Both the parts can be taken as separate capacitors connected in parallel.

So, C=C1+ C2=\(\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} +K_{2}\right)}{2}\)

At t =0 capacitor behaves as closed circuit to the 1000 ohm resistor connected in parallel with capacitor will get short circuited.

current  through the other resistor = 2/1000 = 2mA

At t = infinite capacitor behaves as open circuit so equivalent resistance becomes R=1000+1000 = 2000 Ohm

current  through the  resistor = 2/2000 = 1 mA

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:

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

Let's verify and calculate the correct answer step-by-step:

Given Data:

• Capacitances:
  $C_1 = 2 \, \mu\text{F}, C_2 = 3 \, \mu\text{F}, C_3 = 5 \, \mu\text{F}$ 
• Maximum voltages:
  $V_1 = 3 \, \text{V}, V_2 = 2 \, \text{V}, V_3 = 1 \, \text{V}$ 

Step 1: Maximum charge each capacitor can store:

$$Q_1 = C_1 \cdot V_1 = 2 \cdot 3 = 6 \, \mu\text{C}$$

$$Q_2 = C_2 \cdot V_2 = 3 \cdot 2 = 6 \, \mu\text{C}$$

$$Q_3 = C_3 \cdot V_3 = 5 \cdot 1 = 5 \, \mu\text{C}$$

The capacitor with the minimum charge capacity limits the system. Here,

$Q_{\text{max}} = 5 \, \mu\text{C}$

, dictated by

$C_3$

.

Step 2: Voltage distribution across each capacitor:

In series, charge

$Q$

is the same on all capacitors, and the voltage across each capacitor is:

$$V_1 = \frac{Q}{C_1}, \quad V_2 = \frac{Q}{C_2}, \quad V_3 = \frac{Q}{C_3}$$

Total voltage across the series combination:

$$V_{\text{total}} = V_1 + V_2 + V_3$$

Substitute

$Q = 5 \, \mu\text{C}$

:

$$V_1 = \frac{5}{2} = 2.5 \, \text{V}, \quad V_2 = \frac{5}{3} \approx 1.67 \, \text{V}, \quad V_3 = \frac{5}{5} = 1 \, \text{V}$$

Step 3: Total voltage:

$$V_{\text{total}} = V_1 + V_2 + V_3 = 2.5 + 1.67 + 1 = \frac{15}{6} + \frac{10}{6} + \frac{6}{6} = \frac{31}{6} \, \text{V}.$$

Final Answer:

The maximum voltage the series combination can withstand is:

$$\boxed{\frac{31}{6} \, \text{V}} \approx 5.17 \, \text{V}.$$

The problem involves two spherical conductors

$A_1$

and

$A_2$

connected by a copper wire. Let’s analyze and compute the equivalent capacitance of the system.

Given:

• 
  $A_1$ 

  and $A_2$ 

  are concentric spherical conductors.

• Radii of the spheres:
  $r_1$ 

  (inner) and $r_2$ 

  (outer).

• The medium is air, so the permittivity is
  $\varepsilon_0$ 

  .

Key Concepts:

1. Potential Difference Between the Spheres: The two conductors are connected by a wire, meaning they are at the same potential. As a result, the electric field exists only between the two spheres.
2. Capacitance of a Single Isolated Sphere: If only
   $A_2$ 

   existed as a spherical conductor, its capacitance would be: 

   $$C_{\text{single}} = 4 \pi \varepsilon_0 r_2.$$ 

3. Why the System is Equivalent to an Isolated Sphere: Since
   $A_1$ 

   is connected to $A_2$ 

   via a conducting wire, any charge added to $A_1$ 

   immediately flows to $A_2$ 

   , making the system behave as if there is only one conductor of radius $r_2$ 

   .

Equivalent Capacitance:

Thus, the capacitance of the system is:

$$C_{\text{equivalent}} = 4 \pi \varepsilon_0 r_2.$$

Final Answer:

The equivalent capacitance of the system is:

$$\boxed{4 \pi \varepsilon_0 r_2}.$$

To find the equivalent capacitance for this cubical capacitor network, where each edge of the cube has a capacitance

$C$

, here’s the shortest solution:

Step-by-Step:

1. Symmetry analysis:
   • By symmetry, all corners of the cube can be grouped into equivalent potential nodes.
   • The cube's symmetry allows reduction to a simpler circuit.
2. Key nodes:
   • Node
     $A$ 

     is connected to one corner of the cube.

   • Node
     $B$ 

     is connected to the diagonally opposite corner.

3. Effective connections:
   • Due to symmetry, three capacitors are effectively in parallel between
     $A$ 

     and an intermediate point.

   • Similarly, three capacitors are effectively in parallel between
     $B$ 

     and the same intermediate point.

   • Two capacitors remain directly between
     $A$ 

     and $B$ 

     .

4. Simplification:
   • The three parallel capacitors at each node result in:
     $$C_{\text{parallel}} = 3C$$ 
   • The equivalent circuit becomes two
     $3C$ 

     capacitors in series with a $2C$ 

     capacitor: $$\text{Series combination:} \quad \frac{1}{C_{\text{eq}}} = \frac{1}{3C} + \frac{1}{3C} + \frac{1}{2C}$$ 

5. Calculation:
   • Combine series:
     $$\frac{1}{C_{\text{eq}}} = \frac{2}{3C} + \frac{1}{2C} = \frac{4}{6C} + \frac{3}{6C} = \frac{7}{6C}$$ 
   • Invert to find
     $C_{\text{eq}}$ 

     : $$C_{\text{eq}} = \frac{6C}{7} \times 2 = \frac{12C}{7}$$ 

Thus, the equivalent capacitance is:

$$C_{\text{eq}} = \frac{12C}{7}$$

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:

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