Chapter 1: Electric Charges and Fields

Solution Steps

1. Step 1: Identify Given Quantities

   Given charges: q₁ = 2 × 10⁻⁷ C q₂ = 3 × 10⁻⁷ C
   Distance between charges, r = 30 cm = 0.30 m
   Permittivity constant for air (or vacuum), k = 1 / (4πε₀) = 9 × 10⁹ N m²/C²

2. Step 2: Apply Coulomb's Law

   Coulomb's Law states that the electrostatic force (F) between two point charges is given by:
   F = k |q₁q₂| / r²

3. Step 3: Substitute Values and Calculate

   F = (9 × 10⁹ N m²/C²) × |(2 × 10⁻⁷ C) × (3 × 10⁻⁷ C)| / (0.30 m)²
   F = (9 × 10⁹) × (6 × 10⁻¹⁴) / (0.09)
   F = (54 × 10⁻⁵) / (0.09)
   F = 600 × 10⁻⁵ N
   F = 6 × 10⁻³ N

4. Step 4: Determine Nature of Force

   Since both charges (q₁ and q₂) are positive, they are like charges. Therefore, the force between them is repulsive.

Solution Steps

1. Step 1: Identify Given Quantities and Convert Units

   Given charges: q₁ = 0.4 μC = 0.4 × 10⁻⁶ C q₂ = –0.8 μC = –0.8 × 10⁻⁶ C
   Magnitude of electrostatic force, F = 0.2 N
   Permittivity constant for air (or vacuum), k = 1 / (4πε₀) = 9 × 10⁹ N m²/C²

2. Step 2: Part (a): Calculate the Distance Between Spheres using Coulomb's Law

   Coulomb's Law: F = k |q₁q₂| / r²
   We need to find r, so rearrange the formula: r² = k |q₁q₂| / F r = √(k |q₁q₂| / F)

   Substitute the values: r = √ [ (9 × 10⁹ N m²/C²) × |(0.4 × 10⁻⁶ C) × (–0.8 × 10⁻⁶ C)| / (0.2 N) ] r = √ [ (9 × 10⁹) × (0.32 × 10⁻¹²) / (0.2) ] r = √ [ (9 × 10⁹) × (3.2 × 10⁻¹³) / (0.2) ] r = √ [ (28.8 × 10⁻⁴) / (0.2) ] r = √ [ 144 × 10⁻⁴ ] r = √(0.0144) r = 0.12 m

3. Step 3: Part (b): Determine the Force on the Second Sphere

   According to Newton's Third Law of Motion, the force exerted by the first sphere on the second sphere is equal in magnitude and opposite in direction to the force exerted by the second sphere on the first sphere.

   Given that the force on the first sphere (q₁) due to the second sphere (q₂) is 0.2 N.
   Therefore, the force on the second sphere (q₂) due to the first sphere (q₁) is also 0.2 N.

   Since q₁ is positive and q₂ is negative, they are unlike charges, and the force between them is attractive. So, the force on the second sphere will be 0.2 N, directed towards the first sphere.

Solution Steps

1. Step 1: Identify Constants and Their Dimensions

   We need the dimensions of the following constants:
   1. k (Coulomb's constant): [k] = [F r² / q²] = [M L T⁻² L² / I² T²] = [M L³ T⁻⁴ I⁻²]
   2. e (elementary charge): [e] = [I T]
   3. G (Gravitational constant): [G] = [F r² / m²] = [M L T⁻² L² / M²] = [M⁻¹ L³ T⁻²]
   4. m_e (mass of electron): [m_e] = [M]
   5. m_p (mass of proton): [m_p] = [M]

2. Step 2: Determine Dimensions of the Numerator (ke²)

   [ke²] = [M L³ T⁻⁴ I⁻²] × [I T]²
   [ke²] = [M L³ T⁻⁴ I⁻²] × [I² T²]
   [ke²] = [M L³ T⁻²]

3. Step 3: Determine Dimensions of the Denominator (G m_e m_p)

   [G m_e m_p] = [M⁻¹ L³ T⁻²] × [M] × [M]
   [G m_e m_p] = [M⁻¹ L³ T⁻² M²]
   [G m_e m_p] = [M L³ T⁻²]

4. Step 4: Check if the Ratio is Dimensionless

   Ratio = [ke²] / [G m_e m_p]
   Ratio = [M L³ T⁻²] / [M L³ T⁻²]
   Ratio = [M⁰ L⁰ T⁰]
   Since the dimensions cancel out, the ratio is dimensionless.

5. Step 5: Look Up Physical Constants

   Using standard values: k = 9 × 10⁹ N m²/C² e = 1.602 × 10⁻¹⁹ C
   G = 6.674 × 10⁻¹¹ N m²/kg² m_e = 9.109 × 10⁻³¹ kg m_p = 1.672 × 10⁻²⁷ kg

6. Step 6: Calculate the Value of the Ratio

   Numerator: ke² = (9 × 10⁹) × (1.602 × 10⁻¹⁹)² ke² = (9 × 10⁹) × (2.566404 × 10⁻³⁸) ke² ≈ 2.30976 × 10⁻²⁸ N m²

   Denominator: G m_e m_p = (6.674 × 10⁻¹¹) × (9.109 × 10⁻³¹) × (1.672 × 10⁻²⁷)
   G m_e m_p = (6.674 × 9.109 × 1.672) × 10⁻¹¹⁻³¹⁻²⁷
   G m_e m_p ≈ 101.5 × 10⁻⁶⁹
   G m_e m_p ≈ 1.015 × 10⁻⁶⁷ N m²

   Ratio = ke² / G m_e m_p = (2.30976 × 10⁻²⁸) / (1.015 × 10⁻⁶⁷)
   Ratio ≈ 2.275 × 10³⁹
   (Using more precise values, the ratio is approximately 2.29 × 10³⁹)

7. Step 7: Significance of the Ratio

   This ratio represents the ratio of the electrostatic force to the gravitational force between an electron and a proton. The extremely large value (approximately 2.29 × 10³⁹) signifies that:
   1. Electrostatic force is enormously stronger than gravitational force: For elementary particles like electrons and protons, the electrostatic interaction completely dominates the gravitational interaction.
   2. Dominance in atomic and molecular structures: The stability of atoms and molecules is governed by electrostatic forces, not gravitational forces.
   3. Gravitational force is negligible at the microscopic level: While gravity is the dominant force at macroscopic (planetary, stellar) scales, it is practically insignificant at the atomic and subatomic levels.

Solution Steps

1. Step 1: Part (a): Explanation of Quantisation of Electric Charge

   The principle of quantisation of electric charge states that electric charge is not continuous but comes in discrete packets. This means that the total charge (Q) on any body is always an integral multiple of a basic unit of charge, denoted by 'e'.

   This basic unit of charge is the magnitude of the charge on an electron or a proton, which is approximately 1.602 × 10⁻¹⁹ C.

   Mathematically, this can be expressed as Q = ±ne, where:
   * Q is the total charge on the body.
   * n is an integer (n = 1, 2, 3, ...).
   * e is the elementary charge (1.602 × 10⁻¹⁹ C).

   This implies that charges like 0.5e or 1.7e are not possible. All free charges are integral multiples of 'e'. Quarks, which have fractional charges (e.g., ±e/3, ±2e/3), exist inside protons and neutrons but are not observed as free particles.

2. Step 2: Part (b): Ignoring Quantisation for Macroscopic Charges

   When dealing with macroscopic charges, the quantisation of electric charge can be ignored because the elementary charge 'e' is extremely small (1.602 × 10⁻¹⁹ C).

   Macroscopic charges typically involve a very large number of elementary charges. For example, a charge of 1 microcoulomb (1 μC = 10⁻⁶ C) corresponds to approximately (10⁻⁶ C) / (1.602 × 10⁻¹⁹ C/electron) ≈ 6.24 × 10¹² elementary charges.

