Elements of Electromagnetics – Solutions Manual
(7th Edition) Comprehensive Q&A Study Guide
2026-2027
Chapter 1: Vector Algebra
Q1: Given three vectors A = 3ax + 2ay - az, B = ax - 4ay + 2az, and C = -2ax + ay + 3az, find A + B -
C.
A1:
Add the corresponding x, y, and z components:
x: 3 + 1 - (-2) = 6
y: 2 + (-4) - 1 = -3
z: -1 + 2 - 3 = -2
Result: 6ax - 3ay - 2az
Q2: Find the dot product of A = 2ax + ay and B = ax - 3ay.
A2:
A · B = (2)(1) + (1)(-3) = 2 - 3 = -1
Q3: Determine the angle between vectors A = ax + ay + az and B = -2ax - 2ay + 2az.
A3:
|A| = √(1² + 1² + 1²) = √3
|B| = √((-2)² + (-2)² + 2²) = √12 = 2√3
A · B = (1)(-2) + (1)(-2) + (1)(2) = -2
cos θ = (A · B) / (|A||B|) = -2 / (√3 * 2√3) = -2/6 = -1/3
θ = cos⁻¹(-1/3) ≈ 109.47°
,Q4: Find the cross product A × B if A = ax + 2ay and B = 2ax + ay.
A4:
A × B = ( (2)(1) - (0)(0) )ax - ( (1)(1) - (0)(2) )ay + ( (1)(0) - (2)(2) )az
Result: 2ax - ay - 4az
Q5: Convert the Cartesian point P(1, 2, 3) to cylindrical coordinates.
A5:
ρ = √(x² + y²) = √(1² + 2²) = √5
φ = tan⁻¹(y/x) = tan⁻¹(2/1) ≈ 63.43°
z=3
Result: (√5, 63.43°, 3)
Chapter 2: Coordinate Systems and Transformation
Q6: Convert the vector A = zax - xay + yaz into cylindrical coordinates.
A6:
In cylindrical coordinates, ax = cosφ aρ - sinφ aφ and ay = sinφ aρ + cosφ aφ.
Substitute x = ρcosφ, y = ρsinφ:
A = z(cosφ aρ - sinφ aφ) - ρcosφ(sinφ aρ + cosφ aφ) + ρsinφ az
Result: (zcosφ - ρsinφcosφ)aρ - (zsinφ + ρcos²φ)aφ + ρsinφ az
Q7: Convert the Cartesian point P(2, -2, 1) to spherical coordinates.
A7:
r = √(x² + y² + z²) = √(4 + 4 + 1) = 3
θ = cos⁻¹(z/r) = cos⁻¹(1/3) ≈ 70.53°
φ = tan⁻¹(y/x) = tan⁻¹(-2/2) = tan⁻¹(-1) = -45° (or 315°)
Result: (3, 70.53°, 315°)
,Q8: Given A = 4ar - sinθ aθ + aφ in spherical coordinates, find A · ar at point P(2, π/2, 0).
A8:
The dot product A · ar simply extracts the radial component of A.
Therefore, A · ar = 4 (independent of the point P).
Chapter 3: Vector Calculus
Q9: Find the gradient of the scalar field V = x²y + z at point (1, 2, 3).
A9:
∇V = (∂V/∂x)ax + (∂V/∂y)ay + (∂V/∂z)az
∂V/∂x = 2xy, ∂V/∂y = x², ∂V/∂z = 1
At (1, 2, 3): ∇V = (212)ax + (1²)ay + 1az = 4ax + ay + az
Q10: Calculate the divergence of the vector field D = x²z ax + y²x ay + z²y az.
A10:
∇ · D = ∂(x²z)/∂x + ∂(y²x)/∂y + ∂(z²y)/∂z
∇ · D = 2xz + 2yx + 2zy
Result: 2xz + 2xy + 2yz
Q11: Find the curl of the vector field A = yz ax + xz ay + xy az.
