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TOPICAL PRACTICE TEST
PHYS 112
Electrostatics &
Coulomb's Law
100-Question Practice Test
Answers, calculations & detailed explanations
100 4 A-D
QUESTIONS CORE AREAS FORMAT
FOCUSED COVERAGE
Charge, quantization & Coulomb force
Superposition & vector force addition
Electric field & force on charges
Charging, conductors & electrostatic equilibrium
Focused calculation practice plus conceptual electrostatics review.
PASSPOINTPRO
,PHYS 112 - GENERAL PHYSICS II | ELECTROSTATICS & COULOMB'S LAW PassPointPro
Electrostatics & Coulomb's Law Test Overview
This 100-question PHYS 112 topical test focuses on electrostatics fundamentals and problem solving. Questions mix charge
quantization, Coulomb-force calculations, vector superposition, electric fields, charging methods, conductors, and electrostatic
equilibrium. Every item includes four answer choices, the correct answer, and a detailed explanation immediately below the
question.
Coverage Questions
Charge, Quantization & Coulomb Force 25
Superposition & Net Electrostatic Force 25
Electric Field & Force on Charges 25
Charging, Conductors & Electrostatic Equilibrium 25
Suggested use: Solve each problem independently first. For calculation questions, identify the known quantities, convert units to
SI, choose the governing electrostatics relation, and then check the sign and direction of the result.
Quick Formula Reference
Coulomb force: F = k|q1q2|/r^2 | k = 8.99 x 10^9 N m^2/C^2
Charge quantization: q = ne | e = 1.602 x 10^-19 C
Electric field: E = F/q | Point charge: E = k|Q|/r^2
Vector components: Fx = F cos(theta), Fy = F sin(theta) | Fnet = sqrt(Fx^2 + Fy^2)
Conductor surface field: E = sigma/eps0 | eps0 = 8.854 x 10^-12 C^2/(N m^2)
PHYS 112 | 100-Question Electrostatics Test Page 2
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Charge, Quantization & Coulomb Force
1. Two positive point charges, +4 uC and +2 uC, are separated by 0.25 m. What is the
magnitude of the electrostatic force between them?
A 2.301 N
B 1.151 N
C 0.288 N
D 4.603 N
Answer: B - 1.151 N
Explanation: Use Coulomb's law, F = k|q1q2|/r^2. Substituting gives F = (8.99 x
10^9)|(4.00e-06)(2.00e-06)|/(0.25)^2 = 1.15 N. Because both charges are positive, the force is repulsive.
2. A +3 uC charge and a -6 uC charge are 0.10 m apart. What is the magnitude of the
force on either charge?
A 32.36 N
B 16.18 N
C 4.045 N
D 8.091 N
Answer: B - 16.18 N
Explanation: Coulomb's law uses the magnitude of the charge product: F = k|q1q2|/r^2 = 16.2 N. Opposite signs
mean the force is attractive, but the requested magnitude is positive.
3. An object gains 1.20e+10 electrons. What is its net charge?
A -9.61e-10 C
B -3.84e-09 C
C -1.92e-09 C
D 1.92e-09 C
Answer: C - -1.92e-09 C
Explanation: Each electron carries -e = -1.602 x 10^-19 C. Thus q = -ne = -(1.20e+10)(1.602 x 10^-19) =
-1.922e-09 C.
4. A small body has a positive charge of 1.602e-12 C. Approximately how many electrons
were removed from it?
A 1.00e+07 electrons
B 2.00e+07 electrons
C 5.00e+06 electrons
D 1.00e+08 electrons
Answer: A - 1.00e+07 electrons
Explanation: Charge is quantized: n = |q|/e. Therefore n = (1.602e-12)/(1.602 x 10^-19) = 1.000e+07 electrons.
Removing electrons leaves a positive net charge.
