PHYS 261 MIDTERM EXAM QUESTIONS AND
ANSWERS UPDATED 2026/2027 | 100% PASS
2026/2027 Examination
1|Page
,PHYS 261 MIDTERM EXAM QUESTIONS AND ANSWERS UPDATED 2026/2027 | 100% PASS
2026/2027 Examination
Total Questions: 200
Instructions:
• Answer all questions.
• Select the single best answer.
• Each question has four answer choices: A, B, C, and D.
• Select only ONE answer for each question.
SECTION 1: ELECTROSTATICS AND COULOMB'S LAW
Questions 1–25
Q1. Two identical point charges are separated by a distance r. If the distance between them is
tripled while both charges remain unchanged, by what factor does the electrostatic force between
them change?
A. It increases by a factor of 9
B. It decreases by a factor of 3
C. It decreases by a factor of 9
D. It remains unchanged
Correct Answer: C. It decreases by a factor of 9
Rationale: Coulomb's law states that the electrostatic force is inversely proportional to the
square of the separation distance (F ∝ 1/r²). Tripling the distance means the force becomes 1/(3)² =
1/9 of its original value, so it decreases by a factor of 9.
Q2. A charge of +4 μC is placed 0.2 m from a charge of −9 μC in vacuum. What is the magnitude
of the electrostatic force between them? (k = 9 × 10⁹ N·m²/C²)
A. 1.8 N
B. 8.1 N
C. 18 N
D. 0.81 N
Correct Answer: B. 8.1 N
Rationale: F = k|q₁q₂|/r² = (9 × 10⁹)(4 × 10⁻⁶)(9 × 10⁻⁶)/(0.2)² = (9 × 10⁹)(36 × 10⁻¹²)/0.04 = (324 ×
10⁻³)/0.04 = 8.1 N. The force is attractive because the charges have opposite signs.
Q3. Three point charges are arranged in a straight line: +q at x = 0, −2q at x = d, and +q at x = 2d.
What is the net force on the charge at x = d?
A. Zero
B. Directed toward the positive x-direction
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,C. Directed toward the negative x-direction
D. Directed perpendicular to the line
Correct Answer: A. Zero
Rationale: The charge at x = d experiences equal-magnitude forces from the two +q charges at x
= 0 and x = 2d. Both forces are attractive (toward the +q charges) and equal in magnitude because
the distances are equal and the charges are identical. They cancel exactly, giving a net force of zero.
Q4. The electric field at a point in space is defined as:
A. The force exerted on a test charge multiplied by the test charge
B. The force per unit positive test charge placed at that point
C. The total charge enclosed divided by the permittivity
D. The potential energy per unit charge at that point
Correct Answer: B. The force per unit positive test charge placed at that point
Rationale: The electric field E is defined as E = F/q₀, where F is the force on a small positive test
charge q₀. It represents the force per unit positive charge and is a vector quantity. Choice D
describes electric potential, not field.
Q5. A uniform electric field of 500 N/C points in the +x direction. What is the electric force on a
−3 μC charge placed in this field?
A. 1.5 × 10⁻³ N in the +x direction
B. 1.5 × 10⁻³ N in the −x direction
C. 6.0 × 10⁻³ N in the +x direction
D. 6.0 × 10⁻³ N in the −x direction
Correct Answer: B. 1.5 × 10⁻³ N in the −x direction
Rationale: F = qE = (−3 × 10⁻⁶ C)(500 N/C) = −1.5 × 10⁻³ N. The negative sign indicates the force is
opposite to the field direction, i.e., in the −x direction. A negative charge experiences force opposite
to the electric field.
Q6. Two charges, +Q and −Q, are separated by a distance 2a. At the midpoint between them, the
electric field is:
A. Zero
B. Directed from +Q to −Q
C. Directed from −Q to +Q
D. Perpendicular to the line joining the charges
Correct Answer: B. Directed from +Q to −Q
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, Rationale: At the midpoint, the field due to +Q points away from +Q (toward −Q), and the field
due to −Q points toward −Q (also toward −Q). Both contributions point in the same direction—from
+Q toward −Q—so they add constructively.
Q7. A point charge of +6 nC is at the origin. What is the electric field magnitude at a point 0.3 m
away on the x-axis? (k = 9 × 10⁹ N·m²/C²)
A. 200 N/C
B. 600 N/C
C. 1800 N/C
D. 5400 N/C
Correct Answer: B. 600 N/C
Rationale: E = k|q|/r² = (9 × 10⁹)(6 × 10⁻⁹)/(0.3)² = 54/0.09 = 600 N/C. The field points away from
the positive charge.
Q8. An electric dipole consists of charges +q and −q separated by a small distance d. If the dipole
is placed in a uniform electric field, the net force on the dipole is:
A. qE
B. 2qE
C. Zero
D. qEd
Correct Answer: C. Zero
Rationale: In a uniform field, the forces on the +q and −q charges are equal in magnitude (qE)
but opposite in direction. They cancel, giving zero net force. However, there is a net torque that
tends to align the dipole with the field.
Q9. The electric field inside a conductor in electrostatic equilibrium is:
A. Equal to the applied external field
B. Zero
C. Constant but nonzero
D. Proportional to the surface charge density
Correct Answer: B. Zero
Rationale: In electrostatic equilibrium, charges in a conductor redistribute until the internal field
is zero. Any nonzero internal field would cause charge motion, violating the equilibrium condition.
The excess charge resides entirely on the surface.