   When such a vast number of charges are involved, the difference between (n)e and (n+1)e becomes negligibly small compared to the total charge. It's analogous to how the granular nature of sand is not noticeable when looking at a large beach; from a distance, the beach appears as a continuous surface. Similarly, for large-scale charges, the discrete nature of charge is 'smoothed out', and charge can be treated as a continuous quantity for practical calculations and observations.

   Therefore, for macroscopic charges, we can effectively consider charge to be continuously variable without introducing significant error.

Solution Steps

1. Step 1: Identify Given Data and Convert Units

   Given charges: qA = 2 μC = 2 × 10⁻⁶ C qB = –5 μC = –5 × 10⁻⁶ C qC = 2 μC = 2 × 10⁻⁶ C qD = –5 μC = –5 × 10⁻⁶ C
   Charge at the centre, Q = 1 μC = 1 × 10⁻⁶ C
   Side of the square, a = 10 cm = 0.1 m

   Let the centre of the square be O. The distance from each corner to the centre of the square (r) is half of the diagonal length.
   Diagonal length = a√2 = 0.1√2 m r = (a√2) / 2 = a / √2 = 0.1 / √2 m

2. Step 2: Calculate Magnitude of Force due to each Corner Charge

   According to Coulomb's Law, the magnitude of the force between two point charges q₁ and q₂ separated by a distance r is F = k |q₁q₂| / r², where k = 9 × 10⁹ N m² C⁻².

   Force due to qA on Q (F_OA):
   F_OA = k |qA * Q| / r² = (9 × 10⁹ N m² C⁻²) * |(2 × 10⁻⁶ C) * (1 × 10⁻⁶ C)| / (0.1/√2 m)²
   F_OA = (9 × 10⁹ * 2 × 10⁻¹²) / (0.01 / 2) = (18 × 10⁻³) / 0.005 = 3.6 N

   Force due to qB on Q (F_OB):
   F_OB = k |qB * Q| / r² = (9 × 10⁹ N m² C⁻²) * |(–5 × 10⁻⁶ C) * (1 × 10⁻⁶ C)| / (0.1/√2 m)²
   F_OB = (9 × 10⁹ * 5 × 10⁻¹²) / (0.01 / 2) = (45 × 10⁻³) / 0.005 = 9.0 N

   Force due to qC on Q (F_OC):
   F_OC = k |qC * Q| / r² = (9 × 10⁹ N m² C⁻²) * |(2 × 10⁻⁶ C) * (1 × 10⁻⁶ C)| / (0.1/√2 m)²
   F_OC = (9 × 10⁹ * 2 × 10⁻¹²) / (0.01 / 2) = (18 × 10⁻³) / 0.005 = 3.6 N

   Force due to qD on Q (F_OD):
   F_OD = k |qD * Q| / r² = (9 × 10⁹ N m² C⁻²) * |(–5 × 10⁻⁶ C) * (1 × 10⁻⁶ C)| / (0.1/√2 m)²
   F_OD = (9 × 10⁹ * 5 × 10⁻¹²) / (0.01 / 2) = (45 × 10⁻³) / 0.005 = 9.0 N

3. Step 3: Determine Direction of Forces and Apply Superposition Principle

   The charge Q (1 μC) is positive. The forces are directed as follows:
   - F_OA: qA is positive, so F_OA is repulsive, directed away from A, along OA.
   - F_OB: qB is negative, so F_OB is attractive, directed towards B, along OB.
   - F_OC: qC is positive, so F_OC is repulsive, directed away from C, along OC.
   - F_OD: qD is negative, so F_OD is attractive, directed towards D, along OD.

   By symmetry, the distance from each corner to the center is the same (r). The charges at opposite corners are qA and qC, and qB and qD.

4. Step 4: Analyze Forces along Diagonals

   Consider the forces along the diagonal AC:
   - F_OA = 3.6 N, directed along OA (away from A).
   - F_OC = 3.6 N, directed along OC (away from C).
   Since A, O, C are collinear and O is the midpoint, the direction 'away from A' (along OA) is opposite to the direction 'away from C' (along OC). Therefore, F_OA and F_OC are equal in magnitude and opposite in direction.
   Net force along diagonal AC = F_OA + F_OC = 3.6 N (along OA) + 3.6 N (along OC) = 3.6 N - 3.6 N = 0.

   Consider the forces along the diagonal BD:
   - F_OB = 9.0 N, directed along OB (towards B).
   - F_OD = 9.0 N, directed along OD (towards D).
   Since B, O, D are collinear and O is the midpoint, the direction 'towards B' (along OB) is opposite to the direction 'towards D' (along OD). Therefore, F_OB and F_OD are equal in magnitude and opposite in direction.
   Net force along diagonal BD = F_OB + F_OD = 9.0 N (along OB) + 9.0 N (along OD) = 9.0 N - 9.0 N = 0.

5. Step 5: Calculate Total Net Force

   The total net force on the charge Q at the centre is the vector sum of all individual forces:
   F_net = F_OA + F_OB + F_OC + F_OD
   Since the net force along diagonal AC is zero and the net force along diagonal BD is zero, the total net force on the charge Q at the centre of the square is zero.
   F_net = 0.

Solution Steps

1. Step 1: Identify Given Data

   Given:
   Electric dipole moment, p = 4 × 10⁻⁹ C m
   Angle between the dipole moment and electric field, θ = 30°
   Magnitude of uniform electric field, E = 5 × 10⁴ N C⁻¹

2. Step 2: Recall Formula for Torque on an Electric Dipole

   The magnitude of the torque (τ) acting on an electric dipole placed in a uniform electric field is given by the formula:
   τ = pE sinθ

3. Step 3: Substitute Values and Calculate

   Substitute the given values into the formula:
   τ = (4 × 10⁻⁹ C m) × (5 × 10⁴ N C⁻¹) × sin(30°)
   We know that sin(30°) = 0.5
   τ = (4 × 10⁻⁹) × (5 × 10⁴) × 0.5
   τ = (20 × 10⁻⁵) × 0.5
   τ = 10 × 10⁻⁵ N m
   τ = 1.0 × 10⁻⁴ N m

Solution Steps

1. Step 1: Identify Given Parameters

   Radius of the spherical conductor, R = 12 cm = 0.12 m.
   Charge on the sphere, Q = 1.6 × 10⁻⁷ C.
   Coulomb's constant, k = 1 / (4πε₀) = 9 × 10⁹ N m²/C².

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

   For any conductor in electrostatic equilibrium, the electric field inside its volume is always zero. This is a fundamental property of conductors, as charges redistribute themselves to cancel out any internal field.
   Therefore, E_inside = 0 N/C.

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

   Just outside the sphere means at its surface, so the distance from the center is r = R = 0.12 m. For points outside a uniformly charged spherical conductor, the electric field can be calculated as if all the charge is concentrated at its center.
   The formula for electric field is E = kQ / r².
   Substitute the values: E_outside = (9 × 10⁹ N m²/C²) × (1.6 × 10⁻⁷ C) / (0.12 m)².
   E_outside = (14.4 × 10²) / (0.0144) = 100000 N/C = 1.0 × 10⁵ N/C.

4. Step 4: Calculate Electric Field at 18 cm from the Centre (c)

   The distance from the center is r = 18 cm = 0.18 m. This point is outside the sphere.
   Use the same formula: E = kQ / r².
   Substitute the values: E = (9 × 10⁹ N m²/C²) × (1.6 × 10⁻⁷ C) / (0.18 m)².
   E = (14.4 × 10²) / (0.0324) = 44444.44 N/C.
   Round to two significant figures: E ≈ 4.44 × 10⁴ N/C.

