(a) Consider a point charge of 5 nC placed at a distance of 1 m from a perfect conducting plane (z = 0) of infinite extent. Find…

(a)

Choose the origin at the foot of the perpendicular from the point charge to the plane z = 0. Since the charge is 1 m above the plane, take the charge at (0, 0, 1) m. The observation point is P = (2, 2, 0) m.

For a perfect conducting infinite plane, use the method of images. The plane z = 0 is replaced by an image charge q′ = −q at (0, 0, −1) m, and the field in z > 0 is the superposition of the fields of q and q′.

Given q = 5 nC = 5 × 10⁻⁹ C, and 1/(4π ε₀) = 9 × 10⁹ N m² C⁻².

Vector from real charge at (0, 0, 1) to P: r₁ = (2, 2, −1) m, |r₁| = √(2² + 2² + (−1)²) = √9 = 3 m.

Field due to real charge: E₁ = (1/(4π ε₀)) q r₁/|r₁|³ = (9 × 10⁹)(5 × 10⁻⁹)/27 × (2, 2, −1) = (5/3)(2, 2, −1) = (10/3, 10/3, −5/3) N C⁻¹.

Vector from image charge at (0, 0, −1) to P: r₂ = (2, 2, 1) m, |r₂| = 3 m.

Field due to image charge q′ = −5 × 10⁻⁹ C: E₂ = (1/(4π ε₀)) q′ r₂/|r₂|³ = (9 × 10⁹)(−5 × 10⁻⁹)/27 × (2, 2, 1) = −(5/3)(2, 2, 1) = (−10/3, −10/3, −5/3) N C⁻¹.

Total field at P: E = E₁ + E₂ = (0, 0, −10/3) N C⁻¹.

Thus E = −(10/3) k̂ N C⁻¹ = (0, 0, −3.33) N C⁻¹, where k̂ is the unit vector normal to the plane z = 0. Since E has only a z-component, its tangential component on the plane is zero. Hence the electric field is normal to the conducting plane, as required. At the plane itself, this is the external field just above z = 0; the field inside the conductor is zero.

(b)

For a flat coil in a magnetic field, the magnetic dipole moment has magnitude μ = N I A, where N is the number of turns, I is the current, and A is the area.

Area of the rectangular coil: A = 0.5 m × 0.4 m = 0.20 m².

Therefore, μ = 50 × 1.5 A × 0.20 m² = 15 A m².

The torque magnitude on a magnetic dipole is τ = μ B sin θ, where θ is the angle between the magnetic dipole moment and B. The dipole moment is normal to the plane of the coil. The magnetic field makes 60° with the plane of the coil, so the angle between B and the normal is θ = 90° − 60° = 30°.

Thus, τ = 15 A m² × 0.1 T × sin 30° = 15 × 0.1 × 0.5 N m = 0.75 N m.

Torque = 0.75 N m, directed perpendicular to both the coil normal and B; its sense is given by the right-hand rule from μ⃗ × B⃗.

(c)

Kirchhoff’s current law (KCL): At any junction or node in an electrical network, the algebraic sum of currents entering the node is zero, or the total current entering the node equals the total current leaving it. That is, ∑ I = 0 at a node, taking currents entering as positive and leaving as negative.

Explanation and derivation from charge conservation: Consider a node. If currents I₁, I₂, … flow into it, the net rate at which charge enters the node is ∑ I. Charge conservation requires ∑ I = dq/dt, where q is the charge accumulated at the node. In the lumped-circuit approximation, an ideal node has negligible capacitance and cannot store charge, so dq/dt = 0. Hence ∑ I = 0. More generally, if charge can accumulate, ∑ I_in = ∑ I_out + dq/dt. For steady currents, this reduces exactly to KCL.

Kirchhoff’s voltage law (KVL): Around any closed loop in a network, the algebraic sum of all potential differences is zero, or the sum of electromotive forces equals the sum of potential drops. That is, ∑ V = 0 around a closed loop.

