PHY 102 WEEK 4 EXAM SOLUTION | 80%
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Section 1: Electrostatics & Electric Forces (Q1-12)
Q1. Two point charges, q₁ = +3.0 μC and q₂ = -5.0 μC, are separated by a distance of
0.40 m. What is the magnitude of the electrostatic force between them?
A. 0.84 N
B. 0.42 N
C. 0.21 N
D. 1.68 N
Correct Answer: A
Rationale: Using Coulomb's Law: F = k|q₁q₂|/r² = (8.99 × 10⁹ N·m²/C²)(3.0 × 10⁻⁶
C)(5.0 × 10⁻⁶ C)/(0.40 m)² = (8.99 × 10⁹)(15 × 10⁻¹²)/0.16 = 0.84 N. Option B uses r =
0.40 m instead of r². Option C uses r = 0.80 m. Option D forgets to square the
distance. Physics Principle: Coulomb's Law F = k|q₁q₂|/r²; like charges repel, unlike
charges attract. Unit Handling: Convert μC to C (10⁻⁶); distance in meters.
Q2. An electron is placed in a uniform electric field of magnitude 500 N/C directed to
the right. What is the magnitude and direction of the force on the electron?
A. 8.0 × 10⁻¹⁷ N to the right
B. 8.0 × 10⁻¹⁷ N to the left
,C. 1.6 × 10⁻¹⁹ N to the right
D. 1.6 × 10⁻¹⁹ N to the left
Correct Answer: B
Rationale: F = qE = (1.6 × 10⁻¹⁹ C)(500 N/C) = 8.0 × 10⁻¹⁷ N. Since the electron is
negative, the force is opposite to the field direction (to the left). Option A has correct
magnitude but wrong direction. Options C and D use the elementary charge
magnitude incorrectly. Physics Principle: F = qE; negative charges experience force
opposite to E-field direction. Unit Handling: Electron charge = -1.6 × 10⁻¹⁹ C;
direction matters for vector quantities.
Q3. Three point charges are placed on the x-axis: q₁ = +2.0 μC at x = 0, q₂ = -4.0 μC
at x = 0.30 m, and q₃ = +3.0 μC at x = 0.60 m. What is the net electrostatic force on
q₂?
A. 0.80 N to the left
B. 1.20 N to the right
C. 0.40 N to the right
D. 0.40 N to the left
Correct Answer: C
Rationale: Force on q₂ from q₁: F₁₂ = k|q₁q₂|/r² = (8.99 × 10⁹)(2.0 × 10⁻⁶)(4.0 ×
10⁻⁶)/(0.30)² = 0.80 N (attractive, so q₂ pulled left). Force on q₂ from q₃: F₃₂ =
k|q₃q₂|/r² = (8.99 × 10⁹)(3.0 × 10⁻⁶)(4.0 × 10⁻⁶)/(0.30)² = 1.20 N (attractive, so q₂
pulled right). Net force = 1.20 N right - 0.80 N left = 0.40 N to the right. Option A is
just F₁₂. Option B is just F₃₂. Option D has wrong direction. Physics Principle:
Superposition principle—net force is vector sum of individual forces. Unit Handling:
Careful with direction signs; attractive forces point toward the other charge.
Q4. What is the electric field at a point 0.50 m from a point charge of +6.0 μC?
, A. 2.16 × 10⁵ N/C directed away from the charge
B. 1.08 × 10⁵ N/C directed toward the charge
C. 2.16 × 10⁵ N/C directed toward the charge
D. 1.08 × 10⁵ N/C directed away from the charge
Correct Answer: A
Rationale: E = k|q|/r² = (8.99 × 10⁹)(6.0 × 10⁻⁶)/(0.50)² = (8.99 × 10⁹)(6.0 × 10⁻⁶)/0.25
= 2.16 × 10⁵ N/C. Positive charge: E-field points away. Option B has wrong
magnitude and direction. Option C has wrong direction. Option D has wrong
magnitude. Physics Principle: E = kq/r² for point charge; direction away from
positive, toward negative. Unit Handling: r must be in meters; field direction
depends on source charge sign.
Q5. Two parallel plates are separated by 2.0 mm and have a potential difference of
500 V. What is the magnitude of the electric field between the plates?
A. 250 N/C
B. 2.5 × 10⁵ N/C
C. 1000 N/C
D. 1.0 × 10⁶ N/C
Correct Answer: B
Rationale: For uniform field between parallel plates: E = ΔV/d = 500 V / (2.0 × 10⁻³
m) = 2.5 × 10⁵ N/C. Option A uses d in mm without conversion. Option C uses d =
0.5 m. Option D uses d = 0.5 mm. Physics Principle: E = ΔV/d for uniform field; valid
for parallel plate capacitor. Unit Handling: Convert mm to m (2.0 mm = 2.0 × 10⁻³
m).
Q6. A proton is moved from a point where the electric potential is +100 V to a point
where the potential is +50 V. What is the change in electric potential energy of the
proton?