A11:
∇ × A = (∂(xy)/∂y - ∂(xz)/∂z)ax - (∂(xy)/∂x - ∂(yz)/∂z)ay + (∂(xz)/∂x - ∂(yz)/∂y)az
∇ × A = (x - x)ax - (y - y)ay + (z - z)az = 0
Q12: Given the vector field A = ρ aρ in cylindrical coordinates, find ∇ · A.
A12:
, In cylindrical coordinates: ∇ · A = (1/ρ) ∂(ρ Aρ)/∂ρ + (1/ρ) ∂Aφ/∂φ + ∂Az/∂z
∇ · A = (1/ρ) ∂(ρ * ρ)/∂ρ = (1/ρ) ∂(ρ²)/∂ρ = (1/ρ)(2ρ) = 2
Q13: State the divergence theorem.
A13:
The divergence theorem states that the volume integral of the divergence of a vector field A
over a volume v is equal to the closed surface integral of A over the boundary surface S of that
volume:
∫_v (∇ · A) dv = ∮_S A · dS
Q14: State Stokes's theorem.
A14:
Stokes's theorem states that the surface integral of the curl of a vector field A over an open
surface S is equal to the closed line integral of A around the boundary curve L of that surface:
∫_S (∇ × A) · dS = ∮_L A · dl
Q15: If ∇ · A = 0, what type of vector field is A?
A15:
A is a solenoidal vector field.
Chapter 4: Electrostatic Fields
Q16: Two point charges, Q1 = 2 nC and Q2 = -4 nC, are separated by a distance of 10 cm in free
space. Find the force exerted on Q1 by Q2.
A16:
Using Coulomb's Law: F = (k * Q1 * Q2) / R²
k = 9 × 10⁹ N·m²/C², R = 0.1 m
F = (9 × 10⁹)(2 × 10⁻⁹)(-4 × 10⁻⁹) / (0.1)² = -72 × 10⁻⁹ / 0.01 = -7.2 × 10⁻⁶ N
(7th Edition) Comprehensive Q&A Study Guide
2026-2027
Chapter 1: Vector Algebra
Q1: Given three vectors A = 3ax + 2ay - az, B = ax - 4ay + 2az, and C = -2ax + ay + 3az, find A + B -
C.
A1:
Add the corresponding x, y, and z components:
x: 3 + 1 - (-2) = 6
y: 2 + (-4) - 1 = -3
z: -1 + 2 - 3 = -2
Result: 6ax - 3ay - 2az
Q2: Find the dot product of A = 2ax + ay and B = ax - 3ay.
A2:
A · B = (2)(1) + (1)(-3) = 2 - 3 = -1
Q3: Determine the angle between vectors A = ax + ay + az and B = -2ax - 2ay + 2az.
A3:
|A| = √(1² + 1² + 1²) = √3
|B| = √((-2)² + (-2)² + 2²) = √12 = 2√3
A · B = (1)(-2) + (1)(-2) + (1)(2) = -2
cos θ = (A · B) / (|A||B|) = -2 / (√3 * 2√3) = -2/6 = -1/3
θ = cos⁻¹(-1/3) ≈ 109.47°
,Q4: Find the cross product A × B if A = ax + 2ay and B = 2ax + ay.
A4:
A × B = ( (2)(1) - (0)(0) )ax - ( (1)(1) - (0)(2) )ay + ( (1)(0) - (2)(2) )az
Result: 2ax - ay - 4az
Q5: Convert the Cartesian point P(1, 2, 3) to cylindrical coordinates.
A5:
ρ = √(x² + y²) = √(1² + 2²) = √5
φ = tan⁻¹(y/x) = tan⁻¹(2/1) ≈ 63.43°
z=3
Result: (√5, 63.43°, 3)
Chapter 2: Coordinate Systems and Transformation
Q6: Convert the vector A = zax - xay + yaz into cylindrical coordinates.
