RIO SALADO PHYSICS 112 MIDTERM
ACTUAL EXAM PREP 2026 ( 2 CURRENTLY
TESTING VERSIONS ) ALL QUESTIONS AND
CORRECT DETAILED ANSWERS ALREADY
A GRADED WITH EXPERT FEEDBACK |NEW
AND REVISED
1. Two equal positive point charges are placed a distance (d) apart. The electric
potential at the midpoint between them is:
A. Zero
B. Negative
C. Positive
D. Infinite
Rationale: Potentials from each positive charge add algebraically, so the
midpoint has positive potential (sum of two positive contributions).
2. A point charge +Q is brought from infinity to a point near another +Q. Work
done by an external agent (positive charge brought slowly) is:
A. Negative (work done by field)
B. Positive (work done against repulsive force)
C. Zero
D. Infinite
Rationale: Because like charges repel, an external agent must do positive
work to bring them together slowly.
3. Electric field (\mathbf{E}) and electric potential (V) are related by:
A. (\mathbf{E} = \nabla V)
B. (\mathbf{E} = -\nabla V)
C. (V = \nabla \cdot \mathbf{E})
D. (V = \mathbf{E} \times \nabla)
Rationale: The electric field is the negative gradient of the potential.
4. Inside a conducting hollow spherical shell at electrostatic equilibrium, the
electric field is:
A. Dependent on external fields
B. Zero
C. Nonzero and uniform
,2|Page
D. Infinite at center
Rationale: The interior of a conductor in electrostatic equilibrium has E =
0 everywhere.
5. A charge (q) moves between two equipotential surfaces of difference (\Delta
V). The change in potential energy (\Delta U) is:
A. (q/\Delta V)
B. (q\Delta V)
C. (-q\Delta V)
D. Independent of q
Rationale: Change in electric potential energy is charge times potential
difference: (\Delta U = q\Delta V).
6. The capacitance of a parallel-plate capacitor is directly proportional to:
A. Plate separation (d)
B. Inverse of area (1/A)
C. Plate area (A)
D. Dielectric constant inverse
Rationale: (C=\varepsilon_0\varepsilon_r A/d), so capacitance ∝ area.
7. Two capacitors, (C_1) and (C_2), are connected in series. The equivalent
capacitance (C_{eq}) is:
A. (C_1 + C_2)
B. (\displaystyle\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}})
C. (C_1C_2)
D. (|C_1-C_2|)
Rationale: Series capacitors combine via reciprocal sum analogous to
resistors in parallel.
8. When a capacitor is fully charged through a resistor by a DC source, the
final current through the resistor is:
A. Maximum value (V/R)
B. Zero
C. Constant nonzero
D. Oscillatory
Rationale: Once the capacitor is fully charged, no DC current flows;
current decays to zero.
9. Ohm’s law relates voltage, current, and resistance as:
A. (V = IR^2)
B. (V = IR)
C. (I = VR)
D. (R = VI)
Rationale: Ohm’s law states V equals current times resistance.
,3|Page
10.Two resistors of equal resistance are connected: which configuration gives
half the equivalent resistance of one resistor?
A. Series
B. Parallel
C. Neither
D. Both
Rationale: Two equal resistors (R) in parallel give (R/2); in series give
(2R).
11.Kirchhoff’s current law (KCL) states that at any junction:
A. The voltage sum is zero
B. The algebraic sum of currents is zero
C. The resistance sum is zero
D. Power is conserved only
Rationale: KCL is conservation of charge: sum of currents entering equals
sum leaving.
12.In an RLC series circuit driven at the resonant frequency, the impedance is:
A. Maximum
B. Purely resistive (minimum impedance equal to R)
C. Purely inductive
D. Infinite
Rationale: At resonance, inductive and capacitive reactances cancel
leaving only R.
13.The reactance of a capacitor (X_C) at angular frequency (\omega) is:
A. (X_C = \omega C)
B. (X_C = 1/(\omega C))
C. (X_C = \omega / C)
D. (X_C = \omega^2 C)
Rationale: Capacitive reactance decreases with frequency: (X_C =
1/\omega C).
14.A charged particle with charge (q) moving with velocity (\mathbf{v}) in a
magnetic field (\mathbf{B}) experiences magnetic force
(\mathbf{F}=q\mathbf{v}\times\mathbf{B}). The force is zero when:
A. v is perpendicular to B
B. v is parallel to B
C. q = 0 only
D. B = 0 only
Rationale: Cross product is zero when vectors are parallel (or
antiparallel).
15.The magnetic field at the center of a circular loop of radius (R) carrying
current (I) is proportional to:
, 4|Page
A. (I/R^2)
B. (I/R)
C. (I R)
D. (I^2 R)
Rationale: For a single loop, (B=\mu_0 I/(2R)), so B ∝ I/R.
16.Faraday’s law of induction states the induced emf equals:
A. Sum of fields
B. Negative rate of change of magnetic flux: (\mathcal{E} = -d\Phi_B/dt)
C. Product of voltage and current
D. Zero in all closed loops
Rationale: Induced emf opposes flux change (Lenz’s law included as
negative sign).
17.A coil of N turns experiences a changing flux (\Phi_B). The induced emf is:
A. (d\Phi_B/dt)
B. (\Phi_B/N)
C. (-N, d\Phi_B/dt)
D. (N/\Phi_B)
Rationale: Faraday’s law for N-turn coil: (\mathcal{E} = -N, d\Phi_B/dt).
18.Lenz’s law gives the direction of induced current such that it:
A. Enhances the change in flux
B. Opposes the change in flux
C. Is always clockwise
D. Is independent of flux change
Rationale: Lenz’s law ensures energy conservation by opposing flux
change.
19.An ideal transformer with primary voltage (V_p) and turns ratio (N_p:N_s)
gives secondary voltage (V_s) equal to:
A. (V_p (N_p/N_s))
B. (V_p (N_s/N_p))
C. (V_p N_p N_s)
D. Independent of turns ratio
Rationale: Transformer voltages scale with turns ratio: (V_s/V_p =
N_s/N_p).
20.In electromagnetic waves in vacuum, the relationship among electric field
amplitude (E), magnetic field amplitude (B), and speed of light (c) is:
A. (E = c B)
B. (E = cB) (same as A — reaffirming)
C. (E = B/c)
D. (E) independent of (B)
Rationale: In free space, (E) and (B) amplitudes are related by (E=cB).