PASSPOINTPRO • PHYS 112
Electric Fields &
Electric Potential
100-Question Practice Test
100 80 20
QUESTIONS CALCULATIONS CONCEPTUAL
Electric-field magnitude • superposition • force on charge • potential • potential energy • work • uniform
fields • equipotentials
Answers + detailed explanations after every question
,PHYS 112 - ELECTRIC FIELDS & ELECTRIC POTENTIAL
Formula Reference
Point-charge field E = k|q| / r^2
Force in a field F = qE
Point-charge potential V = kq / r
Potential energy U = qV
Potential-energy change Delta U = q Delta V
Electric work W_field = -Delta U = -q Delta V
Uniform field |Delta V| = Ed (parallel displacement)
Field-potential relation E = -dV/dx
Coulomb constant k = 8.99 x 10^9 N*m^2/C^2
Elementary charge e = 1.602 x 10^-19 C
Use SI units unless otherwise stated. Direction matters in electric-field and force reasoning, while electric potential is a scalar and
adds algebraically.
PassPointPro | General Physics II Practice Test Page 2
,PHYS 112 - ELECTRIC FIELDS & ELECTRIC POTENTIAL
QUESTION 1 Electric-field superposition
At a point between two positive charges, charge q1 = 2 uC is 0.2 m away and charge q2 = 3 uC
is 0.3 m away. The fields from the two charges point in opposite directions at the point. What is
the net electric-field magnitude there?
A 1.5 x 10^5 N/C
B 5.99 x 10^5 N/C
C 3 x 10^5 N/C
D 1.5 x 10^4 N/C
Answer: A - 1.5 x 10^5 N/C
Explanation: Use superposition: compute each field separately and then combine with direction. E1 =
kq1/r1^2 = 4.49 x 10^5 N/C, while E2 = kq2/r2^2 = 3 x 10^5 N/C. Because the two fields oppose
each other, the net magnitude is |E1 - E2| = 1.5 x 10^5 N/C.
QUESTION 2 Conceptual foundations
At a point where the electric potential is zero, which statement must be true?
A The electric field must also be zero
B The electric field may still be nonzero
C A positive charge there has zero force
D No charges can be nearby
Answer: B - The electric field may still be nonzero
Explanation: Electric potential is a scalar reference quantity, while electric field depends on how
potential changes with position. Contributions from charges can cancel in potential without canceling
in field. Therefore V = 0 does not necessarily imply E = 0.
PassPointPro | General Physics II Practice Test Page 3
, PHYS 112 - ELECTRIC FIELDS & ELECTRIC POTENTIAL
QUESTION 3 Conceptual foundations
When several point charges contribute to electric potential at one location, the total potential is
found by:
A Adding the individual potentials algebraically
B Adding only the magnitudes
C Multiplying the individual potentials
D Taking a vector sum
Answer: A - Adding the individual potentials algebraically
Explanation: Electric potential is a scalar, so contributions from each point charge add algebraically
with their signs. No vector components are needed. Positive and negative contributions can partially
or completely cancel.
QUESTION 4 Potential energy and work
A positive charge with magnitude 0.8 uC moves through a potential change of +35 V. What is
the change in its electric potential energy, Delta U?
A 2.8 x 10^-5 J
B 1.4 x 10^-5 J
C 5.6 x 10^-5 J
D -2.8 x 10^-5 J
Answer: A - 2.8 x 10^-5 J
Explanation: Potential-energy change is Delta U = q Delta V. Here q = 0.8 x 10^-6 C and Delta V =
+35 V, so Delta U = 2.8 x 10^-5 J. A negative result means the electric potential energy decreases; a
positive result means it increases.
PassPointPro | General Physics II Practice Test Page 4
Electric Fields &
Electric Potential
100-Question Practice Test
100 80 20
QUESTIONS CALCULATIONS CONCEPTUAL
Electric-field magnitude • superposition • force on charge • potential • potential energy • work • uniform
fields • equipotentials
Answers + detailed explanations after every question
,PHYS 112 - ELECTRIC FIELDS & ELECTRIC POTENTIAL
Formula Reference
Point-charge field E = k|q| / r^2
Force in a field F = qE
Point-charge potential V = kq / r
Potential energy U = qV
Potential-energy change Delta U = q Delta V
Electric work W_field = -Delta U = -q Delta V
Uniform field |Delta V| = Ed (parallel displacement)
Field-potential relation E = -dV/dx
Coulomb constant k = 8.99 x 10^9 N*m^2/C^2
Elementary charge e = 1.602 x 10^-19 C
Use SI units unless otherwise stated. Direction matters in electric-field and force reasoning, while electric potential is a scalar and
adds algebraically.
PassPointPro | General Physics II Practice Test Page 2
,PHYS 112 - ELECTRIC FIELDS & ELECTRIC POTENTIAL
QUESTION 1 Electric-field superposition
At a point between two positive charges, charge q1 = 2 uC is 0.2 m away and charge q2 = 3 uC
is 0.3 m away. The fields from the two charges point in opposite directions at the point. What is
the net electric-field magnitude there?
A 1.5 x 10^5 N/C
B 5.99 x 10^5 N/C
C 3 x 10^5 N/C
D 1.5 x 10^4 N/C
Answer: A - 1.5 x 10^5 N/C
Explanation: Use superposition: compute each field separately and then combine with direction. E1 =
kq1/r1^2 = 4.49 x 10^5 N/C, while E2 = kq2/r2^2 = 3 x 10^5 N/C. Because the two fields oppose
each other, the net magnitude is |E1 - E2| = 1.5 x 10^5 N/C.
QUESTION 2 Conceptual foundations
At a point where the electric potential is zero, which statement must be true?
A The electric field must also be zero
B The electric field may still be nonzero
C A positive charge there has zero force
D No charges can be nearby
Answer: B - The electric field may still be nonzero
Explanation: Electric potential is a scalar reference quantity, while electric field depends on how
potential changes with position. Contributions from charges can cancel in potential without canceling
in field. Therefore V = 0 does not necessarily imply E = 0.
PassPointPro | General Physics II Practice Test Page 3
, PHYS 112 - ELECTRIC FIELDS & ELECTRIC POTENTIAL
QUESTION 3 Conceptual foundations
When several point charges contribute to electric potential at one location, the total potential is
found by:
A Adding the individual potentials algebraically
B Adding only the magnitudes
C Multiplying the individual potentials
D Taking a vector sum
Answer: A - Adding the individual potentials algebraically
Explanation: Electric potential is a scalar, so contributions from each point charge add algebraically
with their signs. No vector components are needed. Positive and negative contributions can partially
or completely cancel.
QUESTION 4 Potential energy and work
A positive charge with magnitude 0.8 uC moves through a potential change of +35 V. What is
the change in its electric potential energy, Delta U?
A 2.8 x 10^-5 J
B 1.4 x 10^-5 J
C 5.6 x 10^-5 J
D -2.8 x 10^-5 J
Answer: A - 2.8 x 10^-5 J
Explanation: Potential-energy change is Delta U = q Delta V. Here q = 0.8 x 10^-6 C and Delta V =
+35 V, so Delta U = 2.8 x 10^-5 J. A negative result means the electric potential energy decreases; a
positive result means it increases.
PassPointPro | General Physics II Practice Test Page 4