A6:
In cylindrical coordinates, ax = cosφ aρ - sinφ aφ and ay = sinφ aρ + cosφ aφ.
Substitute x = ρcosφ, y = ρsinφ:
A = z(cosφ aρ - sinφ aφ) - ρcosφ(sinφ aρ + cosφ aφ) + ρsinφ az
Result: (zcosφ - ρsinφcosφ)aρ - (zsinφ + ρcos²φ)aφ + ρsinφ az
Q7: Convert the Cartesian point P(2, -2, 1) to spherical coordinates.
A7:
r = √(x² + y² + z²) = √(4 + 4 + 1) = 3
θ = cos⁻¹(z/r) = cos⁻¹(1/3) ≈ 70.53°
φ = tan⁻¹(y/x) = tan⁻¹(-2/2) = tan⁻¹(-1) = -45° (or 315°)
Result: (3, 70.53°, 315°)
,Q8: Given A = 4ar - sinθ aθ + aφ in spherical coordinates, find A · ar at point P(2, π/2, 0).
A8:
The dot product A · ar simply extracts the radial component of A.
Therefore, A · ar = 4 (independent of the point P).
Chapter 3: Vector Calculus
Q9: Find the gradient of the scalar field V = x²y + z at point (1, 2, 3).
A9:
∇V = (∂V/∂x)ax + (∂V/∂y)ay + (∂V/∂z)az
∂V/∂x = 2xy, ∂V/∂y = x², ∂V/∂z = 1
At (1, 2, 3): ∇V = (212)ax + (1²)ay + 1az = 4ax + ay + az
Q10: Calculate the divergence of the vector field D = x²z ax + y²x ay + z²y az.
A10:
∇ · D = ∂(x²z)/∂x + ∂(y²x)/∂y + ∂(z²y)/∂z
∇ · D = 2xz + 2yx + 2zy
Result: 2xz + 2xy + 2yz
Q11: Find the curl of the vector field A = yz ax + xz ay + xy az.
A11:
∇ × A = (∂(xy)/∂y - ∂(xz)/∂z)ax - (∂(xy)/∂x - ∂(yz)/∂z)ay + (∂(xz)/∂x - ∂(yz)/∂y)az
∇ × A = (x - x)ax - (y - y)ay + (z - z)az = 0
Q12: Given the vector field A = ρ aρ in cylindrical coordinates, find ∇ · A.
A12:
, In cylindrical coordinates: ∇ · A = (1/ρ) ∂(ρ Aρ)/∂ρ + (1/ρ) ∂Aφ/∂φ + ∂Az/∂z
∇ · A = (1/ρ) ∂(ρ * ρ)/∂ρ = (1/ρ) ∂(ρ²)/∂ρ = (1/ρ)(2ρ) = 2
Q13: State the divergence theorem.
A13:
The divergence theorem states that the volume integral of the divergence of a vector field A
over a volume v is equal to the closed surface integral of A over the boundary surface S of that
volume:
∫_v (∇ · A) dv = ∮_S A · dS
Q14: State Stokes's theorem.
A14:
Stokes's theorem states that the surface integral of the curl of a vector field A over an open
surface S is equal to the closed line integral of A around the boundary curve L of that surface:
∫_S (∇ × A) · dS = ∮_L A · dl
Q15: If ∇ · A = 0, what type of vector field is A?
A15:
A is a solenoidal vector field.
Chapter 4: Electrostatic Fields
Q16: Two point charges, Q1 = 2 nC and Q2 = -4 nC, are separated by a distance of 10 cm in free
space. Find the force exerted on Q1 by Q2.
A16:
Using Coulomb's Law: F = (k * Q1 * Q2) / R²
k = 9 × 10⁹ N·m²/C², R = 0.1 m
F = (9 × 10⁹)(2 × 10⁻⁹)(-4 × 10⁻⁹) / (0.1)² = -72 × 10⁻⁹ / 0.01 = -7.2 × 10⁻⁶ N