Solution Steps

1. Step 1: Identify Given Parameters

   Point charge, q = 2.0 μC = 2.0 × 10⁻⁶ C.
   Side of the cubic Gaussian surface = 9.0 cm. (Note: The dimensions and shape of the Gaussian surface are irrelevant for the total flux, as long as it encloses the charge).

2. Step 2: Recall Gauss's Law

   Gauss's Law states that the total electric flux (Φ) through any closed surface is equal to the net charge (q) enclosed within that surface divided by the permittivity of free space (ε₀).
   The formula is: Φ = q / ε₀.
   The value of permittivity of free space, ε₀ = 8.854 × 10⁻¹² C²/N m².

3. Step 3: Substitute Values and Calculate Flux

   Substitute the given charge and the value of ε₀ into Gauss's Law:
   Φ = (2.0 × 10⁻⁶ C) / (8.854 × 10⁻¹² C²/N m²)
   Φ = (2.0 / 8.854) × 10⁶ N m²/C
   Φ ≈ 0.22588 × 10⁶ N m²/C
   Φ ≈ 2.26 × 10⁵ N m²/C.

Solution Steps

1. Step 1: Identify Given Parameters and Understand the Setup

   Point charge, q = 10 μC = 10 × 10⁻⁶ C = 10⁻⁵ C.
   Side of the square, L = 10 cm.
   Distance of the charge from the center of the square, d = 5 cm.
   The hint suggests imagining the square as one face of a cube with edge 10 cm. Since the charge is 5 cm above the center of the 10 cm square, it means the charge is exactly at the center of a cube of side 10 cm, with the square being one of its faces. This is a crucial symmetry argument.

2. Step 2: Apply Gauss's Law for the Entire Cube

   If the charge 'q' is at the center of a closed cubic surface (Gaussian surface), then according to Gauss's Law, the total electric flux (Φ_total) through this entire closed surface is:
   Φ_total = q / ε₀ where ε₀ is the permittivity of free space (8.854 × 10⁻¹² C²/N m²).

3. Step 3: Calculate Flux Through One Face (the Square)

   Due to the symmetry of the cube and the central placement of the charge, the electric flux will be distributed equally among all six faces of the cube.
   Therefore, the flux through one face (the given square) is:
   Φ_square = Φ_total / 6 = (q / ε₀) / 6 = q / (6ε₀).

4. Step 4: Substitute Values and Compute

   Substitute the given charge and the value of ε₀:
   Φ_square = (10⁻⁵ C) / (6 × 8.854 × 10⁻¹² C²/N m²)
   Φ_square = (10⁻⁵) / (53.124 × 10⁻¹²)
   Φ_square = (1 / 5.3124) × 10⁶ N m²/C
   Φ_square ≈ 0.1882 × 10⁶ N m²/C
   Φ_square ≈ 1.88 × 10⁵ N m²/C.

Solution Steps

1. Step 1: Identify Given Parameters

   Diameter of the conducting sphere, D = 2.4 m.
   Surface charge density, σ = 80.0 μC/m² = 80.0 × 10⁻⁶ C/m².
   Permittivity of free space, ε₀ = 8.854 × 10⁻¹² C²/N m².

2. Step 2: Calculate Radius of the Sphere

   Radius, R = D / 2 = 2.4 m / 2 = 1.2 m.

3. Step 3: Calculate Surface Area of the Sphere

   The surface area of a sphere is given by the formula A = 4πR².
   A = 4π (1.2 m)² = 4π (1.44 m²) = 5.76π m².

4. Step 4: Calculate Total Charge on the Sphere (a)

   Surface charge density (σ) is defined as the total charge (Q) divided by the surface area (A): σ = Q / A.
   Rearranging for Q: Q = σ × A.
   Q = (80.0 × 10⁻⁶ C/m²) × (5.76π m²).
   Q = 460.8π × 10⁻⁶ C.
   Using π ≈ 3.14159, Q ≈ 460.8 × 3.14159 × 10⁻⁶ C ≈ 1447.64 × 10⁻⁶ C.
   Q ≈ 1.448 × 10⁻³ C (or approximately 1.45 mC).

5. Step 5: Calculate Total Electric Flux (b)

   According to Gauss's Law, the total electric flux (Φ) leaving a closed surface enclosing a net charge 'Q' is given by Φ = Q / ε₀.
   Substitute the calculated charge Q and the value of ε₀:
   Φ = (1.448 × 10⁻³ C) / (8.854 × 10⁻¹² C²/N m²).
   Φ = (1.448 / 8.854) × 10⁹ N m²/C.
   Φ ≈ 0.1635 × 10⁹ N m²/C.
   Φ ≈ 1.635 × 10⁸ N m²/C (or approximately 1.64 × 10⁸ N m²/C).

Solution Steps

1. Step 1: Identify Given Parameters

   Radius of the spherical conductor, R = 12 cm = 0.12 m
   Charge on the conductor, Q = 1.6 × 10⁻⁷ C
   Permittivity of free space, ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²
   Coulomb's constant, k = 1 / (4πε₀) = 9 × 10⁹ N m² C⁻²

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

   For a spherical conductor, the charge resides entirely on its outer surface. According to Gauss's Law, for any point inside the conductor (r < R), the Gaussian surface encloses no charge. Therefore, the electric field inside a charged spherical conductor is always zero.
   E_inside = 0 N/C

3. Step 3: Calculate Electric Field (b) Just outside the sphere

   Just outside the sphere, the distance from the center is r = R. The electric field at the surface of a spherical conductor is given by:
   E_surface = kQ / R²
   E_surface = (9 × 10⁹ N m² C⁻²) × (1.6 × 10⁻⁷ C) / (0.12 m)²
   E_surface = (14.4 × 10²) / (0.0144)
   E_surface = 100000 N/C = 1.0 × 10⁵ N/C

4. Step 4: Calculate Electric Field (c) At a point 18 cm from the centre of the sphere

   The distance from the center is r = 18 cm = 0.18 m. Since r > R, the sphere behaves like a point charge located at its center. The electric field is given by:
   E_outside = kQ / r²
   E_outside = (9 × 10⁹ N m² C⁻²) × (1.6 × 10⁻⁷ C) / (0.18 m)²
   E_outside = (14.4 × 10²) / (0.0324)
   E_outside = 44444.44 N/C ≈ 4.44 × 10⁴ N/C

Solution Steps

1. Step 1: Identify Given Parameters

   Electric field, E = 9 × 10⁴ N/C
   Distance from the line charge, r = 2 cm = 0.02 m
   Permittivity of free space, ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²
   Coulomb's constant, k = 1 / (4πε₀) = 9 × 10⁹ N m² C⁻²

2. Step 2: Recall Formula for Electric Field due to Infinite Line Charge

   The electric field (E) at a distance (r) from an infinite line charge with linear charge density (λ) is given by Gauss's Law as:
   E = λ / (2πε₀r)
   This can also be written using Coulomb's constant k as:
   E = (2kλ) / r

3. Step 3: Rearrange Formula to Solve for Linear Charge Density (λ)

   From E = (2kλ) / r, we can solve for λ:
   λ = (E × r) / (2k)

4. Step 4: Substitute Values and Calculate λ

   λ = (9 × 10⁴ N/C × 0.02 m) / (2 × 9 × 10⁹ N m² C⁻²)
   λ = (0.18 × 10⁴) / (18 × 10⁹)
   λ = (18 × 10⁻²) × 10⁴ / (18 × 10⁹)
   λ = 10⁻² × 10⁴ × 10⁻⁹
   λ = 10⁻⁷ C/m

Solution Steps

1. Step 1: Identify Given Parameters and Setup

   Magnitude of surface charge density, σ = 17.0 × 10⁻²² C/m².
   Let plate 1 have +σ and plate 2 have -σ.
   Permittivity of free space, ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².

