Solution:

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:

$$W = \frac{\varepsilon_0 A V^2}{2d}$$

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Solution:

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:

$$\boxed{10V}$$

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Solution:

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:

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

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Solution:

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:

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

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Solution:

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:

$$\boxed{2U}$$

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Solution:

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%.

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Solution:

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:

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

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Solution:

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

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Solution:

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:

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

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