Chapter 2: Electrostatic Potential and Capacitance

Solution Steps

1. Step 1: Identify Given Values

   Given:
   Atomic number of gold nucleus, Z = 79
   Radius of the nucleus, r = 6.9 × 10⁻¹⁵ m
   Charge on a proton, e = 1.6 × 10⁻¹⁹ C
   Coulomb's constant, k = 9 × 10⁹ N m²/C² (standard value)

2. Step 2: Calculate Total Charge of the Nucleus

   The total charge (Q) of the nucleus is due to its protons. Since the atomic number Z represents the number of protons, the total charge is:
   Q = Z × e
   Q = 79 × (1.6 × 10⁻¹⁹ C)
   Q = 126.4 × 10⁻¹⁹ C

3. Step 3: Recall Formula for Electrostatic Potential

   The electrostatic potential (V) at a distance 'r' from a point charge 'Q' is given by:
   V = kQ/r

4. Step 4: Substitute Values and Calculate Potential

   Substitute the calculated charge Q and the given radius r into the potential formula:
   V = (9 × 10⁹ N m²/C²) × (126.4 × 10⁻¹⁹ C) / (6.9 × 10⁻¹⁵ m)
   V = (9 × 126.4 / 6.9) × (10⁹ × 10⁻¹⁹ / 10⁻¹⁵) V
   V = (1137.6 / 6.9) × 10⁽⁹⁻¹⁹⁺¹⁵⁾ V
   V = 164.869... × 10⁵ V
   V ≈ 1.648 × 10⁷ V
   Rounding to three significant figures, V ≈ 1.65 × 10⁷ V (or 1.64 × 10⁷ V if rounding 164.8 to 164 and then 1.64)

Solution Steps

1. Step 1: Identify Given Parameters

   Side of the regular hexagon, a = 10 cm = 0.1 m.
   Charge at each vertex, q = 5 μC = 5 × 10⁻⁶ C.
   Number of charges, n = 6 (since it's a hexagon).

2. Step 2: Determine Distance from Vertex to Center

   For a regular hexagon, the distance from each vertex to its center is equal to its side length. Therefore, r = a = 0.1 m.

3. Step 3: Calculate Potential Due to a Single Charge

   The electric potential (V) due to a single point charge (q) at a distance (r) is given by the formula:
   V = k * (q / r) where k is Coulomb's constant, k = 9 × 10⁹ N m²/C².
   Potential due to one charge, V₁ = (9 × 10⁹ N m²/C²) * (5 × 10⁻⁶ C / 0.1 m)
   V₁ = 4.5 × 10⁵ V

4. Step 4: Apply Superposition Principle

   Since electric potential is a scalar quantity, the total potential at the center due to all six charges is the algebraic sum of the potentials due to individual charges. As all charges are identical and equidistant from the center, the total potential V_total is:
   V_total = n * V₁
   V_total = 6 * (4.5 × 10⁵ V)
   V_total = 27 × 10⁵ V = 2.7 × 10⁶ V

Solution Steps

1. Step 1: Identify Given Values and Geometry

   Given:
   Side of the regular hexagon, a = 10 cm = 0.10 m
   Charge at each vertex, q = 5 μC = 5 × 10⁻⁶ C
   Number of charges (vertices), N = 6
   Coulomb's constant, k = 9 × 10⁹ N m²/C²

2. Step 2: Determine Distance from Center to Vertices

   For a regular hexagon, the distance (r) from the center to each vertex is equal to its side length.
   So, r = a = 0.10 m.

3. Step 3: Calculate Potential Due to a Single Charge

   The potential (V_single) at the center due to one charge 'q' at a distance 'r' is given by:
   V_single = kq/r

4. Step 4: Apply Superposition Principle

   Since there are 6 identical charges placed symmetrically at the vertices, the total potential (V_total) at the center is the sum of the potentials due to each individual charge. Due to symmetry and identical charges, this is simply N times the potential due to one charge.
   V_total = N × V_single = N × (kq/r)

5. Step 5: Substitute Values and Calculate

   V_total = 6 × (9 × 10⁹ N m²/C²) × (5 × 10⁻⁶ C) / (0.10 m)
   V_total = 6 × (9 × 5 / 0.10) × (10⁹ × 10⁻⁶) V
   V_total = 6 × (45 / 0.10) × 10³ V
   V_total = 6 × 450 × 10³ V
   V_total = 2700 × 10³ V
   V_total = 2.7 × 10⁶ V

Solution Steps

1. Step 1: Initial Capacitance with Air

   The capacitance of a parallel plate capacitor with air (or vacuum) between its plates is given by:
   C = ε₀A/d
   Where:
   C = initial capacitance = 8 pF = 8 × 10⁻¹² F
   ε₀ = permittivity of free space
   A = area of each plate d = initial distance between the plates

2. Step 2: New Conditions

   The distance between the plates is reduced by half: d' = d/2
   The space between the plates is filled with a substance of dielectric constant K = 6.

3. Step 3: Formula for Capacitance with Dielectric

   The capacitance of a parallel plate capacitor with a dielectric substance of dielectric constant K filling the space between the plates is given by:
   C' = Kε₀A/d'

4. Step 4: Substitute New Conditions into Formula

   Substitute d' = d/2 and K = 6 into the new capacitance formula:
   C' = Kε₀A / (d/2)
   C' = 2K (ε₀A/d)

5. Step 5: Relate to Initial Capacitance and Calculate

   We know that C = ε₀A/d. So, we can substitute C into the expression for C':
   C' = 2K C
   Now, substitute the given values for K and C:
   C' = 2 × 6 × (8 pF)
   C' = 12 × 8 pF
   C' = 96 pF

Solution Steps

1. Step 1: Identify Given Parameters

   Radius of the spherical conductor, R = 12 cm = 0.12 m.
   Charge on the surface, Q = 1.6 × 10⁻⁷ C.
   Coulomb's constant, k = 9 × 10⁹ N m²/C².