   Electric field due to a single infinite plane sheet of charge is E = σ / (2ε₀). The direction is away from a positive sheet and towards a negative sheet.

2. Step 2: Analyze Electric Field (a) In the outer region of the first plate

   Consider a point P₁ to the left of plate 1.
   Electric field due to plate 1 (+σ): E₁ = σ / (2ε₀), directed away from plate 1 (leftward).
   Electric field due to plate 2 (-σ): E₂ = σ / (2ε₀), directed towards plate 2 (rightward).

   Net electric field E_a = E₁ + E₂ (vector sum).
   Since E₁ and E₂ are equal in magnitude and opposite in direction, they cancel out.
   E_a = (σ / (2ε₀)) - (σ / (2ε₀)) = 0 N/C.

3. Step 3: Analyze Electric Field (b) In the outer region of the second plate

   Consider a point P₂ to the right of plate 2.
   Electric field due to plate 1 (+σ): E₁ = σ / (2ε₀), directed away from plate 1 (rightward).
   Electric field due to plate 2 (-σ): E₂ = σ / (2ε₀), directed towards plate 2 (leftward).

   Net electric field E_b = E₁ + E₂ (vector sum).
   Since E₁ and E₂ are equal in magnitude and opposite in direction, they cancel out.
   E_b = (σ / (2ε₀)) - (σ / (2ε₀)) = 0 N/C.

4. Step 4: Analyze Electric Field (c) Between the plates

   Consider a point P₃ between the plates.
   Electric field due to plate 1 (+σ): E₁ = σ / (2ε₀), directed away from plate 1 (rightward).
   Electric field due to plate 2 (-σ): E₂ = σ / (2ε₀), directed towards plate 2 (rightward).

   Net electric field E_c = E₁ + E₂ (vector sum).
   Since both fields are in the same direction (rightward), they add up.
   E_c = (σ / (2ε₀)) + (σ / (2ε₀)) = σ / ε₀.

   Substitute the values:
   E_c = (17.0 × 10⁻²² C/m²) / (8.854 × 10⁻¹² C² N⁻¹ m⁻²)
   E_c ≈ 1.9200 × 10⁻¹⁰ N/C
   E_c ≈ 1.92 × 10⁻¹⁰ N/C

Solution Steps

1. Step 1: Identify Given Parameters and Initial Conditions

   Mass of the particle = m
   Charge of the particle = -q
   Uniform electric field = E
   Initial velocity (released from rest) = u = 0
   Time = t
   Neglect gravity.

2. Step 2: Calculate the Electric Force on the Particle

   The force (F) experienced by a charge (Q) in an electric field (E) is given by F = QE.
   Here, Q = -q, so the force on the particle is F = (-q)E.
   The magnitude of the force is F = qE. Since the charge is negative, the force will be in the direction opposite to the electric field E.

3. Step 3: Calculate the Acceleration of the Particle

   According to Newton's second law, F = ma.
   So, ma = qE
   Acceleration, a = qE / m

4. Step 4: Calculate the Velocity of the Particle after time t

   Since the particle is released from rest (u=0) and moves with constant acceleration (a), we can use the kinematic equation: v = u + at v = 0 + (qE / m)t v = (qEt) / m

5. Step 5: Calculate the Kinetic Energy of the Particle

   The kinetic energy (KE) of a particle is given by KE = (1/2)mv².
   Substitute the expression for velocity v:
   KE = (1/2)m [(qEt) / m]²
   KE = (1/2)m [q²E²t² / m²]
   KE = (1/2) [q²E²t² / m]
   KE = (q²E²t²) / (2m)

Solution Steps

1. Step 1: Identify Given Parameters and Convert Units

   Number of excess electrons, n = 12
   Charge of an electron, e = 1.60 × 10⁻¹⁹ C
   Electric field, E = 2.55 × 10⁴ N C⁻¹
   Density of oil, ρ = 1.26 g cm⁻³ = 1.26 × (10⁻³ kg) / (10⁻² m)³ = 1.26 × 10³ kg m⁻³
   Acceleration due to gravity, g = 9.81 m s⁻²

2. Step 2: Calculate the Total Charge on the Oil Drop

   The total charge (q) on the oil drop is due to the excess electrons: q = n × e q = 12 × (1.60 × 10⁻¹⁹ C) q = 19.2 × 10⁻¹⁹ C

3. Step 3: Apply Equilibrium Condition

   For the oil drop to be held stationary, the upward electric force (F_e) must balance the downward gravitational force (F_g).
   F_e = F_g

4. Step 4: Express Forces in Terms of Given Parameters

   Electric force, F_e = qE
   Gravitational force, F_g = mg
   Where m is the mass of the oil drop. The mass can be expressed using density (ρ) and volume (V) of the spherical drop (V = (4/3)πr³): m = ρV = ρ(4/3)πr³
   So, F_g = ρ(4/3)πr³g

5. Step 5: Set Up the Equilibrium Equation and Solve for Radius (r)

   qE = ρ(4/3)πr³g
   Rearrange to solve for r³: r³ = (3qE) / (4πρg)
   Substitute the calculated and given values: r³ = (3 × 19.2 × 10⁻¹⁹ C × 2.55 × 10⁴ N C⁻¹) / (4 × 3.14159 × 1.26 × 10³ kg m⁻³ × 9.81 m s⁻²) r³ = (146.88 × 10⁻¹⁵) / (155.02 × 10³) r³ ≈ 0.9474 × 10⁻¹⁸ m³

6. Step 6: Calculate the Radius (r)

   r = (0.9474 × 10⁻¹⁸)^(1/3) m r ≈ (0.9474)^(1/3) × (10⁻¹⁸)^(1/3) m r ≈ 0.9822 × 10⁻⁶ m r ≈ 9.82 × 10⁻⁷ m

Solution Steps

1. Step 1: Identify Given Parameters

   Given: Radius of the spherical conductor, R = 12 cm = 0.12 m. Charge on the conductor, Q = 1.6 × 10⁻⁷ C. Coulomb's constant, k = 1/(4πε₀) = 9 × 10⁹ Nm²/C².

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

   For a spherical conductor, the electric field inside the conductor is always zero. This is a fundamental property of conductors in electrostatic equilibrium, as all excess charge resides on the surface, and the field lines originate/terminate on charges, resulting in no net field inside. Therefore, E_inside = 0 N/C.

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

   The electric field just outside the surface of a spherical conductor is given by the formula E = kQ / R², where R is the radius of the sphere. Substitute the given values:
   E_outside = (9 × 10⁹ Nm²/C²) × (1.6 × 10⁻⁷ C) / (0.12 m)²
   E_outside = (14.4 × 10²) / (0.0144) N/C
   E_outside = 100000 N/C = 1.0 × 10⁵ N/C.

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

   The distance from the centre is r = 18 cm = 0.18 m. Since r > R, the point is outside the sphere. For an external point, a uniformly charged spherical conductor behaves as if all its charge is concentrated at its centre. The electric field is given by E = kQ / r².
   E = (9 × 10⁹ Nm²/C²) × (1.6 × 10⁻⁷ C) / (0.18 m)²
   E = (14.4 × 10²) / (0.0324) N/C
   E = 1440 / 0.0324 N/C
   E ≈ 44444.44 N/C ≈ 4.44 × 10⁴ N/C.

Solution Steps

1. Step 1: Identify Given Parameters and Understand the Setup

   Given: Point charge q = +10 μC = +10 × 10⁻⁶ C. Side of the square a = 10 cm. Distance of the charge from the centre of the square d = 5 cm. The hint suggests considering the square as one face of a cube of side 10 cm. Since the charge is 5 cm above the center of a 10 cm square, it means the charge is exactly at the geometric center of a cube formed by extending the square to make a cube of side 10 cm. This is crucial for applying symmetry.