Explanation and derivation from energy conservation: Let a unit positive charge be carried once around a closed loop. In the electrostatic or lumped-circuit approximation, the electric field is conservative, so the work done per unit charge around a closed path is ∮ E⃗ · dl⃗ = 0. Equivalently, the potential at the starting point equals the potential at the ending point after one complete traversal. Therefore, every potential rise must be balanced by an equal potential drop. If sources of emf are present, energy conservation gives ∑ emf = ∑ IR. If a changing magnetic flux links the loop, KVL is modified by Faraday’s law: the induced emf −dΦ/dt must be included, and then ∑ V = −dΦ/dt. In ordinary lumped-circuit analysis, this induced effect is either negligible or represented by an explicit emf source. Thus KCL follows from conservation of charge and KVL from conservation of energy.

(d)

Inside a charging parallel-plate capacitor, the electric field is axial and changing. The changing electric field produces a displacement current density J_d = ε₀ ∂E/∂t.

Use the Ampère–Maxwell law: ∮ H⃗ · dl⃗ = ∫ (J⃗ + ∂D⃗/∂t) · dA⃗. Inside the capacitor gap, conduction current J = 0, and D = ε₀E. Choose a circular Amperian loop of radius r centred on the axis, with r < R. By symmetry, H is azimuthal and has constant magnitude on this loop. Therefore, H (2πr) = ε₀ (dE/dt) (πr²).

Hence, |H⃗| = (ε₀ r/2)(dE/dt).

Given dE/dt = 5.0 V m⁻¹ s⁻¹, ε₀ = 8.854 × 10⁻¹² F m⁻¹, and plate radius R = 10 cm = 0.10 m. At the rim just inside the capacitor, r = R = 0.10 m: |H⃗| = (8.854 × 10⁻¹² × 0.10 × 5.0)/2 A m⁻¹ = 2.2135 × 10⁻¹² A m⁻¹.

Thus |H⃗| at r = 0.10 m is about 2.21 × 10⁻¹² A m⁻¹. More generally, at radial distance r from the axis, |H⃗| = (ε₀ r/2)(dE/dt), so the field is zero on the axis and maximum at the plate edge.

(e)

Let Q₁₂₀₀ = +1200 kJ be the heat absorbed by the engine from the 1200 K reservoir. The engine delivers work W = +400 kJ. Let Q₄₀₀ and Q₃₀₀ be the heat interactions with the 400 K and 300 K reservoirs, taken positive when absorbed by the engine.

By the first law for a cyclic engine, Q₁₂₀₀ + Q₄₀₀ + Q₃₀₀ = W. Therefore, 1200 + Q₄₀₀ + Q₃₀₀ = 400, so Q₄₀₀ + Q₃₀₀ = −800 kJ. … (1)

For a reversible engine, the total entropy change of the engine over a cycle is zero, and the entropy changes of the reservoirs sum to zero: Q₁₂₀₀/T₁ + Q₄₀₀/T₂ + Q₃₀₀/T₃ = 0. Thus, 1200/1200 + Q₄₀₀/400 + Q₃₀₀/300 = 0, or 1 + Q₄₀₀/400 + Q₃₀₀/300 = 0. … (2)

From (1), Q₄₀₀ = −800 − Q₃₀₀. Substitute in (2): 1 + (−800 − Q₃₀₀)/400 + Q₃₀₀/300 = 0. This gives 1 − 2 − Q₃₀₀/400 + Q₃₀₀/300 = 0, so −1 + Q₃₀₀(1/300 − 1/400) = 0. Since 1/300 − 1/400 = 1/1200, Q₃₀₀ = 1200 kJ. Then Q₄₀₀ = −800 − 1200 = −2000 kJ.

Therefore, the engine absorbs 1200 kJ from the 300 K reservoir and rejects 2000 kJ to the 400 K reservoir. The 1200 K reservoir supplies 1200 kJ, and the net heat absorbed is 1200 + 1200 − 2000 = 400 kJ, which equals the work delivered. Check entropy: 1200/1200 + 1200/300 − 2000/400 = 1 + 4 − 5 = 0, confirming reversibility.