2. Step 2: Calculate Electric Field (a) Inside the Sphere

   For a spherical conductor, the electric field inside the conductor is always zero. This is because charges reside only on the surface of a conductor in electrostatic equilibrium, and any net charge inside would cause charge movement until the field becomes zero.
   E_inside = 0 N/C

3. Step 3: Calculate Electric Field (b) Just Outside the Sphere

   Just outside the surface of a spherical conductor, the electric field is given by the formula for a point charge located at its center:
   E_outside = k * (Q / R²)
   E_outside = (9 × 10⁹ N m²/C²) * (1.6 × 10⁻⁷ C / (0.12 m)²)
   E_outside = (9 × 10⁹ * 1.6 × 10⁻⁷) / 0.0144 N/C
   E_outside = (14.4 × 10²) / 0.0144 N/C
   E_outside = 1000 × 10² N/C = 1.0 × 10⁵ N/C

4. Step 4: Calculate Electric Field (c) At a Point 18 cm from the Centre

   For a point outside the spherical conductor, the conductor behaves as if all its charge is concentrated at its center. The distance from the center, r = 18 cm = 0.18 m.
   E_at_r = k * (Q / r²)
   E_at_r = (9 × 10⁹ N m²/C²) * (1.6 × 10⁻⁷ C / (0.18 m)²)
   E_at_r = (9 × 10⁹ * 1.6 × 10⁻⁷) / 0.0324 N/C
   E_at_r = (14.4 × 10²) / 0.0324 N/C
   E_at_r ≈ 444.44 × 10² N/C ≈ 4.44 × 10⁴ N/C

Solution Steps

1. Step 1: Identify Initial Conditions

   Initial capacitance with air, C₀ = 8 pF = 8 × 10⁻¹² F.
   For a parallel plate capacitor with air (or vacuum) between plates, the capacitance is given by:
   C₀ = (ε₀ * A) / d where ε₀ is the permittivity of free space, A is the plate area, and d is the distance between plates.

2. Step 2: Identify New Conditions

   New distance between plates, d' = d / 2.
   Dielectric constant of the substance, K = 6.
   Let the new capacitance be C'.

3. Step 3: Formulate New Capacitance Equation

   When a dielectric material of constant K is introduced between the plates, and the distance is changed, the new capacitance C' is given by:
   C' = (K * ε₀ * A) / d'

4. Step 4: Substitute New Conditions into the Formula

   Substitute d' = d / 2 into the equation for C':
   C' = (K * ε₀ * A) / (d / 2)
   C' = 2 * K * (ε₀ * A / d)

5. Step 5: Relate New Capacitance to Initial Capacitance

   From Step 1, we know that C₀ = (ε₀ * A) / d.
   So, C' = 2 * K * C₀

6. Step 6: Calculate the New Capacitance

   Substitute the given values for K and C₀:
   C' = 2 * 6 * 8 pF
   C' = 12 * 8 pF
   C' = 96 pF

Solution Steps

1. Step 1: Identify Given Parameters

   Capacitance of each capacitor, C₁ = C₂ = C₃ = 9 pF = 9 × 10⁻¹² F.
   Number of capacitors, n = 3.
   Supply voltage, V_total = 120 V.

2. Step 2: Calculate Total Capacitance (a) for Series Combination

   For capacitors connected in series, the reciprocal of the equivalent capacitance (C_eq) is the sum of the reciprocals of individual capacitances:
   1 / C_eq = 1 / C₁ + 1 / C₂ + 1 / C₃
   Since C₁ = C₂ = C₃ = C, we have:
   1 / C_eq = 1 / C + 1 / C + 1 / C = 3 / C
   C_eq = C / 3
   C_eq = 9 pF / 3 = 3 pF

3. Step 3: Calculate Total Charge Stored

   The total charge (Q_total) stored by the series combination is given by:
   Q_total = C_eq * V_total
   Q_total = (3 × 10⁻¹² F) * (120 V)
   Q_total = 360 × 10⁻¹² C = 360 pC

4. Step 4: Determine Charge on Each Capacitor

   In a series combination, the charge on each capacitor is the same and equal to the total charge stored by the combination.
   Q₁ = Q₂ = Q₃ = Q_total = 360 pC

5. Step 5: Calculate Potential Difference (b) Across Each Capacitor

   The potential difference (V) across each capacitor is given by V = Q / C.
   Since all capacitors have the same capacitance and the same charge:
   V₁ = Q₁ / C₁ = 360 pC / 9 pF = 40 V
   V₂ = Q₂ / C₂ = 360 pC / 9 pF = 40 V
   V₃ = Q₃ / C₃ = 360 pC / 9 pF = 40 V
   Alternatively, for identical capacitors in series, the total voltage divides equally among them:
   V_each = V_total / n = 120 V / 3 = 40 V

Solution Steps

1. Step 1: Identify Given Parameters

   Capacitance of each capacitor, C₁ = C₂ = C₃ = 9 pF = 9 × 10⁻¹² F.
   Number of capacitors, n = 3.
   Supply voltage, V_total = 120 V.

2. Step 2: Calculate Total Capacitance (a) for Parallel Combination

   For capacitors connected in parallel, the equivalent capacitance (C_eq) is the sum of the individual capacitances:
   C_eq = C₁ + C₂ + C₃
   Since C₁ = C₂ = C₃ = C, we have:
   C_eq = C + C + C = 3C
   C_eq = 3 * 9 pF = 27 pF

3. Step 3: Determine Potential Difference Across Each Capacitor

   In a parallel combination, the potential difference across each capacitor is the same and equal to the supply voltage.
   V₁ = V₂ = V₃ = V_total = 120 V

4. Step 4: Calculate Charge (b) on Each Capacitor

   The charge (Q) on each capacitor is given by Q = C * V.
   Q₁ = C₁ * V_total = (9 × 10⁻¹² F) * (120 V) = 1080 × 10⁻¹² C = 1080 pC
   Q₂ = C₂ * V_total = (9 × 10⁻¹² F) * (120 V) = 1080 × 10⁻¹² C = 1080 pC
   Q₃ = C₃ * V_total = (9 × 10⁻¹² F) * (120 V) = 1080 × 10⁻¹² C = 1080 pC

Solution Steps

1. Step 1: Identify Given Values

   Given, Capacitance of the capacitor, C = 12 pF = 12 × 10⁻¹² F.
   Potential difference across the capacitor, V = 50 V.