2. Step 2: Apply Gauss's Law for the Entire Cube

   According to Gauss's Law, the total electric flux (Φ_total) through any closed surface enclosing a net charge q is given by Φ_total = q / ε₀, where ε₀ is the permittivity of free space (ε₀ = 8.854 × 10⁻¹² C²/Nm²).
   For the entire cube, which encloses the charge q:
   Φ_total = (10 × 10⁻⁶ C) / (8.854 × 10⁻¹² C²/Nm²)
   Φ_total ≈ 1.1294 × 10⁶ Nm²/C.

3. Step 3: Calculate Flux Through One Face Using Symmetry

   Due to the symmetry of the cube and the charge being at its exact center, the electric flux will be equally distributed through all six faces of the cube. Therefore, the electric flux through one face (the given square) will be one-sixth of the total flux:
   Φ_square = Φ_total / 6
   Φ_square = (1.1294 × 10⁶ Nm²/C) / 6
   Φ_square ≈ 0.18823 × 10⁶ Nm²/C
   Φ_square ≈ 1.88 × 10⁵ Nm²/C.

Solution Steps

1. Step 1: Identify Given Parameters

   Given: Point charge q = 2.0 μC = 2.0 × 10⁻⁶ C. The Gaussian surface is a cube of side 9.0 cm. The charge is at the center of the cube. The permittivity of free space ε₀ = 8.854 × 10⁻¹² C²/Nm².

2. Step 2: Apply Gauss's Law

   Gauss's Law states that the total electric flux (Φ) through any closed surface is directly proportional to the net electric charge (q) enclosed within that surface, divided by the permittivity of free space (ε₀). The shape and size of the Gaussian surface are irrelevant as long as it encloses the charge. The formula is Φ = q / ε₀.

3. Step 3: Calculate the Net Electric Flux

   Substitute the given charge into Gauss's Law:
   Φ = (2.0 × 10⁻⁶ C) / (8.854 × 10⁻¹² C²/Nm²)
   Φ ≈ 0.22588 × 10⁶ Nm²/C
   Φ ≈ 2.26 × 10⁵ Nm²/C.
   The net electric flux through the surface is approximately 2.26 × 10⁵ Nm²/C.

Solution Steps

1. Step 1: Identify Given Parameters

   Given: Initial electric flux Φ = –1.0 × 10³ Nm²/C. Initial radius of spherical Gaussian surface r₁ = 10.0 cm. Permittivity of free space ε₀ = 8.854 × 10⁻¹² C²/Nm².

2. Step 2: Determine Flux if Radius is Doubled (a)

   Gauss's Law states that the total electric flux through a closed surface depends only on the net charge enclosed within the surface and is independent of the size or shape of the Gaussian surface. Since the point charge remains the same and is still enclosed by the Gaussian surface (even if its radius is doubled), the net electric flux passing through the surface will remain unchanged. Therefore, the flux will still be –1.0 × 10³ Nm²/C.

3. Step 3: Calculate the Value of the Point Charge (b)

   Using Gauss's Law, Φ = q / ε₀. We can rearrange this to solve for the charge q: q = Φ × ε₀.
   Substitute the given flux and the value of ε₀: q = (–1.0 × 10³ Nm²/C) × (8.854 × 10⁻¹² C²/Nm²) q = –8.854 × 10⁻⁹ C.
   This can also be expressed as q = –8.854 nC. The negative sign indicates that the charge is negative, which is consistent with the negative flux (implying field lines pointing inwards).

Solution Steps

1. Step 1: Identify Given Parameters and Determine Charge Sign

   Given: Radius of the conducting sphere R = 10 cm = 0.10 m. Distance from the centre r = 20 cm = 0.20 m. Electric field E = 1.5 × 10³ N/C. The electric field points radially inward. Since electric field lines point towards negative charges, the net charge on the sphere must be negative.

2. Step 2: Recall Electric Field Formula for an External Point

   For a conducting sphere with charge Q, at an external point (r > R), the electric field is given by the formula E = k |Q| / r², where k = 1 / (4πε₀) = 9 × 10⁹ Nm²/C² and |Q| is the magnitude of the charge. We use |Q| because the formula gives the magnitude of the field, and the direction is determined separately.

3. Step 3: Rearrange Formula and Calculate Magnitude of Charge

   Rearrange the formula to solve for the magnitude of the charge |Q|: |Q| = E × r² / k.
   Substitute the given values:
   |Q| = (1.5 × 10³ N/C) × (0.20 m)² / (9 × 10⁹ Nm²/C²)
   |Q| = (1.5 × 10³) × (0.04) / (9 × 10⁹) C
   |Q| = (0.06 × 10³) / (9 × 10⁹) C
   |Q| = (6 × 10⁻² × 10³) / (9 × 10⁹) C
   |Q| = (6 × 10¹) / (9 × 10⁹) C
   |Q| = (2/3) × 10⁻⁸ C ≈ 0.6667 × 10⁻⁸ C ≈ 6.67 × 10⁻⁹ C.

4. Step 4: State the Net Charge with Correct Sign

   Since the electric field points radially inward, the net charge on the sphere must be negative. Therefore, the net charge on the sphere, Q = –6.67 × 10⁻⁹ C (or –6.67 nC).

Solution Steps

1. Step 1: Identify Given Parameters and Convert Units

   Given diameter D = 2.4 m, so radius R = D/2 = 1.2 m. Surface charge density σ = 80.0 µC/m² = 80.0 × 10⁻⁶ C/m². Permittivity of free space ε₀ = 8.854 × 10⁻¹² C²/Nm².

2. Step 2: Calculate Surface Area of the Sphere

   The surface area of a sphere is given by A = 4πR². Substitute the value of R: A = 4π(1.2 m)² = 4π(1.44 m²) = 5.76π m².

3. Step 3: Calculate Charge on the Sphere (Part a)

   The total charge q on the sphere is the product of its surface charge density and its surface area: q = σ × A. Substitute the values: q = (80.0 × 10⁻⁶ C/m²) × (5.76π m²) = 460.8π × 10⁻⁶ C. Using π ≈ 3.14159, q ≈ 1447.64 × 10⁻⁶ C ≈ 1.45 × 10⁻³ C.

4. Step 4: Apply Gauss's Law for Electric Flux (Part b)

   According to Gauss's Law, the total electric flux Φ_E leaving a closed surface is given by Φ_E = q / ε₀, where q is the total charge enclosed by the surface. In this case, the entire charge of the sphere is enclosed.

5. Step 5: Calculate Total Electric Flux

   Substitute the calculated charge q and the value of ε₀ into Gauss's Law: Φ_E = (1.44764 × 10⁻³ C) / (8.854 × 10⁻¹² C²/Nm²). Φ_E ≈ 1.635 × 10⁸ Nm²/C. Rounding to three significant figures, Φ_E ≈ 1.63 × 10⁸ Nm²/C.

Solution Steps

1. Step 1: Identify Given Parameters and Convert Units

   Given electric field E = 9 × 10⁴ N/C. Distance r = 2 cm = 0.02 m. The constant k = 1/(4πε₀) = 9 × 10⁹ Nm²/C².

2. Step 2: Recall Formula for Electric Field due to Infinite Line Charge

   The electric field E at a distance r from an infinite line charge with linear charge density λ is given by E = λ / (2πε₀r). This can also be written as E = (2kλ) / r, where k = 1/(4πε₀).

3. Step 3: Rearrange Formula to Solve for Linear Charge Density (λ)

   From E = (2kλ) / r, we can write λ = (E × r) / (2k).

