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 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 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:

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

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:

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