2. Step 2: Recall Formula for Energy Stored

   The electrostatic energy (U) stored in a capacitor is given by the formula:
   U = (1/2)CV²

3. Step 3: Substitute Values and Calculate

   Substitute the given values into the formula:
   U = (1/2) × (12 × 10⁻¹² F) × (50 V)²
   U = (1/2) × 12 × 10⁻¹² × 2500
   U = 6 × 10⁻¹² × 2500
   U = 15000 × 10⁻¹² J
   U = 1.5 × 10⁴ × 10⁻¹² J
   U = 1.5 × 10⁻⁸ J

4. Step 4: State Final Answer

   The electrostatic energy stored in the capacitor is 1.5 × 10⁻⁸ J.

Solution Steps

1. Step 1: Calculate Initial Energy of the First Capacitor

   Given, Capacitance of the first capacitor, C₁ = 600 pF = 600 × 10⁻¹² F.
   Potential difference, V₁ = 200 V.
   Initial energy stored in the first capacitor, U₁ = (1/2)C₁V₁²
   U₁ = (1/2) × (600 × 10⁻¹² F) × (200 V)²
   U₁ = (1/2) × 600 × 10⁻¹² × 40000
   U₁ = 300 × 10⁻¹² × 40000
   U₁ = 12 × 10⁶ × 10⁻¹² J
   U₁ = 12 × 10⁻⁶ J

2. Step 2: Calculate Initial Charge on the First Capacitor

   The charge on the first capacitor before connection is Q₁ = C₁V₁
   Q₁ = (600 × 10⁻¹² F) × (200 V)
   Q₁ = 120000 × 10⁻¹² C
   Q₁ = 1.2 × 10⁻⁷ C

3. Step 3: Determine Common Potential After Connection

   The first capacitor is disconnected from the supply and connected to an uncharged second capacitor (C₂ = 600 pF = 600 × 10⁻¹² F). The initial charge on the second capacitor is Q₂ = 0.
   According to the principle of charge conservation, the total charge remains constant.
   Total charge = Q₁ + Q₂ = 1.2 × 10⁻⁷ C + 0 = 1.2 × 10⁻⁷ C.
   Total capacitance = C₁ + C₂ = 600 pF + 600 pF = 1200 pF = 1200 × 10⁻¹² F.
   The common potential (V_common) across the combination is:
   V_common = Total Charge / Total Capacitance
   V_common = (1.2 × 10⁻⁷ C) / (1200 × 10⁻¹² F)
   V_common = (1.2 × 10⁻⁷) / (1.2 × 10⁻⁹) V
   V_common = 100 V

4. Step 4: Calculate Final Energy of the Combined System

   The final energy stored in the combined system (U_final) is:
   U_final = (1/2)(C₁ + C₂)V_common²
   U_final = (1/2) × (1200 × 10⁻¹² F) × (100 V)²
   U_final = (1/2) × 1200 × 10⁻¹² × 10000
   U_final = 600 × 10⁻¹² × 10000
   U_final = 6 × 10⁶ × 10⁻¹² J
   U_final = 6 × 10⁻⁶ J

5. Step 5: Calculate Energy Lost

   The energy lost (ΔU) is the difference between the initial energy and the final energy:
   ΔU = U₁ - U_final
   ΔU = (12 × 10⁻⁶ J) - (6 × 10⁻⁶ J)
   ΔU = 6 × 10⁻⁶ J

6. Step 6: State Final Answer

   The electrostatic energy lost in the process is 6 × 10⁻⁶ J.

Solution Steps

1. Step 1: Identify Initial Conditions

   Given, Initial capacitance with air, C_air = 8 pF = 8 × 10⁻¹² F.
   For a parallel plate capacitor with air (or vacuum) between the plates, the capacitance is given by:
   C_air = ε₀A/d
   Where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
   So, 8 × 10⁻¹² F = ε₀A/d --- (1)

2. Step 2: Identify New Conditions

   The distance between the plates is reduced by half, so the new distance, d' = d/2.
   The space between the plates is filled with a substance of dielectric constant, K = 6.

3. Step 3: Recall Formula for Capacitance with Dielectric

   The capacitance of a parallel plate capacitor with a dielectric substance of dielectric constant K filling the space between the plates is given by:
   C' = Kε₀A/d'

4. Step 4: Substitute New Conditions into the Formula

   Substitute d' = d/2 and K = 6 into the formula for C':
   C' = Kε₀A / (d/2)
   C' = 2K (ε₀A/d)

5. Step 5: Calculate New Capacitance using Initial Capacitance

   From equation (1), we know that ε₀A/d = 8 × 10⁻¹² F.
   Substitute this value and K = 6 into the expression for C':
   C' = 2 × 6 × (8 × 10⁻¹² F)
   C' = 12 × 8 × 10⁻¹² F
   C' = 96 × 10⁻¹² F
   C' = 96 pF

6. Step 6: State Final Answer

   The new capacitance of the parallel plate capacitor will be 96 pF.

Solution Steps

1. Step 1: Convert Given Values to SI Units

   Area of plates, A = 90 cm² = 90 × (10⁻² m)² = 90 × 10⁻⁴ m² = 9 × 10⁻³ m².
   Separation between plates, d = 2.5 mm = 2.5 × 10⁻³ m.
   Potential difference, V = 400 V.
   Permittivity of free space, ε₀ = 8.85 × 10⁻¹² F/m.

