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PHYS 112 – General Physics II Exam 2 | 200-Question Practice Exam with Answers, Calculations & Detailed Explanations

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Strengthen your preparation for PHYS 112 – General Physics II Exam 2 with this 200-question practice exam featuring both conceptual and calculation-based problems. Topics include electric fields, electric potential, capacitance, current, resistance, circuits, Kirchhoff’s rules, and related problem-solving skills. Every question includes the correct answer, detailed explanation, and worked calculations where needed.

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PASSPOINTPRO


EXAM 2 PRACTICE




PHYS 112
General Physics II
200-Question Practice Exam
Answers, worked calculations & detailed explanations




200 8 A-D
QUESTIONS CORE AREAS FORMAT




TYPICAL EXAM 2 COVERAGE

Electric potential & energy | Capacitance & dielectrics

Current, resistance & power | Series/parallel circuits

Kirchhoff rules & RC circuits | Magnetic forces & fields

Electromagnetic induction | AC & electromagnetic waves




Designed for focused practice, calculation fluency and concept review.



PASSPOINTPRO

,PHYS 112 - GENERAL PHYSICS II | EXAM 2 PRACTICE PassPointPro




Exam 2 Practice Overview
This 200-question PHYS 112 practice exam focuses on common second-unit General Physics II material. Questions combine
conceptual reasoning with multi-step calculations. Each item includes four answer choices, the correct answer, and a detailed
explanation immediately below the question.

Coverage Questions

Electric Potential & Electric Potential Energy 25

Capacitance & Dielectrics 25

Current, Resistance & Electrical Power 25

Series & Parallel DC Circuits 25

Kirchhoff Rules & RC Circuits 25

Magnetic Fields & Magnetic Forces 25

Electromagnetic Induction & Inductance 25

AC, Electromagnetic Waves & Mixed Applications 25


Suggested use: Work each problem without looking at the explanation, then use the rationale to identify the equation, sign
convention, or reasoning step that controls the answer.


Quick Formula Reference
Electric potential: V = kq/r | Potential energy: U = qV


Capacitance: C = Q/V | Parallel plate: C = eps0 A/d | Stored energy: U = (1/2)CV^2


Ohm's law: V = IR | Power: P = IV = I^2R = V^2/R


RC time constant: tau = RC | Charging: Vc = V(1 - e^(-t/tau))


Magnetic force: F = qvB sin(theta) | Wire force: F = ILB sin(theta)


Long straight wire: B = mu0 I/(2 pi r) | Solenoid: B = mu0 nI


Faraday: |emf| = N|Delta Phi|/Delta t | Flux: Phi = BA cos(theta)


AC: Vpeak = sqrt(2) Vrms | XL = 2 pi fL | XC = 1/(2 pi fC) | c = f lambda




PHYS 112 | 200-Question Exam 2 Practice Exam Page 2

,PHYS 112 - GENERAL PHYSICS II | EXAM 2 PRACTICE PassPointPro



Electric Potential & Electric Potential Energy


1. A point charge of +4 uC is isolated in space. What electric potential does it create at a
point 0.25 m away?

A 1.44e+04 V
B 7.19e+04 V
C 1.44e+05 V
D 2.88e+05 V


Answer: C - 1.44e+05 V

Explanation: For a point charge, V = kq/r. Substituting k = 8.99 x 10^9 N m^2/C^2, q = 4.00e-06 C, and r =
0.25 m gives V = 1.438e+05 V. Electric potential is a scalar, so only the sign of q determines the sign here.


2. A +3 uC charge is placed at a location where the electric potential is 240 V. What is its
electric potential energy?

A 7.20e-04 J
B 7.20e-05 J
C 1.44e-03 J
D 7.20e-03 J


Answer: A - 7.20e-04 J

Explanation: Potential energy is U = qV. Using q = 3.00e-06 C and V = 240 V gives U = 7.200e-04 J. Because
both q and V are positive, U is positive.


3. A +6 uC point charge is fixed. What is the change in electric potential, V_final -
V_initial, when moving from 0.30 m to 0.60 m from the charge?

A -4.50e+04 V
B -8.99e+04 V
C 8.99e+04 V
D -1.80e+05 V


Answer: B - -8.99e+04 V

Explanation: For a point charge, Delta V = kq(1/r_f - 1/r_i). The bracket is negative because the final point is
farther away. Substitution gives Delta V = -8.990e+04 V, so the potential decreases.


4. A charge of +4 uC moves through a potential difference of +80 V. How much work is
done by the electric field?

A -1.60e-04 J
B -3.20e-04 J
C -6.40e-04 J
D 3.20e-04 J


Answer: B - -3.20e-04 J

Explanation: The change in potential energy is Delta U = q Delta V = (4.00e-06)(80) J. Work done by the electric
field is W_field = -Delta U, giving -3.200e-04 J. The sign indicates whether the field gives energy to or removes
energy from the charge.




PHYS 112 | 200-Question Exam 2 Practice Exam Page 3

, PHYS 112 - GENERAL PHYSICS II | EXAM 2 PRACTICE PassPointPro



5. In a uniform electric field of 600 N/C directed to the right, a point moves 0.08 m to the
right. What is V_final - V_initial?

A -96 V
B -24 V
C -48 V
D 48 V


Answer: C - -48 V

Explanation: For displacement parallel to a uniform field, Delta V = -E d. Thus Delta V = -(600)(0.08) = -48.00 V.
Electric potential decreases in the direction of the electric field.


6. Which statement about electric potential is correct?

A Electric potential is a scalar quantity measured in volts.
B Electric potential is a vector that always points with the electric field.
C Electric potential is measured in newtons per coulomb.
D Electric potential can never be negative.


Answer: A - Electric potential is a scalar quantity measured in volts.

Explanation: Electric potential is potential energy per unit charge, V = U/q, and is measured in joules per
coulomb, or volts. Unlike the electric field, potential has no direction and is therefore a scalar.


7. A positive test charge is released from rest in an electrostatic field. If only the electric
force acts, it naturally tends to move toward:

A Lower electric potential and lower electric potential energy.
B A region of zero electric field regardless of potential.
C Higher electric potential and higher potential energy.
D Higher electric potential but lower potential energy.


Answer: A - Lower electric potential and lower electric potential energy.

Explanation: For a positive charge, U = qV, so lowering V lowers U. A conservative electric force accelerates the
charge in the direction that reduces its potential energy.


8. At point P, a +3 uC charge is 0.30 m away and a -1 uC charge is 0.30 m away. What is
the net electric potential at P?

A 3.00e+04 V
B 1.20e+05 V
C 5.99e+04 V
D -5.99e+04 V


Answer: C - 5.99e+04 V

Explanation: Electric potential adds algebraically because it is a scalar: V = kq1/r1 + kq2/r2. Substitution gives V
= 5.993e+04 V. The negative charge contributes a negative term.




PHYS 112 | 200-Question Exam 2 Practice Exam Page 4

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Subido en
14 de septiembre de 2026
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