Q10. A Gaussian surface encloses two charges: +5 μC and −3 μC. What is the net electric flux
through the surface? (ε₀ = 8.85 × 10⁻¹² C²/N·m²)
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ANSWERS UPDATED 2026/2027 | 100% PASS
2026/2027 Examination
1|Page
,PHYS 261 MIDTERM EXAM QUESTIONS AND ANSWERS UPDATED 2026/2027 | 100% PASS
2026/2027 Examination
Total Questions: 200
Instructions:
• Answer all questions.
• Select the single best answer.
• Each question has four answer choices: A, B, C, and D.
• Select only ONE answer for each question.
SECTION 1: ELECTROSTATICS AND COULOMB'S LAW
Questions 1–25
Q1. Two identical point charges are separated by a distance r. If the distance between them is
tripled while both charges remain unchanged, by what factor does the electrostatic force between
them change?
A. It increases by a factor of 9
B. It decreases by a factor of 3
C. It decreases by a factor of 9
D. It remains unchanged
Correct Answer: C. It decreases by a factor of 9
Rationale: Coulomb's law states that the electrostatic force is inversely proportional to the
square of the separation distance (F ∝ 1/r²). Tripling the distance means the force becomes 1/(3)² =
1/9 of its original value, so it decreases by a factor of 9.
Q2. A charge of +4 μC is placed 0.2 m from a charge of −9 μC in vacuum. What is the magnitude
of the electrostatic force between them? (k = 9 × 10⁹ N·m²/C²)
A. 1.8 N
B. 8.1 N
C. 18 N
D. 0.81 N
Correct Answer: B. 8.1 N
Rationale: F = k|q₁q₂|/r² = (9 × 10⁹)(4 × 10⁻⁶)(9 × 10⁻⁶)/(0.2)² = (9 × 10⁹)(36 × 10⁻¹²)/0.04 = (324 ×
10⁻³)/0.04 = 8.1 N. The force is attractive because the charges have opposite signs.
Q3. Three point charges are arranged in a straight line: +q at x = 0, −2q at x = d, and +q at x = 2d.
What is the net force on the charge at x = d?
A. Zero
B. Directed toward the positive x-direction
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,C. Directed toward the negative x-direction
D. Directed perpendicular to the line
Correct Answer: A. Zero
Rationale: The charge at x = d experiences equal-magnitude forces from the two +q charges at x
= 0 and x = 2d. Both forces are attractive (toward the +q charges) and equal in magnitude because
the distances are equal and the charges are identical. They cancel exactly, giving a net force of zero.
Q4. The electric field at a point in space is defined as:
A. The force exerted on a test charge multiplied by the test charge
B. The force per unit positive test charge placed at that point
C. The total charge enclosed divided by the permittivity
D. The potential energy per unit charge at that point
Correct Answer: B. The force per unit positive test charge placed at that point
Rationale: The electric field E is defined as E = F/q₀, where F is the force on a small positive test
charge q₀. It represents the force per unit positive charge and is a vector quantity. Choice D
describes electric potential, not field.
Q5. A uniform electric field of 500 N/C points in the +x direction. What is the electric force on a
−3 μC charge placed in this field?
A. 1.5 × 10⁻³ N in the +x direction
B. 1.5 × 10⁻³ N in the −x direction
C. 6.0 × 10⁻³ N in the +x direction
D. 6.0 × 10⁻³ N in the −x direction
Correct Answer: B. 1.5 × 10⁻³ N in the −x direction
Rationale: F = qE = (−3 × 10⁻⁶ C)(500 N/C) = −1.5 × 10⁻³ N. The negative sign indicates the force is
opposite to the field direction, i.e., in the −x direction. A negative charge experiences force opposite
to the electric field.
Q6. Two charges, +Q and −Q, are separated by a distance 2a. At the midpoint between them, the
electric field is:
A. Zero
B. Directed from +Q to −Q
C. Directed from −Q to +Q
D. Perpendicular to the line joining the charges
Correct Answer: B. Directed from +Q to −Q
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, Rationale: At the midpoint, the field due to +Q points away from +Q (toward −Q), and the field
due to −Q points toward −Q (also toward −Q). Both contributions point in the same direction—from
+Q toward −Q—so they add constructively.
Q7. A point charge of +6 nC is at the origin. What is the electric field magnitude at a point 0.3 m
away on the x-axis? (k = 9 × 10⁹ N·m²/C²)
A. 200 N/C
B. 600 N/C
C. 1800 N/C
D. 5400 N/C
Correct Answer: B. 600 N/C
Rationale: E = k|q|/r² = (9 × 10⁹)(6 × 10⁻⁹)/(0.3)² = 54/0.09 = 600 N/C. The field points away from
the positive charge.
Q8. An electric dipole consists of charges +q and −q separated by a small distance d. If the dipole
is placed in a uniform electric field, the net force on the dipole is:
A. qE
B. 2qE
C. Zero
D. qEd
Correct Answer: C. Zero
Rationale: In a uniform field, the forces on the +q and −q charges are equal in magnitude (qE)
but opposite in direction. They cancel, giving zero net force. However, there is a net torque that
tends to align the dipole with the field.
Q9. The electric field inside a conductor in electrostatic equilibrium is:
A. Equal to the applied external field
B. Zero
C. Constant but nonzero
D. Proportional to the surface charge density
Correct Answer: B. Zero
Rationale: In electrostatic equilibrium, charges in a conductor redistribute until the internal field
is zero. Any nonzero internal field would cause charge motion, violating the equilibrium condition.
The excess charge resides entirely on the surface.
Q10. A Gaussian surface encloses two charges: +5 μC and −3 μC. What is the net electric flux
through the surface? (ε₀ = 8.85 × 10⁻¹² C²/N·m²)
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