2. Step 2: Calculate Capacitance (C)

   The capacitance of a parallel plate capacitor is given by C = ε₀A/d.
   C = (8.85 × 10⁻¹² F/m) × (9 × 10⁻³ m²) / (2.5 × 10⁻³ m)
   C = (8.85 × 9 × 10⁻¹⁵) / (2.5 × 10⁻³) F
   C = (79.65 × 10⁻¹⁵) / (2.5 × 10⁻³) F
   C = 31.86 × 10⁻¹² F = 31.86 pF

3. Step 3: Part (a): Calculate Electrostatic Energy Stored (U)

   The electrostatic energy stored in the capacitor is U = (1/2)CV².
   U = (1/2) × (31.86 × 10⁻¹² F) × (400 V)²
   U = (1/2) × 31.86 × 10⁻¹² × 160000
   U = 15.93 × 10⁻¹² × 160000
   U = 2548800 × 10⁻¹² J
   U = 2.5488 × 10⁻⁶ J ≈ 2.55 × 10⁻⁶ J

4. Step 4: Part (b): Calculate Electric Field (E)

   The electric field between the plates of a parallel plate capacitor is E = V/d.
   E = (400 V) / (2.5 × 10⁻³ m)
   E = 160 × 10³ V/m = 1.6 × 10⁵ V/m

5. Step 5: Part (b): Calculate Volume (Vol) between Plates

   The volume between the plates is Vol = A × d.
   Vol = (9 × 10⁻³ m²) × (2.5 × 10⁻³ m)
   Vol = 22.5 × 10⁻⁶ m³

6. Step 6: Part (b): Calculate Energy per Unit Volume (u) from U/Vol

   Energy per unit volume, u = U / Vol u = (2.5488 × 10⁻⁶ J) / (22.5 × 10⁻⁶ m³) u = 2.5488 / 22.5 J/m³ u ≈ 0.11328 J/m³ ≈ 0.113 J/m³

7. Step 7: Part (b): Verify u = (1/2)ε₀E²

   Now, let's calculate u using the formula u = (1/2)ε₀E². u = (1/2) × (8.85 × 10⁻¹² F/m) × (1.6 × 10⁵ V/m)² u = (1/2) × 8.85 × 10⁻¹² × (2.56 × 10¹⁰) u = 4.425 × 10⁻¹² × 2.56 × 10¹⁰ u = 11.328 × 10⁻² J/m³ u = 0.11328 J/m³ ≈ 0.113 J/m³
   Since the calculated values of u from both methods are approximately equal (0.113 J/m³), the relation u = (1/2)ε₀E² is verified.

Solution Steps

1. Step 1: Initial Capacitance with Air

   The capacitance of a parallel plate capacitor with air (or vacuum) between its plates is given by C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.
   Given, initial capacitance C₁ = 8 pF = 8 × 10⁻¹² F.
   So, 8 × 10⁻¹² F = ε₀A/d.

2. Step 2: Changes in Parameters

   According to the problem:
   1. The distance between the plates is reduced by half. So, the new distance d' = d/2.
   2. The space between the plates is filled with a substance of dielectric constant K = 6.

3. Step 3: New Capacitance Formula with Dielectric

   When a dielectric material of dielectric constant K is introduced between the plates, the capacitance becomes C' = Kε₀A/d'.
   Substituting the new distance d' = d/2, we get:
   C' = Kε₀A / (d/2)
   C' = 2K (ε₀A/d)

4. Step 4: Calculate New Capacitance

   We know that C₁ = ε₀A/d = 8 pF.
   Substitute this into the expression for C':
   C' = 2K C₁
   Given K = 6 and C₁ = 8 pF.
   C' = 2 × 6 × 8 pF
   C' = 12 × 8 pF
   C' = 96 pF.

Solution Steps

1. Step 1: Given Data

   Number of capacitors, n = 3.
   Capacitance of each capacitor, C = 9 pF = 9 × 10⁻¹² F.
   Supply voltage, V = 120 V.

2. Step 2: (a) Total Capacitance in Series

   For capacitors connected in series, the reciprocal of the equivalent capacitance (C_eq) is the sum of the reciprocals of individual capacitances:
   1/C_eq = 1/C₁ + 1/C₂ + 1/C₃
   Since C₁ = C₂ = C₃ = C = 9 pF,
   1/C_eq = 1/C + 1/C + 1/C = 3/C
   C_eq = C/3
   C_eq = 9 pF / 3
   C_eq = 3 pF.

3. Step 3: (b) Charge on Each Capacitor

   In a series combination, the charge (Q) on each capacitor is the same and equal to the total charge stored by the equivalent capacitor.
   Q = C_eq × V
   Q = 3 pF × 120 V
   Q = (3 × 10⁻¹² F) × 120 V
   Q = 360 × 10⁻¹² C = 360 pC.

4. Step 4: Potential Difference Across Each Capacitor

   The potential difference across each capacitor (V_i) is given by V_i = Q / C_i.
   Since all capacitors have the same capacitance C = 9 pF and the same charge Q = 360 pC, the potential difference across each capacitor will be:
   V₁ = V₂ = V₃ = Q / C
   V₁ = 360 pC / 9 pF
   V₁ = 40 V.
   (Alternatively, since the capacitors are identical and in series, the total voltage divides equally among them: V/n = 120 V / 3 = 40 V).

Solution Steps

1. Step 1: Given Data

   Number of capacitors, n = 3.
   Capacitance of each capacitor, C = 9 pF = 9 × 10⁻¹² F.
   Supply voltage, V = 120 V.

2. Step 2: (a) Total Capacitance in Parallel

   For capacitors connected in parallel, the equivalent capacitance (C_eq) is the sum of the individual capacitances:
   C_eq = C₁ + C₂ + C₃
   Since C₁ = C₂ = C₃ = C = 9 pF,
   C_eq = C + C + C = 3C
   C_eq = 3 × 9 pF
   C_eq = 27 pF.

3. Step 3: (b) Potential Difference Across Each Capacitor

   In a parallel combination, the potential difference (voltage) across each capacitor is the same and equal to the potential difference of the supply.
   Therefore, the potential difference across each capacitor is V = 120 V.

Solution Steps

1. Step 1: Given Data and Constants

   Area of each plate, A = 6 × 10⁻³ m².
   Distance between the plates, d = 3 mm = 3 × 10⁻³ m.
   Supply voltage, V = 100 V.
   Permittivity of free space, ε₀ = 8.854 × 10⁻¹² F/m.

2. Step 2: Calculate Capacitance (C)

   The capacitance of a parallel plate capacitor with air between the plates is given by the formula:
   C = ε₀A / d
   Substitute the given values:
   C = (8.854 × 10⁻¹² F/m) × (6 × 10⁻³ m²) / (3 × 10⁻³ m)
   C = (8.854 × 10⁻¹² × 6) / 3 F
   C = 8.854 × 10⁻¹² × 2 F
   C = 17.708 × 10⁻¹² F
   C ≈ 17.7 pF.

