EEE 241 — FUNDAMENTALS OF ELECTROMAGNETICS QUIZ 3
PRACTICE EXAM — 200 QUESTIONS WITH ANSWERS & RATIONALES
|ARIZONA STATE UNIVERSITY | 2026 UPDATE |
100% CORRECT
SECTION 1 — POSTULATES OF ELECTROSTATICS (Questions 1–40)
1. Coulomb's law states that the force between two point charges is:
A) Directly proportional to the product of the charges and inversely
proportional to the square of the distance between them
B) Directly proportional to the distance and inversely proportional to the
product of the charges
C) Independent of the medium between the charges
D) Always attractive
Answer: A
Rationale: Coulomb's law: F = (1/4πε₀) × (q₁q₂/r²). The force is directly
proportional to the product of the charges and inversely proportional to the
square of the distance between them.
2. The electric field intensity E at a point is defined as:
A) The force per unit charge exerted on a positive test charge at that point
B) The force per unit mass on a test charge
C) The potential difference per unit length
D) The charge density per unit volume
Answer: A
Rationale: E = F/q, where F is the force on a positive test charge q placed at
the point. The direction of E is the direction of the force on a positive test
charge.
3. Gauss's law states that the total electric flux through a closed surface is
equal to:
A) The total charge enclosed divided by ε₀
B) The total charge enclosed multiplied by ε₀
C) The product of E and the area
D) Zero always
Answer: A
1
,Rationale: ∮ D · dS = Q_enclosed. In free space, ∮ E · dS = Q_enclosed/ε₀.
The total flux depends only on the enclosed charge.
4. What is the electric potential V at a distance r from a point charge q?
A) V = q/(4πε₀r)
B) V = q/(4πε₀r²)
C) V = qr/(4πε₀)
D) V = q/(2πε₀r)
Answer: A
Rationale: The electric potential due to a point charge is V = q/(4πε₀r),
which is the work per unit charge to bring a positive test charge from
infinity to distance r.
5. The relationship between electric field intensity E and electric potential
V is:
A) E = -∇V
B) E = ∇V
C) E = V/r
D) E = -∇ × V
Answer: A
Rationale: E = -∇V. The electric field points in the direction of decreasing
potential, and the negative gradient accounts for this direction.
6. If the electric field is zero in some region of space, the electric
potential in that region must be:
A) Constant
B) Zero
C) Increasing linearly
D) Undefined
Answer: A
Rationale: If E = -∇V = 0, then V is constant in that region. The potential
need not be zero, but it must be uniform.
7. The electric flux density D is related to the electric field intensity E by:
2
,A) D = ε₀E
B) D = εE in a dielectric medium, where ε = ε₀ε_r
C) D = E/ε
D) Both A and B
Answer: D
Rationale: In free space, D = ε₀E. In a dielectric medium, D = εE where
ε = ε₀ε_r. Both statements are correct.
8. A point charge q = 2 μC is located at the origin. What is the electric
field magnitude at r = 3 m in free space?
A) 2 × 10³ N/C
B) 2 × 10⁴ N/C
C) 2 × 10² N/C
D) 2 × 10⁵ N/C
Answer: A
Rationale: E = q/(4πε₀r²) = (2 × 10⁻⁶) / (4π × 8.854 × 10⁻¹² × 9)
= (2 × 10⁻⁶) / (1.0 × 10⁻⁹) ≈ 2 × 10³ N/C.
9. The work done in moving a charge q from point A to point B in an electric
field depends on:
A) The path taken
B) Only the initial and final positions
C) The velocity of the charge
D) The mass of the charge
Answer: B
Rationale: The electrostatic field is conservative, so the work done depends
only on the potential difference between the endpoints, not the path.
10. The potential difference between two points A and B is defined as:
A) The work done per unit charge to move a positive charge from A to B
B) The work done to move any charge from A to B
C) The electric field multiplied by the distance
D) The charge divided by the distance
Answer: A
3
, Rationale: V_AB = V_A - V_B = W/q, where W is the work done in moving a
positive test charge from A to B.
11. What is the electric field inside a uniformly charged conducting sphere?
A) Zero
B) q/(4πε₀r²)
C) Constant
D) Infinite
Answer: A
Rationale: Inside a conductor in electrostatic equilibrium, the electric field
is zero. All charge resides on the surface.
12. The potential inside a uniformly charged conducting sphere is:
A) Zero
B) Constant and equal to the surface potential
C) q/(4πε₀r)
D) Linearly decreasing with r
Answer: B
Rationale: Since E = 0 inside the conductor, the potential is constant and
equals the potential at the surface: V = q/(4πε₀R).
13. A parallel-plate capacitor has plates of area A separated by distance d.
What is the capacitance?
A) C = ε₀A/d
B) C = ε₀d/A
C) C = ε₀Ad
D) C = A/(ε₀d)
Answer: A
Rationale: For a parallel-plate capacitor, C = εA/d. In free space,
C = ε₀A/d.
14. What is the electric field between the plates of a parallel-plate
capacitor with surface charge density σ?