PHYS 112 | 100-Question Electrostatics Test Page 3
TOPICAL PRACTICE TEST
PHYS 112
Electrostatics &
Coulomb's Law
100-Question Practice Test
Answers, calculations & detailed explanations
100 4 A-D
QUESTIONS CORE AREAS FORMAT
FOCUSED COVERAGE
Charge, quantization & Coulomb force
Superposition & vector force addition
Electric field & force on charges
Charging, conductors & electrostatic equilibrium
Focused calculation practice plus conceptual electrostatics review.
PASSPOINTPRO
,PHYS 112 - GENERAL PHYSICS II | ELECTROSTATICS & COULOMB'S LAW PassPointPro
Electrostatics & Coulomb's Law Test Overview
This 100-question PHYS 112 topical test focuses on electrostatics fundamentals and problem solving. Questions mix charge
quantization, Coulomb-force calculations, vector superposition, electric fields, charging methods, conductors, and electrostatic
equilibrium. Every item includes four answer choices, the correct answer, and a detailed explanation immediately below the
question.
Coverage Questions
Charge, Quantization & Coulomb Force 25
Superposition & Net Electrostatic Force 25
Electric Field & Force on Charges 25
Charging, Conductors & Electrostatic Equilibrium 25
Suggested use: Solve each problem independently first. For calculation questions, identify the known quantities, convert units to
SI, choose the governing electrostatics relation, and then check the sign and direction of the result.
Quick Formula Reference
Coulomb force: F = k|q1q2|/r^2 | k = 8.99 x 10^9 N m^2/C^2
Charge quantization: q = ne | e = 1.602 x 10^-19 C
Electric field: E = F/q | Point charge: E = k|Q|/r^2
Vector components: Fx = F cos(theta), Fy = F sin(theta) | Fnet = sqrt(Fx^2 + Fy^2)
Conductor surface field: E = sigma/eps0 | eps0 = 8.854 x 10^-12 C^2/(N m^2)
PHYS 112 | 100-Question Electrostatics Test Page 2
, PHYS 112 - GENERAL PHYSICS II | ELECTROSTATICS & COULOMB'S LAW PassPointPro
Charge, Quantization & Coulomb Force
1. Two positive point charges, +4 uC and +2 uC, are separated by 0.25 m. What is the
magnitude of the electrostatic force between them?
A 2.301 N
B 1.151 N
C 0.288 N
D 4.603 N
Answer: B - 1.151 N
Explanation: Use Coulomb's law, F = k|q1q2|/r^2. Substituting gives F = (8.99 x
10^9)|(4.00e-06)(2.00e-06)|/(0.25)^2 = 1.15 N. Because both charges are positive, the force is repulsive.
2. A +3 uC charge and a -6 uC charge are 0.10 m apart. What is the magnitude of the
force on either charge?
A 32.36 N
B 16.18 N
C 4.045 N
D 8.091 N
Answer: B - 16.18 N
Explanation: Coulomb's law uses the magnitude of the charge product: F = k|q1q2|/r^2 = 16.2 N. Opposite signs
mean the force is attractive, but the requested magnitude is positive.
3. An object gains 1.20e+10 electrons. What is its net charge?
A -9.61e-10 C
B -3.84e-09 C
C -1.92e-09 C
D 1.92e-09 C
Answer: C - -1.92e-09 C
Explanation: Each electron carries -e = -1.602 x 10^-19 C. Thus q = -ne = -(1.20e+10)(1.602 x 10^-19) =
-1.922e-09 C.
4. A small body has a positive charge of 1.602e-12 C. Approximately how many electrons
were removed from it?
A 1.00e+07 electrons
B 2.00e+07 electrons
C 5.00e+06 electrons
D 1.00e+08 electrons
Answer: A - 1.00e+07 electrons
Explanation: Charge is quantized: n = |q|/e. Therefore n = (1.602e-12)/(1.602 x 10^-19) = 1.000e+07 electrons.
Removing electrons leaves a positive net charge.
PHYS 112 | 100-Question Electrostatics Test Page 3
