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PHYS 122 MIDTERM EXAM I | QUESTIONS AND ANSWERS | 2026 UPDATE | 100% CORRECT - BINGHAMTON UNIVERSITY.

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PHYS 122 MIDTERM EXAM I | QUESTIONS AND ANSWERS | 2026 UPDATE | 100% CORRECT - BINGHAMTON UNIVERSITY.

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PHYS 122 MIDTERM EXAM I |
QUESTIONS AND ANSWERS | 2026
UPDATE | 100% CORRECT - BINGHAMTON
UNIVERSITY.



If you increase the voltage across a resistor, its resistance:




A) Increases






B) Decreases





C) Remains the same





D) Depends on the material
Answer: C) Remains the same
Rationale: Resistance is a property of the resistor itself determined by
material, length, and cross-sectional area and does not change when
voltage is applied. Ohm's law states V = IR, so increasing voltage
increases current, but resistance remains constant .



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, the net flux through a closed surface equals
Q_enc/ε₀. The flux through the flat base must equal the negative of the
flux through the rounded portion, adjusted for the enclosed charge .



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


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