A) E = σ/ε₀
B) E = σε₀
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PRACTICE EXAM — 200 QUESTIONS WITH ANSWERS & RATIONALES
|ARIZONA STATE UNIVERSITY | 2026 UPDATE |
100% CORRECT
SECTION 1 — POSTULATES OF ELECTROSTATICS (Questions 1–40)
1. Coulomb's law states that the force between two point charges is:
A) Directly proportional to the product of the charges and inversely
proportional to the square of the distance between them
B) Directly proportional to the distance and inversely proportional to the
product of the charges
C) Independent of the medium between the charges
D) Always attractive
Answer: A
Rationale: Coulomb's law: F = (1/4πε₀) × (q₁q₂/r²). The force is directly
proportional to the product of the charges and inversely proportional to the
square of the distance between them.
2. The electric field intensity E at a point is defined as:
A) The force per unit charge exerted on a positive test charge at that point
B) The force per unit mass on a test charge
C) The potential difference per unit length
D) The charge density per unit volume
Answer: A
Rationale: E = F/q, where F is the force on a positive test charge q placed at
the point. The direction of E is the direction of the force on a positive test
charge.
3. Gauss's law states that the total electric flux through a closed surface is
equal to:
A) The total charge enclosed divided by ε₀
B) The total charge enclosed multiplied by ε₀
C) The product of E and the area
D) Zero always
Answer: A
1
,Rationale: ∮ D · dS = Q_enclosed. In free space, ∮ E · dS = Q_enclosed/ε₀.
The total flux depends only on the enclosed charge.
4. What is the electric potential V at a distance r from a point charge q?
A) V = q/(4πε₀r)
B) V = q/(4πε₀r²)
C) V = qr/(4πε₀)
D) V = q/(2πε₀r)
Answer: A
Rationale: The electric potential due to a point charge is V = q/(4πε₀r),
which is the work per unit charge to bring a positive test charge from
infinity to distance r.
5. The relationship between electric field intensity E and electric potential
V is:
A) E = -∇V
B) E = ∇V
C) E = V/r
D) E = -∇ × V
Answer: A
Rationale: E = -∇V. The electric field points in the direction of decreasing
potential, and the negative gradient accounts for this direction.
6. If the electric field is zero in some region of space, the electric
potential in that region must be:
A) Constant
B) Zero
C) Increasing linearly
D) Undefined
Answer: A
Rationale: If E = -∇V = 0, then V is constant in that region. The potential
need not be zero, but it must be uniform.
7. The electric flux density D is related to the electric field intensity E by:
2
,A) D = ε₀E
B) D = εE in a dielectric medium, where ε = ε₀ε_r
C) D = E/ε
D) Both A and B
Answer: D
Rationale: In free space, D = ε₀E. In a dielectric medium, D = εE where
ε = ε₀ε_r. Both statements are correct.
8. A point charge q = 2 μC is located at the origin. What is the electric
field magnitude at r = 3 m in free space?
A) 2 × 10³ N/C
B) 2 × 10⁴ N/C
C) 2 × 10² N/C
D) 2 × 10⁵ N/C
Answer: A
Rationale: E = q/(4πε₀r²) = (2 × 10⁻⁶) / (4π × 8.854 × 10⁻¹² × 9)
= (2 × 10⁻⁶) / (1.0 × 10⁻⁹) ≈ 2 × 10³ N/C.
9. The work done in moving a charge q from point A to point B in an electric
field depends on:
A) The path taken
B) Only the initial and final positions
C) The velocity of the charge
D) The mass of the charge
Answer: B
Rationale: The electrostatic field is conservative, so the work done depends
only on the potential difference between the endpoints, not the path.
10. The potential difference between two points A and B is defined as:
A) The work done per unit charge to move a positive charge from A to B
B) The work done to move any charge from A to B
C) The electric field multiplied by the distance
D) The charge divided by the distance
Answer: A
3
, Rationale: V_AB = V_A - V_B = W/q, where W is the work done in moving a
positive test charge from A to B.
11. What is the electric field inside a uniformly charged conducting sphere?
A) Zero
B) q/(4πε₀r²)
C) Constant
D) Infinite
Answer: A
Rationale: Inside a conductor in electrostatic equilibrium, the electric field
is zero. All charge resides on the surface.
12. The potential inside a uniformly charged conducting sphere is:
A) Zero
B) Constant and equal to the surface potential
C) q/(4πε₀r)
D) Linearly decreasing with r
Answer: B
Rationale: Since E = 0 inside the conductor, the potential is constant and
equals the potential at the surface: V = q/(4πε₀R).
13. A parallel-plate capacitor has plates of area A separated by distance d.
What is the capacitance?
A) C = ε₀A/d
B) C = ε₀d/A
C) C = ε₀Ad
D) C = A/(ε₀d)
Answer: A
Rationale: For a parallel-plate capacitor, C = εA/d. In free space,
C = ε₀A/d.
14. What is the electric field between the plates of a parallel-plate
capacitor with surface charge density σ?
A) E = σ/ε₀
B) E = σε₀
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