Guarantee Complete Guide | Grand Canyon
University | Pass Guaranteed - A+ Graded
Section 1: Electrostatics & Electric Forces (Q1-12)
Q1. Two point charges, q₁ = +3.0 μC and q₂ = -5.0 μC, are separated by a distance of
0.40 m. What is the magnitude of the electrostatic force between them?
A. 0.84 N
B. 0.42 N
C. 0.21 N
D. 1.68 N
Correct Answer: A
Rationale: Using Coulomb's Law: F = k|q₁q₂|/r² = (8.99 × 10⁹ N·m²/C²)(3.0 × 10⁻⁶
C)(5.0 × 10⁻⁶ C)/(0.40 m)² = (8.99 × 10⁹)(15 × 10⁻¹²)/0.16 = 0.84 N. Option B uses r =
0.40 m instead of r². Option C uses r = 0.80 m. Option D forgets to square the
distance. Physics Principle: Coulomb's Law F = k|q₁q₂|/r²; like charges repel, unlike
charges attract. Unit Handling: Convert μC to C (10⁻⁶); distance in meters.
Q2. An electron is placed in a uniform electric field of magnitude 500 N/C directed to
the right. What is the magnitude and direction of the force on the electron?
A. 8.0 × 10⁻¹⁷ N to the right
B. 8.0 × 10⁻¹⁷ N to the left
,C. 1.6 × 10⁻¹⁹ N to the right
D. 1.6 × 10⁻¹⁹ N to the left
Correct Answer: B
Rationale: F = qE = (1.6 × 10⁻¹⁹ C)(500 N/C) = 8.0 × 10⁻¹⁷ N. Since the electron is
negative, the force is opposite to the field direction (to the left). Option A has correct
magnitude but wrong direction. Options C and D use the elementary charge
magnitude incorrectly. Physics Principle: F = qE; negative charges experience force
opposite to E-field direction. Unit Handling: Electron charge = -1.6 × 10⁻¹⁹ C;
direction matters for vector quantities.
Q3. Three point charges are placed on the x-axis: q₁ = +2.0 μC at x = 0, q₂ = -4.0 μC
at x = 0.30 m, and q₃ = +3.0 μC at x = 0.60 m. What is the net electrostatic force on
q₂?
A. 0.80 N to the left
B. 1.20 N to the right
C. 0.40 N to the right
D. 0.40 N to the left
Correct Answer: C
Rationale: Force on q₂ from q₁: F₁₂ = k|q₁q₂|/r² = (8.99 × 10⁹)(2.0 × 10⁻⁶)(4.0 ×
10⁻⁶)/(0.30)² = 0.80 N (attractive, so q₂ pulled left). Force on q₂ from q₃: F₃₂ =
k|q₃q₂|/r² = (8.99 × 10⁹)(3.0 × 10⁻⁶)(4.0 × 10⁻⁶)/(0.30)² = 1.20 N (attractive, so q₂
pulled right). Net force = 1.20 N right - 0.80 N left = 0.40 N to the right. Option A is
just F₁₂. Option B is just F₃₂. Option D has wrong direction. Physics Principle:
Superposition principle—net force is vector sum of individual forces. Unit Handling:
Careful with direction signs; attractive forces point toward the other charge.
Q4. What is the electric field at a point 0.50 m from a point charge of +6.0 μC?
, A. 2.16 × 10⁵ N/C directed away from the charge
B. 1.08 × 10⁵ N/C directed toward the charge
C. 2.16 × 10⁵ N/C directed toward the charge
D. 1.08 × 10⁵ N/C directed away from the charge
Correct Answer: A
Rationale: E = k|q|/r² = (8.99 × 10⁹)(6.0 × 10⁻⁶)/(0.50)² = (8.99 × 10⁹)(6.0 × 10⁻⁶)/0.25
= 2.16 × 10⁵ N/C. Positive charge: E-field points away. Option B has wrong
magnitude and direction. Option C has wrong direction. Option D has wrong
magnitude. Physics Principle: E = kq/r² for point charge; direction away from
positive, toward negative. Unit Handling: r must be in meters; field direction
depends on source charge sign.
Q5. Two parallel plates are separated by 2.0 mm and have a potential difference of
500 V. What is the magnitude of the electric field between the plates?
A. 250 N/C
B. 2.5 × 10⁵ N/C
C. 1000 N/C
D. 1.0 × 10⁶ N/C
Correct Answer: B
Rationale: For uniform field between parallel plates: E = ΔV/d = 500 V / (2.0 × 10⁻³
m) = 2.5 × 10⁵ N/C. Option A uses d in mm without conversion. Option C uses d =
0.5 m. Option D uses d = 0.5 mm. Physics Principle: E = ΔV/d for uniform field; valid
for parallel plate capacitor. Unit Handling: Convert mm to m (2.0 mm = 2.0 × 10⁻³
m).
Q6. A proton is moved from a point where the electric potential is +100 V to a point
where the potential is +50 V. What is the change in electric potential energy of the
proton?