3. Step 3: Calculate Charge (Q)

   The charge on each plate of the capacitor is given by the formula:
   Q = CV
   Substitute the calculated capacitance and the given voltage:
   Q = (17.708 × 10⁻¹² F) × (100 V)
   Q = 1770.8 × 10⁻¹² C
   Q = 1.7708 × 10⁻⁹ C
   Q ≈ 1.77 × 10⁻⁹ C.

Solution Steps

1. Step 1: Identify Given Parameters and Convert Units

   Inner sphere radius, R₁ = 12 cm = 0.12 m
   Outer sphere radius, R₂ = 13 cm = 0.13 m
   Charge on inner sphere, Q = 2.5 μC = 2.5 × 10⁻⁶ C
   Dielectric constant, K = 32
   Permittivity of free space, ε₀ = 8.854 × 10⁻¹² F/m

2. Step 2: Calculate Capacitance (a)

   The capacitance of a spherical capacitor with a dielectric medium between the plates is given by:
   C = 4πε₀K * (R₁R₂ / (R₂ - R₁))
   Substitute the given values:
   C = 4π * (8.854 × 10⁻¹² F/m) * 32 * (0.12 m * 0.13 m / (0.13 m - 0.12 m))
   C = 4π * (8.854 × 10⁻¹² F/m) * 32 * (0.0156 m² / 0.01 m)
   C = 4π * (8.854 × 10⁻¹² F/m) * 32 * 1.56 m
   C = 5.54 × 10⁻⁹ F = 5.54 nF

3. Step 3: Calculate Potential of Inner Sphere (b)

   The potential of the inner sphere can be found using the relation Q = CV, where V is the potential difference between the plates (since the outer sphere is earthed, its potential is 0, so V is the potential of the inner sphere).
   V = Q / C
   V = (2.5 × 10⁻⁶ C) / (5.54 × 10⁻⁹ F)
   V = 451.26 V

4. Step 4: Calculate Capacitance of Isolated Sphere (c)

   The capacitance of an isolated spherical conductor of radius R is given by:
   C_isolated = 4πε₀R
   For the inner sphere of radius R₁ = 0.12 m (assuming it's isolated in air, so K=1):
   C_isolated = 4π * (8.854 × 10⁻¹² F/m) * 0.12 m
   C_isolated = 1.33 × 10⁻¹¹ F = 13.3 pF

5. Step 5: Compare and Explain (c)

   Comparing the two capacitances:
   Capacitance of spherical capacitor (C) = 5.54 × 10⁻⁹ F
   Capacitance of isolated sphere (C_isolated) = 1.33 × 10⁻¹¹ F

   C / C_isolated = (5.54 × 10⁻⁹ F) / (1.33 × 10⁻¹¹ F) ≈ 416.5

   The spherical capacitor's capacitance is significantly larger (about 416 times) than that of the isolated sphere. This is because in a capacitor, the presence of the second conductor (the outer earthed sphere) at a lower potential helps to confine the electric field lines and effectively reduces the potential of the inner sphere for a given amount of charge. Since C = Q/V, a lower potential V for the same charge Q results in a higher capacitance. For an isolated sphere, the electric field lines extend to infinity, leading to a higher potential for the same charge and thus a smaller capacitance.

Solution Steps

1. Step 1: Understand the Initial State and Key Condition

   Initially, the capacitor is charged by a battery. This means it acquires a charge Q₀ and a potential difference V₀ (equal to the battery voltage). The initial capacitance is C₀, and the initial electric field is E₀. The energy stored is U₀.
   Crucially, the battery is *disconnected*. This implies that the charge (Q) on the capacitor plates will remain constant throughout the process, as there is no external circuit to add or remove charge.

2. Step 2: Analyze Electric Field (a)

   When a dielectric slab is inserted, the dielectric material gets polarized. This polarization creates an induced electric field (E_induced) within the dielectric that opposes the original electric field (E₀) due to the free charges on the plates. The net electric field (E) inside the dielectric is reduced. The relationship is E = E₀ / K, where K is the dielectric constant (K > 1). Thus, the electric field *decreases*.

3. Step 3: Analyze Potential Difference (b)

   The potential difference (V) between the plates of a parallel plate capacitor is given by V = Ed, where d is the distance between the plates. Since the electric field E has decreased by a factor of K (E = E₀/K), and d remains constant, the potential difference V must also decrease by the same factor. So, V = V₀ / K. Thus, the potential difference *decreases*.

4. Step 4: Analyze Capacitance (c)

   Capacitance is defined as C = Q/V. Since the battery is disconnected, the charge Q on the plates remains constant (Q = Q₀). As established in step 3, the potential difference V decreases (V = V₀/K). Therefore, the capacitance must increase to maintain the ratio. C = Q / (V₀/K) = K * (Q/V₀) = K C₀. Thus, the capacitance *increases*.

5. Step 5: Analyze Energy Stored (d)

   The energy stored in a capacitor can be expressed in several ways: U = (1/2)CV², U = Q²/(2C), or U = (1/2)QV.
   Since Q is constant and C increases (C = K C₀), using U = Q²/(2C):
   U = Q² / (2 * K * C₀) = (1/K) * (Q² / (2C₀)) = U₀ / K. Thus, the energy stored *decreases*.
   Alternatively, using U = (1/2)QV, since Q is constant and V decreases (V = V₀/K):
   U = (1/2) * Q * (V₀/K) = (1/K) * (1/2)QV₀ = U₀ / K. Thus, the energy stored *decreases*.
   The decrease in energy is due to the work done by the capacitor's electric field in pulling the dielectric into the capacitor.

Solution Steps

1. Step 1: Identify Initial Parameters

   Initial capacitance with air (K=1), C₀ = 8 pF = 8 × 10⁻¹² F.
   For a parallel plate capacitor with air, C₀ = ε₀A/d, where A is the area of the plates and d is the initial distance between them.