4. Step 4: Substitute Values and Calculate λ

   Substitute the given values into the rearranged formula:
   λ = (9 × 10⁴ N/C × 0.02 m) / (2 × 9 × 10⁹ Nm²/C²)
   λ = (0.18 × 10⁴) / (18 × 10⁹) C/m
   λ = (18 × 10²) / (18 × 10⁹) C/m
   λ = 1 × 10⁻⁷ C/m.

Solution Steps

1. Step 1: Identify Given Parameters and Relevant Formulas

   Given magnitude of surface charge density σ = 17.0 × 10⁻²² C/m². Permittivity of free space ε₀ = 8.854 × 10⁻¹² C²/Nm². The electric field due to a single infinite plane sheet of charge is E = σ / (2ε₀).

2. Step 2: Define Electric Fields from Each Plate

   Let the first plate have charge density +σ and the second plate -σ. The electric field due to the positive plate (E₁) is σ / (2ε₀) and points away from it. The electric field due to the negative plate (E₂) is σ / (2ε₀) and points towards it.

3. Step 3: Analyze Electric Field in Outer Region of First Plate (Region I)

   Consider a point to the left of the first plate. E₁ points left, E₂ points right. Since their magnitudes are equal and directions opposite, the net electric field E_net = E₁ - E₂ = (σ / (2ε₀)) - (σ / (2ε₀)) = 0.

4. Step 4: Analyze Electric Field in Outer Region of Second Plate (Region III)

   Consider a point to the right of the second plate. E₁ points right, E₂ points left. Since their magnitudes are equal and directions opposite, the net electric field E_net = E₁ - E₂ = (σ / (2ε₀)) - (σ / (2ε₀)) = 0.

5. Step 5: Analyze Electric Field Between the Plates (Region II)

   Consider a point between the two plates. E₁ points right (away from positive plate), and E₂ also points right (towards negative plate). Since both fields are in the same direction, the net electric field E_net = E₁ + E₂ = (σ / (2ε₀)) + (σ / (2ε₀)) = σ / ε₀.

6. Step 6: Calculate Numerical Value for Electric Field Between Plates

   Substitute the given values into E_net = σ / ε₀:
   E_net = (17.0 × 10⁻²² C/m²) / (8.854 × 10⁻¹² C²/Nm²)
   E_net ≈ 1.920 × 10⁻¹⁰ N/C. Rounding to three significant figures, E_net ≈ 1.92 × 10⁻¹⁰ N/C.

Solution Steps

1. Step 1: List Given Values and Convert Units

   Number of excess electrons n = 12. Charge of electron e = 1.60 × 10⁻¹⁹ C. Electric field E = 2.55 × 10⁴ N/C. Density of oil ρ = 1.26 g/cm³ = 1.26 × 10³ kg/m³. Acceleration due to gravity g = 9.81 m/s².

2. Step 2: Calculate Total Charge on the Oil Drop

   The total charge q on the oil drop is q = n × e. Substitute values: q = 12 × (1.60 × 10⁻¹⁹ C) = 19.2 × 10⁻¹⁹ C.

3. Step 3: Calculate Electric Force on the Oil Drop

   The electric force F_e acting on the oil drop is F_e = qE. Substitute values: F_e = (19.2 × 10⁻¹⁹ C) × (2.55 × 10⁴ N/C) = 48.96 × 10⁻¹⁵ N.

4. Step 4: Express Gravitational Force in terms of Radius and Density

   The gravitational force F_g acting on the oil drop is F_g = mg. The mass m of the drop can be written as m = ρV, where V is the volume. For a spherical drop, V = (4/3)πr³. So, F_g = ρ × (4/3)πr³ × g.

5. Step 5: Apply Equilibrium Condition

   For the oil drop to be stationary, the upward electric force must balance the downward gravitational force: F_e = F_g. Therefore, qE = ρ × (4/3)πr³ × g.

6. Step 6: Solve for the Radius (r)

   Rearrange the equilibrium equation to solve for r³: r³ = (3qE) / (4πρg). Substitute the calculated F_e (which is qE) and other given values: r³ = (3 × 48.96 × 10⁻¹⁵ N) / (4 × 3.14159 × 1.26 × 10³ kg/m³ × 9.81 m/s²) r³ = (146.88 × 10⁻¹⁵) / (155.04 × 10³) m³ r³ = 0.9473 × 10⁻¹⁸ m³ r = (0.9473 × 10⁻¹⁸)^(1/3) m r ≈ 9.82 × 10⁻⁷ m. Rounding to two significant figures, r ≈ 9.8 × 10⁻⁷ m.

Solution Steps

1. Step 1: Identify Given Quantities and Convert Units

   Given:
   Number of excess electrons, n = 12
   Charge of an electron, e = 1.60 × 10⁻¹⁹ C
   Electric field, E = 2.55 × 10⁴ N C⁻¹
   Density of oil, ρ = 1.26 g cm⁻³
   Convert density to SI units: ρ = 1.26 × (10⁻³ kg) / (10⁻² m)³ = 1.26 × 10³ kg m⁻³
   Acceleration due to gravity, g = 9.81 m s⁻²

2. Step 2: Calculate Total Charge on the Oil Drop

   The total charge (q) on the oil drop due to 'n' excess electrons is given by: q = n × e q = 12 × 1.60 × 10⁻¹⁹ C q = 1.92 × 10⁻¹⁸ C

3. Step 3: Apply Equilibrium Condition

   For the oil drop to be held stationary, the upward electric force (F_e) must balance the downward gravitational force (F_g).
   F_e = F_g

4. Step 4: Express Forces in Terms of Given Quantities

   The electric force is given by:
   F_e = qE
   The gravitational force is given by:
   F_g = mg
   Where 'm' is the mass of the oil drop. Assuming the oil drop is spherical with radius 'r', its volume V = (4/3)πr³.
   The mass 'm' can be expressed using density (ρ) and volume (V): m = ρV = ρ(4/3)πr³
   So, F_g = ρ(4/3)πr³g

5. Step 5: Equate Forces and Solve for Radius (r)

   Equating the electric and gravitational forces: qE = ρ(4/3)πr³g
   Substitute the value of q:
   (n e)E = ρ(4/3)πr³g
   Rearrange the equation to solve for r³: r³ = (3neE) / (4πρg)
   Now, take the cube root to find r: r = [(3neE) / (4πρg)]^(1/3)

6. Step 6: Substitute Values and Calculate

   Substitute the numerical values into the equation for r: r³ = (3 × 12 × 1.60 × 10⁻¹⁹ C × 2.55 × 10⁴ N C⁻¹) / (4 × π × 1.26 × 10³ kg m⁻³ × 9.81 m s⁻²)

   Numerator = 3 × 12 × 1.60 × 10⁻¹⁹ × 2.55 × 10⁴ = 1.4688 × 10⁻¹³
   Denominator = 4 × 3.14159 × 1.26 × 10³ × 9.81 = 155609.97

   r³ = (1.4688 × 10⁻¹³) / (155609.97) ≈ 9.4389 × 10⁻¹⁹ m³

   r = (9.4389 × 10⁻¹⁹)^(1/3) m r ≈ 9.809 × 10⁻⁷ m

   Rounding to two significant figures (based on 2.55 and 1.26): r ≈ 9.8 × 10⁻⁷ m or 0.98 µm.

Solution Steps

1. Step 1: Identify Given Quantities and Directions

   Given:
   Electric field E is along the positive z-direction. Its magnitude increases uniformly along the positive z-direction.
   Rate of increase of electric field, dE/dz = 10⁵ N C⁻¹ m⁻¹.
   Electric dipole moment p = 10⁻⁷ C m in the negative z-direction.

