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.

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:

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

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.

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.

Initially, the uncharged ball is attracted to the positive plate by induction. Upon touching, it acquires positive charge and is repelled by the positive plate, moving towards the negative plate. Upon touching the negative plate, it acquires negative charge and is repelled by the negative plate, moving back to the positive plate. This continuous charge transfer and repulsion causes the ball to oscillate between the plates, touching each in turn.

To handle \( 1000\text{V} \), \( N_s = \frac{1000}{250} = 4 \) capacitors must be connected in series. The capacitance of one row is \( C_s = \frac{8}{4} = 2\ \mu\text{F} \). To get a total capacitance of \( 16\ \mu\text{F} \), we need \( N_p = \frac{16}{2} = 8 \) parallel rows. Total number \( = N_s \times N_p = 4 \times 8 = 32 \).