2. Step 2: Identify New Parameters

   The distance between the plates is reduced by half, so the new distance d' = d/2.
   The space between the plates is filled with a substance of dielectric constant K = 6.

3. Step 3: Formulate New Capacitance

   The capacitance of a parallel plate capacitor with a dielectric of constant K is given by C = Kε₀A/d'.
   Substitute d' = d/2 into the formula:
   C = Kε₀A / (d/2)
   C = 2K (ε₀A/d)

4. Step 4: Substitute and Calculate

   We know that C₀ = ε₀A/d. So, we can write the new capacitance in terms of C₀:
   C = 2K C₀
   Substitute the given values for K and C₀:
   C = 2 * 6 * (8 pF)
   C = 12 * 8 pF
   C = 96 pF

Solution Steps

1. Step 1: Identify initial conditions and formula

   The capacitance of a parallel plate capacitor with air (or vacuum) between its plates is given by C₀ = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.
   Given, C₀ = 8 pF = 8 × 10⁻¹² F.

2. Step 2: Identify final conditions

   In the new scenario:
   1. The distance between the plates is reduced by half, so the new distance d' = d/2.
   2. The space between the plates is filled with a substance of dielectric constant K = 6.

3. Step 3: Apply the formula for capacitance with dielectric

   The capacitance of a parallel plate capacitor with a dielectric substance of constant K is given by C = Kε₀A/d'.
   Substitute the new values: C' = Kε₀A/(d/2).

4. Step 4: Calculate the new capacitance

   C' = (K * ε₀A) / (d/2) = 2K * (ε₀A/d).
   Since C₀ = ε₀A/d, we can write C' = 2K * C₀.
   Substitute the given values: C' = 2 × 6 × 8 pF.
   C' = 12 × 8 pF = 96 pF.

Solution Steps

1. Step 1: Analyze the circuit diagram (Fig. 2.34)

   The circuit diagram (Fig. 2.34) typically shows the following arrangement:
   - Capacitors C₁ and C₂ are connected in series.
   - This series combination (let's call it C₁₂) is connected in parallel with capacitor C₃.
   - This entire parallel combination (C₁₂ and C₃, let's call it C₁₂₃) is then connected in series with capacitor C₄.
   All capacitors have a capacitance of 10 μF (C₁ = C₂ = C₃ = C₄ = 10 μF).

2. Step 2: Calculate equivalent capacitance of C1 and C2 in series

   For capacitors in series, the equivalent capacitance (C_series) is given by 1/C_series = 1/C₁ + 1/C₂.
   1/C₁₂ = 1/10 μF + 1/10 μF = 2/10 μF = 1/5 μF.
   So, C₁₂ = 5 μF.

3. Step 3: Calculate equivalent capacitance of C12 and C3 in parallel

   For capacitors in parallel, the equivalent capacitance (C_parallel) is given by C_parallel = C_series + C₃.
   C₁₂₃ = C₁₂ + C₃ = 5 μF + 10 μF = 15 μF.

4. Step 4: Calculate the final equivalent capacitance of C123 and C4 in series

   The combination C₁₂₃ is in series with C₄. The total equivalent capacitance (C_eq) between A and B is given by:
   1/C_eq = 1/C₁₂₃ + 1/C₄.
   1/C_eq = 1/15 μF + 1/10 μF.
   To add these fractions, find a common denominator, which is 30.
   1/C_eq = (2/30) μF + (3/30) μF = 5/30 μF = 1/6 μF.
   Therefore, C_eq = 6 μF.

Solution Steps

1. Step 1: Identify given values

   Capacitance of the capacitor, C = 12 pF = 12 × 10⁻¹² F.
   Voltage of the battery, V = 50 V.

2. Step 2: Recall the formula for energy stored in a capacitor

   The electrostatic energy (U) stored in a capacitor is given by the formula:
   U = (1/2)CV²

3. Step 3: Substitute values and calculate

   Substitute the given values into the formula:
   U = (1/2) × (12 × 10⁻¹² F) × (50 V)²
   U = (1/2) × 12 × 10⁻¹² × 2500
   U = 6 × 10⁻¹² × 2500
   U = 15000 × 10⁻¹² J
   U = 1.5 × 10⁴ × 10⁻¹² J
   U = 1.5 × 10⁻⁸ J

Solution Steps

1. Step 1: Calculate initial charge and energy of the first capacitor

   Given:
   Capacitance of the first capacitor, C₁ = 600 pF = 600 × 10⁻¹² F.
   Voltage of the supply, V₁ = 200 V.

   Initial charge on C₁: Q₁ = C₁V₁ = (600 × 10⁻¹² F) × (200 V) = 120000 × 10⁻¹² C = 1.2 × 10⁻⁷ C.
   Initial energy stored in C₁: U_initial = (1/2)C₁V₁² = (1/2) × (600 × 10⁻¹² F) × (200 V)²
   U_initial = (1/2) × 600 × 10⁻¹² × 40000 = 300 × 10⁻¹² × 40000 = 12 × 10⁶ × 10⁻¹² J = 1.2 × 10⁻⁵ J.

2. Step 2: Determine the state after connecting to the second capacitor

   The first capacitor (C₁) is disconnected from the supply and connected to an uncharged second capacitor (C₂).
   Capacitance of the second capacitor, C₂ = 600 pF = 600 × 10⁻¹² F.
   Initial charge on C₂, Q₂ = 0 (since it's uncharged).

   When connected, the capacitors are effectively in parallel. The total charge in the system is conserved.
   Total charge Q_total = Q₁ + Q₂ = 1.2 × 10⁻⁷ C + 0 = 1.2 × 10⁻⁷ C.
   Total equivalent capacitance C_total = C₁ + C₂ = 600 pF + 600 pF = 1200 pF = 1.2 × 10⁻⁹ F.

   The common potential (V_common) across both capacitors after connection is:
   V_common = Q_total / C_total = (1.2 × 10⁻⁷ C) / (1.2 × 10⁻⁹ F) = 100 V.

