When the distance between the plates of a charged capacitor is decreased after disconnecting the battery, the following happens:

1. Charge remains constant: Since the battery is disconnected, the charge
   $Q$ 

   on the plates does not change.

2. Capacitance increases: The capacitance of a parallel plate capacitor is given by: 

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

   $d$is the distance between the plates. As

   $d$decreases,

   $C$increases.

3. Potential difference decreases: The voltage
   $V$ 

   across the capacitor is related by: 

   $$V = \frac{Q}{C}$$Since

   $Q$is constant and

   $C$increases,

   $V$decreases.

4. Potential energy decreases: The energy stored in a capacitor is: 

   $$U = \frac{1}{2} \frac{Q^2}{C}$$As

   $C$increases,

   $U$decreases because

   $Q^2$is constant.

Thus, the potential energy decreases when the distance between the plates is reduced.

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

$\frac{1}{2}$

.

To solve this, let us analyze the situation and derive the required expression for the charge:

1. Setup of the System

• The parallel plate capacitor is initially charged, and the battery is disconnected.
• One plate of the capacitor is connected to a spring, and the other is attached to a mass
  $m$ 

  .

• The capacitor plates attract each other due to the opposite charges, generating an electrostatic force.

2. Equilibrium Condition

In equilibrium, the upward force due to the spring's tension balances the downward gravitational force acting on the mass

$m$

:

$$T = mg$$

This tension

$T$

is equal to the electrostatic force

$F_{\text{elec}}$

between the plates of the capacitor:

$$F_{\text{elec}} = \frac{Q^2}{2 \epsilon_0 A}$$

3. Force Balance

At equilibrium:

$$F_{\text{elec}} = mg$$

Substitute

$F_{\text{elec}} = \frac{Q^2}{2 \epsilon_0 A}$

:

$$\frac{Q^2}{2 \epsilon_0 A} = mg$$

4. Solve for $Q$

Rearranging for

$Q$

:

$$Q^2 = 2 mg \epsilon_0 A$$

Taking the square root:

$$Q = \sqrt{2 mg \epsilon_0 A}$$

Final Answer:

The magnitude of the charge on one of the capacitor plates is:

$$Q = \sqrt{2mgA\epsilon_0}$$

To solve this problem, let's analyze the situation step by step:

1. Initial Charges on Capacitors

The charge on each capacitor is given by

$Q = CV$

:

• For the first capacitor (
  $C_1 = 2 \, \mu\text{F}, V_1 = 10 \, \text{V}$ 

  ): 

  $$Q_1 = C_1 V_1 = 2 \times 10^{-6} \cdot 10 = 20 \, \mu\text{C}$$ 

• For the second capacitor (
  $C_2 = 3 \, \mu\text{F}, V_2 = 20 \, \text{V}$ 

  ): 

  $$Q_2 = C_2 V_2 = 3 \times 10^{-6} \cdot 20 = 60 \, \mu\text{C}$$ 

2. Connecting Capacitors with Opposite Polarity

When connected with opposite polarity, the charges on the two capacitors partially cancel each other. The net charge is:

$$Q_{\text{net}} = Q_2 - Q_1 = 60 \, \mu\text{C} - 20 \, \mu\text{C} = 40 \, \mu\text{C}$$

The equivalent capacitance of the two capacitors in parallel is:

$$C_{\text{eq}} = C_1 + C_2 = 2 \, \mu\text{F} + 3 \, \mu\text{F} = 5 \, \mu\text{F}$$

3. Final Energy Stored in the System

The energy stored in a capacitor is given by:

$$U = \frac{1}{2} \frac{Q_{\text{net}}^2}{C_{\text{eq}}}$$

Substitute the values:

$$U = \frac{1}{2} \cdot \frac{(40 \times 10^{-6})^2}{5 \times 10^{-6}}$$

$$U = \frac{1}{2} \cdot \frac{1600 \times 10^{-12}}{5 \times 10^{-6}}$$

$$U = \frac{1}{2} \cdot 320 \times 10^{-6} = 160 \times 10^{-6} \, \text{J}$$

$$U = 540 \, \mu\text{J}$$

Final Answer:

The final energy stored in the system is 540 μJ.

To solve for the energy stored in the system of three plates (X, Y, Z), let's break the system into simpler components:

1. System Description

• The arrangement forms two capacitors:
  • Capacitor 1: Between plates X and Y.
  • Capacitor 2: Between plates Y and Z.
• Each capacitor has the same plate area
  $A$ 

  and plate separation $d$ 

  .

The capacitance of a parallel plate capacitor is given by:

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

Thus, the capacitance of each capacitor is:

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

2. Effective Capacitance

The two capacitors are in parallel because plate Y is connected to the battery on one side and plate X and Z are on the opposite side. For capacitors in parallel, the effective capacitance is:

$$C_{\text{eq}} = C_1 + C_2$$

$$C_{\text{eq}} = \frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{d} = \frac{2 \varepsilon_0 A}{d}$$

3. Energy Stored in the System

The energy stored in a capacitor is:

$$U = \frac{1}{2} C_{\text{eq}} V^2$$

Substitute

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

:

$$U = \frac{1}{2} \cdot \frac{2 \varepsilon_0 A}{d} \cdot V^2$$

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

Final Answer:

The energy stored in the system is:

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

To derive the potential difference between the plates when one plate is given a charge

$Q$

and the other plate is uncharged, let's go step by step:

1. Induced Charge on the Opposite Plate

When a charge

$Q$

is placed on one plate, the other uncharged plate will develop an induced charge of

$-Q$

(due to electrostatic induction). This creates an electric field between the plates.

2. Net Capacitance of the System

For a parallel plate capacitor, the capacitance is given by:

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

However, we are not dealing with the conventional capacitor configuration here. Instead, only one plate is directly charged while the other has an induced charge.

The effective charge separation across the plates is the same, but the field contributions are halved because the uncharged plate contributes only via induction. This effectively makes the potential difference behave as if the charge on the capacitor were shared equally across its plates.

3. Potential Difference

The potential difference

$V$

across the plates is related to the charge

$Q$

by:

$$V = \frac{Q_{\text{effective}}}{C}$$

Here, the effective charge separation contributes as if the charge on each plate were effectively halved due to induction:

$$Q_{\text{effective}} = \frac{Q}{2}$$

Substitute

$Q_{\text{effective}}$

into the equation for

$V$

:

$$V = \frac{\frac{Q}{2}}{C}$$

$$V = \frac{Q}{2C}$$

Final Answer:

The potential difference between the plates is:

$$V = \frac{Q}{2C}$$

The Circled capacitors are in Series so , Their equivalent capacitance is 1 μF . Then it is in parallel with 1 μF capacitor. The circuit will keep on reducing.

For Such Question start solving from the farthest point. The circuit will keep on reducing.

Capacitors circled in the diagram are short circuited so, they can be removed from the circuit.