   In vector form:
   Electric field: E = E(z) k̂
   Dipole moment: p = -10⁻⁷ k̂ C m

2. Step 2: Calculate Force on the Electric Dipole

   For an electric dipole in a non-uniform electric field, the net force is given by F = (p ⋅ ∇)E. When the electric field varies only along the direction of the dipole moment, the force simplifies to:
   F = p_z * (dE_z/dz) k̂
   Here, p_z is the component of the dipole moment along the z-axis, which is -10⁻⁷ C m.

   F = (-10⁻⁷ C m) × (10⁵ N C⁻¹ m⁻¹) k̂
   F = -10⁻² N k̂

   Thus, the force experienced by the dipole is 10⁻² N in the negative z-direction.

3. Step 3: Calculate Torque on the Electric Dipole

   The torque (τ) experienced by an electric dipole in an electric field is given by the cross product:
   τ = p × E

   We have p = -10⁻⁷ k̂ C m and E = E(z) k̂.

   τ = (-10⁻⁷ k̂) × (E(z) k̂)
   Since the cross product of two parallel or anti-parallel unit vectors is zero (k̂ × k̂ = 0):
   τ = 0

   Alternatively, the magnitude of torque is given by |τ| = |p||E|sinθ, where θ is the angle between p and E. Here, p is in the negative z-direction and E is in the positive z-direction, so they are anti-parallel. The angle θ = 180°.
   Since sin(180°) = 0, the torque τ = 0.

   Therefore, the torque experienced by the electric dipole is zero.

Solution Steps

1. Step 1: Identify Given Parameters and Setup

   Given: Length L = 1 m, Area of cross-section A = 25 cm² = 25 × 10⁻⁴ m², Uniform electric field E = 10⁵ N C⁻¹. The axis of the cylinder is parallel to the electric field. We need to find (a) net flux and (b) enclosed charge.

2. Step 2: Analyze Flux through Different Surfaces

   A cylinder has three surfaces: two circular end faces and one curved surface. Since the electric field is uniform and parallel to the axis, field lines enter through one end face and exit through the other. No field lines cross the curved surface.

3. Step 3: Calculate Flux through the Entering Face

   For the face where electric field lines enter, the area vector (dA) points opposite to the electric field (E). So, the angle θ = 180°. Flux Φ₁ = E ⋅ A = EA cos(180°) = -EA.

4. Step 4: Calculate Flux through the Exiting Face

   For the face where electric field lines exit, the area vector (dA) points in the same direction as the electric field (E). So, the angle θ = 0°. Flux Φ₂ = E ⋅ A = EA cos(0°) = +EA.

5. Step 5: Calculate Flux through the Curved Surface

   For the curved surface, the area vector (dA) at any point is perpendicular to the electric field (E). So, the angle θ = 90°. Flux Φ₃ = ∫ E ⋅ dA = ∫ EA cos(90°) = 0.

6. Step 6: Calculate Net Flux (Part a)

   The total net flux through the cylinder is the sum of fluxes through all its surfaces: Φ_net = Φ₁ + Φ₂ + Φ₃ = -EA + EA + 0 = 0. Thus, the net flux is 0 N m² C⁻¹.

7. Step 7: Apply Gauss's Law (Part b)

   According to Gauss's Law, the net electric flux through any closed surface is equal to the net charge enclosed within the surface divided by the permittivity of free space (ε₀): Φ_net = Q_enclosed / ε₀.

8. Step 8: Calculate Enclosed Charge (Part b)

   Since we found Φ_net = 0, substituting this into Gauss's Law: 0 = Q_enclosed / ε₀. This implies Q_enclosed = 0 × ε₀ = 0 C. Thus, the charge enclosed by the cylinder is 0 C.

Solution Steps

1. Step 1: Identify Given Parameters

   Given: Initial electric flux Φ = –1.0 × 10³ N m² C⁻¹, Initial radius r₁ = 10.0 cm. We need to find (a) flux if radius is doubled and (b) the value of the point charge.

2. Step 2: Analyze Flux Dependence on Gaussian Surface (Part a)

   Gauss's Law states that the total electric flux through a closed surface is proportional to the net charge enclosed within that surface, and it is independent of the size or shape of the Gaussian surface, as long as the enclosed charge remains the same.

3. Step 3: Determine Flux for Doubled Radius (Part a)

   Since the point charge remains enclosed within the Gaussian surface even if its radius is doubled, the net charge enclosed (Q_enclosed) does not change. Therefore, the electric flux passing through the surface will remain the same. Flux = –1.0 × 10³ N m² C⁻¹.

4. Step 4: Recall Gauss's Law for Charge Calculation (Part b)

   Gauss's Law relates electric flux (Φ) to the enclosed charge (Q_enclosed) and the permittivity of free space (ε₀): Φ = Q_enclosed / ε₀.

5. Step 5: Substitute Values and Calculate Charge (Part b)

   Rearrange Gauss's Law to solve for Q_enclosed: Q_enclosed = Φ × ε₀. Substitute the given flux Φ = –1.0 × 10³ N m² C⁻¹ and the value of ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².
   Q_enclosed = (–1.0 × 10³) × (8.854 × 10⁻¹²) = –8.854 × 10⁻⁹ C.

6. Step 6: State Final Charge Value

   The value of the point charge is –8.854 × 10⁻⁹ C or –8.854 nC. The negative sign indicates a negative charge, consistent with the negative (inward) flux.

Solution Steps

1. Step 1: Identify Given Parameters

   Given: Radius of sphere R = 10 cm = 0.10 m. Distance from center r = 20 cm = 0.20 m. Electric field E = 1.5 × 10³ N C⁻¹. Direction of E is radially inward. We need to find the net charge Q on the sphere.

2. Step 2: Recall Electric Field Formula for a Charged Sphere

   For a conducting sphere, all charge resides on its surface. For points outside the sphere (r > R), the electric field behaves as if all the charge is concentrated at its center. The formula for the electric field at a distance r from a point charge Q is E = (1 / 4πε₀) * (Q / r²).

3. Step 3: Rearrange Formula to Solve for Charge Q

   From E = (1 / 4πε₀) * (Q / r²), we can write Q = E * r² * (4πε₀) or Q = E * r² / (1 / 4πε₀).

4. Step 4: Substitute Known Values

   Substitute E = 1.5 × 10³ N C⁻¹, r = 0.20 m, and the Coulomb's constant k = (1 / 4πε₀) = 9 × 10⁹ N m² C⁻².
   Q = (1.5 × 10³) × (0.20)² / (9 × 10⁹).

5. Step 5: Perform Calculation

   Q = (1.5 × 10³) × (0.04) / (9 × 10⁹)
   Q = (0.06 × 10³) / (9 × 10⁹)
   Q = (6 × 10⁻² × 10³) / (9 × 10⁹)
   Q = (6 × 10¹) / (9 × 10⁹)
   Q = (2/3) × 10⁻⁸ C
   Q ≈ 0.666... × 10⁻⁸ C ≈ 6.67 × 10⁻⁹ C.

6. Step 6: Determine Sign of the Charge

   The problem states that the electric field points radially inward. Electric field lines point away from positive charges and towards negative charges. Therefore, an inward electric field implies that the net charge on the sphere is negative.

7. Step 7: State Final Answer with Sign

   The net charge on the sphere is –6.67 × 10⁻⁹ C (or –6.67 nC).

Solution Steps

1. Step 1: Identify Given Parameters and Convert Units

   Given: Diameter D = 2.4 m, so Radius R = D/2 = 1.2 m. Surface charge density σ = 80.0 µC/m² = 80.0 × 10⁻⁶ C/m². We need to find (a) total charge Q and (b) total electric flux Φ.