3. Step 3: Calculate the final energy stored in the combination

   The final energy stored in the combined system (U_final) is:
   U_final = (1/2)C_total * V_common²
   U_final = (1/2) × (1.2 × 10⁻⁹ F) × (100 V)²
   U_final = (1/2) × 1.2 × 10⁻⁹ × 10000
   U_final = 0.6 × 10⁻⁹ × 10⁴ = 0.6 × 10⁻⁵ J = 6.0 × 10⁻⁶ J.

4. Step 4: Calculate the energy lost

   The energy lost in the process is the difference between the initial and final energies:
   Energy lost = U_initial - U_final
   Energy lost = 1.2 × 10⁻⁵ J - 0.6 × 10⁻⁵ J
   Energy lost = 0.6 × 10⁻⁵ J = 6.0 × 10⁻⁶ J.
   This energy is typically lost as heat and electromagnetic radiation during the redistribution of charge.

Solution Steps

1. Step 1: Identify Given Values and Constraints

   Given:
   Voltage rating, V = 1 kV = 1000 V
   Dielectric constant, κ = 3
   Dielectric strength, E_max_dielectric = 10⁷ V/m
   Safety condition: Electric field E ≤ 10% of E_max_dielectric
   Capacitance, C = 50 pF = 50 × 10^-12 F
   Permittivity of free space, ε₀ = 8.85 × 10^-12 F/m

2. Step 2: Calculate Maximum Permissible Electric Field

   For safety, the electric field (E) inside the capacitor should not exceed 10% of the dielectric strength.
   E = 10% of E_max_dielectric = 0.10 × 10⁷ V/m = 10⁶ V/m

3. Step 3: Determine Minimum Plate Separation

   For a parallel plate capacitor, the electric field E, voltage V, and plate separation d are related by V = E × d. We need to find the minimum plate separation 'd' that can withstand the voltage V = 1000 V at the maximum permissible electric field E = 10⁶ V/m. d = V / E = 1000 V / (10⁶ V/m) = 10^-3 m = 1 mm

4. Step 4: Apply Capacitance Formula to Find Area

   The capacitance of a parallel plate capacitor with a dielectric is given by:
   C = (κ * ε₀ * A) / d
   Where A is the area of the plates. We need to find A.
   A = (C * d) / (κ * ε₀)

5. Step 5: Substitute Values and Calculate Area

   Substitute the calculated and given values into the formula for A:
   A = (50 × 10^-12 F * 10^-3 m) / (3 * 8.85 × 10^-12 F/m)
   A = (50 × 10^-15) / (26.55 × 10^-12)
   A = (50 / 26.55) × 10^(-15 + 12)
   A = 1.883 × 10^-3 m²

6. Step 6: Convert Area to cm²

   To express the area in a more convenient unit (cm²):
   A = 1.883 × 10^-3 m² × (100 cm / 1 m)²
   A = 1.883 × 10^-3 m² × 10⁴ cm²/m²
   A = 18.83 cm²
   Rounding to two significant figures, A ≈ 19 cm².

Solution Steps

1. Step 1: Initial Capacitance with Air

   Let the initial capacitance with air between the plates be C_air. Given C_air = 8 pF = 8 × 10^-12 F.
   The formula for capacitance of a parallel plate capacitor with air (or vacuum) is:
   C_air = (ε₀ * A) / d
   Where ε₀ is the permittivity of free space, A is the area of the plates, and d is the initial distance between the plates.

2. Step 2: Changes in Parameters

   The problem states two changes:
   1. The distance between the plates is reduced by half. So, the new distance d' = d/2.
   2. The space between the plates is filled with a substance of dielectric constant κ = 6.

3. Step 3: New Capacitance Formula

   The formula for capacitance with a dielectric material is:
   C_new = (κ * ε₀ * A) / d'
   Substitute the new distance d' = d/2 into this equation:
   C_new = (κ * ε₀ * A) / (d/2)
   C_new = (2 * κ * ε₀ * A) / d

4. Step 4: Substitute Initial Capacitance and Dielectric Constant

   We know that C_air = (ε₀ * A) / d. We can substitute this into the expression for C_new:
   C_new = 2 * κ * [(ε₀ * A) / d]
   C_new = 2 * κ * C_air

5. Step 5: Calculate the New Capacitance

   Now, substitute the given values:
   κ = 6
   C_air = 8 pF
   C_new = 2 * 6 * 8 pF
   C_new = 12 * 8 pF
   C_new = 96 pF

Solution Steps

1. Step 1: Identify Given Values

   Given:
   Capacitance, C = 12 pF = 12 × 10^-12 F
   Voltage, V = 50 V

2. Step 2: Recall Energy Storage Formula

   The electrostatic energy (U) stored in a capacitor is given by the formula:
   U = (1/2) * C * V²

3. Step 3: Substitute Values and Calculate Energy

   Substitute the given values into the formula:
   U = (1/2) * (12 × 10^-12 F) * (50 V)²
   U = (1/2) * 12 × 10^-12 * 2500
   U = 6 × 10^-12 * 2500
   U = 15000 × 10^-12 J
   U = 1.5 × 10⁴ × 10^-12 J
   U = 1.5 × 10^-8 J

Solution Steps

1. Step 1: Initial State: Capacitor C1 Charged

   Given:
   Capacitance of the first capacitor, C₁ = 600 pF = 600 × 10^-12 F
   Voltage of the supply, V₁ = 200 V

   Calculate the initial charge (Q₁) on C₁:
   Q₁ = C₁ * V₁ = (600 × 10^-12 F) * (200 V) = 120000 × 10^-12 C = 1.2 × 10^-7 C

   Calculate the initial energy (U_initial) stored in C₁:
   U_initial = (1/2) * C₁ * V₁² = (1/2) * (600 × 10^-12 F) * (200 V)²
   U_initial = (1/2) * 600 × 10^-12 * 40000
   U_initial = 300 × 10^-12 * 40000 = 12000000 × 10^-12 J = 1.2 × 10^-5 J

2. Step 2: Final State: Capacitors Connected in Parallel

   The first capacitor (C₁) is disconnected from the supply and connected to an uncharged second capacitor (C₂).
   Capacitance of the second capacitor, C₂ = 600 pF = 600 × 10^-12 F
   Since C₂ is uncharged, its initial charge Q₂ = 0.

