where:

• 
  $Q(t)$ 

  is the charge on the capacitor at time $t$ 

  ,

• 
  $Q_{\text{max}}$ 

  is the maximum (steady-state) charge the capacitor can hold,

• 
  $\tau$ 

  is the time constant, $\tau = R \cdot C$ 

  , where $R$ 

  is the resistance and $C$ 

  is the capacitance,

• 
  $t$ 

  is the time.

Step 1: Given condition (10% of steady charge)

We are told that at time

$t$

, the capacitor has collected 10% of the steady charge, so:

$$Q(t) = 0.1 \cdot Q_{\text{max}}$$

Step 2: Substitute into the charging equation

Substitute

$Q(t) = 0.1 \cdot Q_{\text{max}}$

into the charging formula:

$$0.1 \cdot Q_{\text{max}} = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Cancel

$Q_{\text{max}}$

from both sides:

$$0.1 = 1 - e^{-t/\tau}$$

Step 3: Solve for $t$

Rearrange the equation to solve for

$e^{-t/\tau}$

:

$$e^{-t/\tau} = 1 - 0.1 = 0.9$$

Take the natural logarithm of both sides:

$$-\frac{t}{\tau} = \ln(0.9)$$

$$t = -\tau \ln(0.9)$$

Using the fact that

$\ln(0.9) = -\ln(10/9)$

:

$$t = \tau \ln\left(\frac{10}{9}\right)$$

Final Answer:

The time at which the capacitor has collected 10% of the steady charge is

$t = \tau \ln\left(\frac{10}{9}\right)$

.

When Switch is Open Equivalent Capacitance is C_eq=C/2

So, Charge on Both the capacitors in CE/2.

When Switch is Closed 1st Capacitor is short circuited .Now  Equivalent Capacitance is C_eq=C.

So, Charge on  the capacitors in CE.

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.

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:

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

The incorrect statement is "The charge on the capacitor is not conserved".

Here’s the explanation:

1. Initially, when the capacitor is connected to the battery with emf
   $V$ 

   , it stores a charge

   $Q_{\text{initial}} = C \times V$ 

   , where

   $C$ 

   is the capacitance of the air capacitor.

2. After disconnecting the capacitor from the battery, the charge on the capacitor is conserved because the capacitor is isolated. No charge can flow in or out.
3. When the dielectric slab of dielectric constant
   $K$ 

   is inserted, the capacitance of the capacitor increases to

   $K \times C$ 

   , but the charge remains the same because the capacitor is disconnected from the battery. The charge is now distributed on the new capacitance, and the voltage across the plates decreases.

4. The statement "charge is not conserved" is incorrect because charge is conserved in this isolated system. The only change is in the voltage across the capacitor due to the increased capacitance.

Thus, charge on the capacitor is conserved and the incorrect statement is the one claiming otherwise.