PHYS 122 MIDTERM EXAM II | QUESTIONS AND ANSWERS |
2026 UPDATE | 100% CORRECT - BINGHAMTON UNIVERSITY.
EXAM INSTRUCTIONS
Total Questions: 50 multiple-choice questions
Time Allotted: 80 minutes
Materials Allowed: Non-programmable scientific calculator,
one 3×5 index card with formulas
Scoring: Each question is worth 2 points (100 points total)
Instructions: Select the single best answer for each question.
Show all work on the exam booklet for partial credit
consideration. Answers marked with multiple selections will
be marked incorrect.
,A 2-μF and a 1-μF capacitor are connected in parallel. When a
potential difference is applied, the 2-μF capacitor has:
A) Twice the charge of the 1-μF capacitor
B) Half the charge of the 1-μF capacitor
C) The same charge as the 1-μF capacitor
D) Four times the charge of the 1-μF capacitor
Answer: A) Twice the charge of the 1-μF capacitor
Rationale: In parallel, capacitors have the same voltage. Since
Q = CV, charge is proportional to capacitance. The 2-μF
capacitor has twice the capacitance, so it stores twice the
charge.
A 3.5-cm radius hemisphere contains a total charge of 6.6 ×
10⁻⁷ C. The flux through the rounded portion is 9.8 × 10⁴
N·m²/C. The flux through the flat base is:
A) -2.3 × 10⁴ N·m²/C
B) +2.3 × 10⁴ N·m²/C
,C) -9.8 × 10⁴ N·m²/C
D) +9.8 × 10⁴ N·m²/C
Answer: A) -2.3 × 10⁴ N·m²/C
Rationale: By Gauss's law, net flux = Q_enc/ε₀. Total flux =
(6.6×10⁻⁷)/(8.85×10⁻¹²) = 7.46×10⁴ N·m²/C. Flux through flat
base = 7.46×10⁴ - 9.8×10⁴ = -2.34×10⁴ N·m²/C.
A conducting sphere of radius 0.01 m has a charge of 1.0 ×
10⁻⁹ C. The electric field just outside the surface is:
A) 9,000 N/C
B) 90,000 N/C
C) 900,000 N/C
D) 9 × 10⁶ N/C
Answer: B) 90,000 N/C
Rationale: E = kQ/R² = (8.99×10⁹)(1.0×10⁻⁹)/(0.01)² = 89,900
N/C ≈ 90,000 N/C.
A voltmeter has an internal resistance of 10,000 Ω and a
range of 0 to 100 V. To extend the range to 0 to 1000 V, one
should connect:
,
A) 90,000 Ω in series
B) 10,000 Ω in series
C) 100,000 Ω in parallel
D) 90,000 Ω in parallel
Answer: A) 90,000 Ω in series
Rationale: R_series = (V_new/V_old - 1) × R_internal =
(1000/100 - 1) × 10,000 = 90,000 Ω in series.
A 5-cm radius conducting sphere has a surface charge density
of 2 × 10⁻⁶ C/m². Its electric potential, relative to potential far
away, is:
A) 1.1 × 10⁴ V
B) 2.2 × 10⁴ V
C) 5.5 × 10³ V
2026 UPDATE | 100% CORRECT - BINGHAMTON UNIVERSITY.
EXAM INSTRUCTIONS
Total Questions: 50 multiple-choice questions
Time Allotted: 80 minutes
Materials Allowed: Non-programmable scientific calculator,
one 3×5 index card with formulas
Scoring: Each question is worth 2 points (100 points total)
Instructions: Select the single best answer for each question.
Show all work on the exam booklet for partial credit
consideration. Answers marked with multiple selections will
be marked incorrect.
,A 2-μF and a 1-μF capacitor are connected in parallel. When a
potential difference is applied, the 2-μF capacitor has:
A) Twice the charge of the 1-μF capacitor
B) Half the charge of the 1-μF capacitor
C) The same charge as the 1-μF capacitor
D) Four times the charge of the 1-μF capacitor
Answer: A) Twice the charge of the 1-μF capacitor
Rationale: In parallel, capacitors have the same voltage. Since
Q = CV, charge is proportional to capacitance. The 2-μF
capacitor has twice the capacitance, so it stores twice the
charge.
A 3.5-cm radius hemisphere contains a total charge of 6.6 ×
10⁻⁷ C. The flux through the rounded portion is 9.8 × 10⁴
N·m²/C. The flux through the flat base is:
A) -2.3 × 10⁴ N·m²/C
B) +2.3 × 10⁴ N·m²/C
,C) -9.8 × 10⁴ N·m²/C
D) +9.8 × 10⁴ N·m²/C
Answer: A) -2.3 × 10⁴ N·m²/C
Rationale: By Gauss's law, net flux = Q_enc/ε₀. Total flux =
(6.6×10⁻⁷)/(8.85×10⁻¹²) = 7.46×10⁴ N·m²/C. Flux through flat
base = 7.46×10⁴ - 9.8×10⁴ = -2.34×10⁴ N·m²/C.
A conducting sphere of radius 0.01 m has a charge of 1.0 ×
10⁻⁹ C. The electric field just outside the surface is:
A) 9,000 N/C
B) 90,000 N/C
C) 900,000 N/C
D) 9 × 10⁶ N/C
Answer: B) 90,000 N/C
Rationale: E = kQ/R² = (8.99×10⁹)(1.0×10⁻⁹)/(0.01)² = 89,900
N/C ≈ 90,000 N/C.
A voltmeter has an internal resistance of 10,000 Ω and a
range of 0 to 100 V. To extend the range to 0 to 1000 V, one
should connect:
,
A) 90,000 Ω in series
B) 10,000 Ω in series
C) 100,000 Ω in parallel
D) 90,000 Ω in parallel
Answer: A) 90,000 Ω in series
Rationale: R_series = (V_new/V_old - 1) × R_internal =
(1000/100 - 1) × 10,000 = 90,000 Ω in series.
A 5-cm radius conducting sphere has a surface charge density
of 2 × 10⁻⁶ C/m². Its electric potential, relative to potential far
away, is:
A) 1.1 × 10⁴ V
B) 2.2 × 10⁴ V
C) 5.5 × 10³ V