2. Step 2: Recall Formula for Total Charge (Part a)

   For a uniformly charged sphere, the total charge Q is the product of the surface charge density (σ) and the total surface area (A) of the sphere. Q = σ × A.

3. Step 3: Calculate Surface Area of the Sphere (Part a)

   The surface area of a sphere is A = 4πR². Substitute R = 1.2 m. A = 4π(1.2)² = 4π(1.44) = 5.76π m².

4. Step 4: Calculate Total Charge Q (Part a)

   Q = σ × A = (80.0 × 10⁻⁶ C/m²) × (5.76π m²)
   Q = 80.0 × 10⁻⁶ × 5.76 × 3.14159
   Q = 1447.64 × 10⁻⁶ C ≈ 1.448 × 10⁻³ C. Rounding to three significant figures, Q ≈ 1.45 × 10⁻³ C.

5. Step 5: Recall Gauss's Law for Electric Flux (Part b)

   Gauss's Law states that the total electric flux (Φ) leaving a closed surface is equal to the net charge (Q) enclosed within the surface divided by the permittivity of free space (ε₀). Φ = Q / ε₀.

6. Step 6: Substitute Values and Calculate Total Electric Flux (Part b)

   Use the calculated charge Q = 1.44764 × 10⁻³ C (using the unrounded value for better accuracy in subsequent calculation) and ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².
   Φ = (1.44764 × 10⁻³ C) / (8.854 × 10⁻¹² C² N⁻¹ m⁻²)
   Φ ≈ 0.16350 × 10⁹ N m² C⁻¹
   Φ ≈ 1.635 × 10⁸ N m² C⁻¹. Rounding to three significant figures, Φ ≈ 1.64 × 10⁸ N m² C⁻¹.

Solution Steps

1. Step 1: Electric Field Outside a Conductor

   Recall that the electric field just outside the surface of a charged conductor in electrostatic equilibrium is given by $E_{out}=\frac{\sigma }{{ϵ}_0}\hat{n}$, where $\sigma$ is the surface charge density and $\hat{n}$ is the unit vector in the outward normal direction. This field is the resultant of all charges on the conductor.

2. Step 2: Applying Superposition Principle (Outside)

   Imagine the conductor's surface charge is divided into two parts: a small 'disc' of charge (which would have been at the location of the hole) and the 'rest' of the conductor. Let $E_1$ be the field due to the small disc and $E_2$ be the field due to the rest of the conductor. At a point P just outside the conductor (near the hole), both $E_1$ and $E_2$ point outwards. Thus, the total field is $E_{out}=E_1+E_2=\frac{\sigma }{{ϵ}_0}\hat{n}$.

3. Step 3: Applying Superposition Principle (Inside)

   Now consider a point Q just inside the conductor (near the hole). The total electric field inside a conductor in electrostatic equilibrium is zero. At point Q, the field $E_1$ (due to the imaginary disc) still points outwards. For the net field to be zero, the field $E_2$ (due to the rest of the conductor) must point inwards and be equal in magnitude to $E_1$. So, $E_{in}=E_1-E_2=0$, which implies $E_1=E_2$.

4. Step 4: Calculating Field in the Hole

   The electric field *in the hole* is the field at a point where the 'disc' of charge is absent. Therefore, the field in the hole is solely due to the 'rest' of the conductor, which is $E_2$. Using the results from Step 2 and Step 3:
   Substitute $E_1=E_2$ into the equation from Step 2:
   $E_2+E_2=\frac{\sigma }{{ϵ}_0}\hat{n}$
   $2E_2=\frac{\sigma }{{ϵ}_0}\hat{n}$
   $E_2=\frac{\sigma }{2{ϵ}_0}\hat{n}$
   Thus, the electric field in the hole is $E_{hole}=\frac{\sigma }{2{ϵ}_0}\hat{n}$.

Solution Steps

1. Step 1: Setup and Differential Charge Element

   Consider a long straight wire placed along the x-axis. Let the point P where the electric field is to be calculated be on the y-axis at a perpendicular distance 'r' from the wire. The linear charge density of the wire is $\lambda$. Take a small differential element of length $dx$ on the wire at a distance $x$ from the origin. The charge on this element is $dq=\lambda dx$.

2. Step 2: Electric Field due to Differential Element

   The distance from the charge element $dq$ to point P is $R=\sqrt{x^2+r^2}$. According to Coulomb's law, the magnitude of the electric field $dE$ at P due to $dq$ is $dE=\frac{1}{4\pi {ϵ}_0}\frac{dq}{R^2}=\frac{1}{4\pi {ϵ}_0}\frac{\lambda dx}{x^2+r^2}$.

3. Step 3: Resolve into Components and Apply Symmetry

   The electric field $dE$ has components $d{E}_x$ (along x-axis) and $d{E}_y$ (along y-axis). Let $heta$ be the angle between $R$ and the y-axis. Then $d{E}_x=dE\sin heta$ and $d{E}_y=dE\cos heta$. From the geometry, $\cos heta=\frac{r}{R}=\frac{r}{\sqrt{x^2+r^2}}$. Due to the infinite length of the wire, for every element at $x$, there's a symmetric element at $-x$. The x-components ($d{E}_x$) will cancel out, leaving only the y-component ($d{E}_y$). Thus, the total electric field $E$ will be in the y-direction (radially outward).

4. Step 4: Set up the Integral for the Perpendicular Component

   The perpendicular component is $dE_y = dE \cos heta = \frac{λ dx}{4π\epsilon₀ (x² + r²)} \left( \frac{r}{√(x² + r²)}
   ight) = \frac{λ r}{4π\epsilon₀} \frac{dx}{(x² + r²)^3/2}$.Tofindthetotalelectricfield$E_y$,integratethisexpressionfrom$x = -\infty$to$x = +\infty$:$E_y = \int_-\infty^{+\infty} \frac{λ r}{4π\epsilon₀} \frac{dx}{(x² + r²)^3/2}$.

5. Step 5: Evaluate the Integral using Trigonometric Substitution

   Let $x=ranhet{a}^{\prime }$. Then $dx=r{\sec }^{2}het{a}^{\prime }dhet{a}^{\prime }$. Also, $x^2+r^2=r^{2}a{n}^{2}het{a}^{\prime }+r^2=r^2{\sec }^{2}het{a}^{\prime }$, so $(x^2+r^2{)}^{3/2}=r^3{\sec }^{3}het{a}^{\prime }$. The limits of integration change from $x=-\infty$ to $x=+\infty$ to $het{a}^{\prime }=-\pi /2$ to $het{a}^{\prime }=+\pi /2$.
   Substituting these into the integral:
   $E_y=\frac{\lambda r}{4\pi {ϵ}_0}{\int }_{-\pi /2}^{\pi /2}\frac{r{\sec }^{2}het{a}^{\prime }dhet{a}^{\prime }}{r^3{\sec }^{3}het{a}^{\prime }}=\frac{\lambda }{4\pi {ϵ}_{0}r}{\int }_{-\pi /2}^{\pi /2}\cos het{a}^{\prime }dhet{a}^{\prime }$.

6. Step 6: Final Calculation

   Evaluate the definite integral:
   $E_y=\frac{\lambda }{4\pi {ϵ}_{0}r}[\sin het{a}^{\prime }{]}_{-\pi /2}^{\pi /2}=\frac{\lambda }{4\pi {ϵ}_{0}r}[\sin (\pi /2)-\sin (-\pi /2)]=\frac{\lambda }{4\pi {ϵ}_{0}r}[1-(-1)]=\frac{\lambda }{4\pi {ϵ}_{0}r}[2]$.
   Therefore, the electric field is $E=E_y=\frac{2\lambda }{4\pi {ϵ}_{0}r}=\frac{\lambda }{2\pi {ϵ}_{0}r}$. The direction is radially outward from the wire.