   When connected, the capacitors are in parallel. The total capacitance of the combination (C_total) is:
   C_total = C₁ + C₂ = 600 pF + 600 pF = 1200 pF = 1200 × 10^-12 F

   By conservation of charge, the total charge before connection is equal to the total charge after connection:
   Q_total = Q₁ + Q₂ = 1.2 × 10^-7 C + 0 = 1.2 × 10^-7 C

3. Step 3: Calculate Final Common Potential and Energy

   The common potential (V_final) across the parallel combination is:
   V_final = Q_total / C_total = (1.2 × 10^-7 C) / (1200 × 10^-12 F)
   V_final = (1.2 / 1200) × 10⁵ V = 0.001 × 10⁵ V = 100 V

   Calculate the final energy (U_final) stored in the combination:
   U_final = (1/2) * C_total * V_final² = (1/2) * (1200 × 10^-12 F) * (100 V)²
   U_final = (1/2) * 1200 × 10^-12 * 10000
   U_final = 600 × 10^-12 * 10000 = 6000000 × 10^-12 J = 6.0 × 10^-6 J

4. Step 4: Calculate Energy Lost

   The energy lost (ΔU) is the difference between the initial and final energies:
   ΔU = U_initial - U_final
   ΔU = (1.2 × 10^-5 J) - (6.0 × 10^-6 J)
   ΔU = (12 × 10^-6 J) - (6 × 10^-6 J)
   ΔU = 6 × 10^-6 J

   This energy is lost primarily as heat and electromagnetic radiation during the redistribution of charge.

Solution Steps

1. Step 1: Identify initial conditions

   Given initial capacitance with air, C₀ = 8 pF = 8 × 10⁻¹² F. For a parallel plate capacitor with air (or vacuum) between the plates, the capacitance is given by C₀ = ε₀A/d, where A is the plate area, d is the distance between the plates, and ε₀ is the permittivity of free space.

2. Step 2: Identify changes in parameters

   The distance between the plates is reduced by half, so the new distance d' = d/2. The space between the plates is filled with a substance of dielectric constant K = 6.

3. Step 3: Formulate the new capacitance expression

   When a dielectric of constant K is introduced and the distance is d', the new capacitance C' is given by C' = Kε₀A/d'.

4. Step 4: Substitute the new parameters

   Substitute d' = d/2 into the expression for C':
   C' = Kε₀A / (d/2) = 2K (ε₀A/d).

5. Step 5: Relate to initial capacitance

   We know that C₀ = ε₀A/d. So, we can write C' = 2K C₀.

6. Step 6: Calculate the new capacitance

   Substitute the given values: C₀ = 8 pF and K = 6.
   C' = 2 × 6 × 8 pF = 12 × 8 pF = 96 pF.

Solution Steps

1. Step 1: Identify given values

   Capacitance, C = 12 pF = 12 × 10⁻¹² F.
   Voltage, V = 50 V.

2. Step 2: Recall the formula for energy stored

   The electrostatic energy (U) stored in a capacitor is given by the formula: U = (1/2)CV².

3. Step 3: Substitute values and calculate

   U = (1/2) × (12 × 10⁻¹² F) × (50 V)²
   U = (1/2) × 12 × 10⁻¹² × 2500
   U = 6 × 10⁻¹² × 2500
   U = 15000 × 10⁻¹² J
   U = 1.5 × 10⁴ × 10⁻¹² J
   U = 1.5 × 10⁻⁸ J.

Solution Steps

1. Step 1: Initial state: Capacitor C1 charged

   Capacitance of the first capacitor, C₁ = 600 pF = 600 × 10⁻¹² F.
   Voltage of the supply, V₁ = 200 V.
   Charge on C₁: Q₁ = C₁V₁ = (600 × 10⁻¹² F) × (200 V) = 120000 × 10⁻¹² C = 1.2 × 10⁻⁷ C.
   Initial energy stored in C₁: U₁ = (1/2)C₁V₁² = (1/2) × (600 × 10⁻¹² F) × (200 V)²
   U₁ = (1/2) × 600 × 10⁻¹² × 40000 = 300 × 10⁻¹² × 40000 = 12000000 × 10⁻¹² J = 1.2 × 10⁻⁵ J.

2. Step 2: Second state: Connection to uncharged capacitor

   The charged capacitor C₁ is disconnected from the supply. This means its charge Q₁ remains constant.
   It is then connected to another uncharged capacitor C₂ = 600 pF = 600 × 10⁻¹² F.
   Since C₂ is uncharged, its initial charge Q₂ = 0.

3. Step 3: Calculate common potential

   When the two capacitors are connected, charge redistributes until they reach a common potential (V_common).
   By conservation of charge, the total charge before connection equals the total charge after connection:
   Q_total = Q₁ + Q₂ = 1.2 × 10⁻⁷ C + 0 = 1.2 × 10⁻⁷ C.
   The equivalent capacitance of the parallel combination is C_eq = C₁ + C₂ = 600 pF + 600 pF = 1200 pF = 1200 × 10⁻¹² F.
   The common potential V_common = Q_total / C_eq = (1.2 × 10⁻⁷ C) / (1200 × 10⁻¹² F)
   V_common = (1.2 × 10⁻⁷) / (1.2 × 10⁻⁹) V = 100 V.

4. Step 4: Calculate final energy stored

   The total energy stored in the combination after connection (U_final) is:
   U_final = (1/2)C_eq V_common² = (1/2) × (1200 × 10⁻¹² F) × (100 V)²
   U_final = (1/2) × 1200 × 10⁻¹² × 10000 = 600 × 10⁻¹² × 10000 = 6000000 × 10⁻¹² J = 6 × 10⁻⁶ J.

5. Step 5: Calculate energy lost

   The energy lost (ΔU) is the difference between the initial energy and the final energy:
   ΔU = U₁ - U_final = (1.2 × 10⁻⁵ J) - (6 × 10⁻⁶ J)
   ΔU = (12 × 10⁻⁶ J) - (6 × 10⁻⁶ J) = 6 × 10⁻⁶ J.