Solutions_ Spring 2026.
Physics 240 — Winter 2026
Quiz #1
Jan. 26, 2026
This is a closed-book exam lasting 45 minutes. Calculators and a single 3x5-inch
notecard are allowed. Do all problems. See the grading algorithm below.
Explanation of Multiple-Choice Grading and Partial Credit
(Please read carefully!)
• Questions with 5 choices are worth 6 points. If you bubble in one answer, and
it’s correct, you’ll receive 6 points. If you bubble in TWO answers, and one of
them is correct, you’ll receive 3 points. If you bubble in more than two answers,
or none of your answers is correct, you’ll receive zero points. By electing to
bubble in two answers, your score for the problem could go up or down!
• Questions with 4 choices are worth 4 points. If you bubble in one answer, and
it’s correct, you’ll receive 4 points. If you bubble in TWO answers, and one of
them is correct, you’ll receive 2 points. If you bubble in more than two answers,
or none of your answers is correct, you’ll receive zero points.
• Questions with 2 or 3 choices are worth 2 and 3 points respectively. There is
no partial credit on these problems.
Useful Physical Constants:
Proton charge e = 1.602 × 10−19 C
Permittivity of free space 0 = 8.85 × 10−12 C2 /(N·m2 )
Coulomb’s law constant k = 1/(4π0 ) = 8.99 × 109 N·m2 / C2
Permeability of free space µ0 = 4π × 10−7 T·m/A
Gravitational acceleration g = 9.8 m/s2
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, Your exam version will be determined by the first two bubbles. This allows me
to put 25 versions on the exam. You will need to fill in the first two bubbles
(for problem 1 and problem 2) exactly as indicated on this page.
1. Bubble in b
2. Bubble in l
,3. A positive point charge q sits at the center of a sphere of radius a as shown:
a
q
Which of the following actions would cause the electric flux through the sphere
to stay the same?
I. Increase he radius of the sphere by a factor of two.
II. Increase the charge to 2q
III. Decrease the radius of the sphere by a factor of two.
IV. Displace the charge a distance a/2 to the right.
A) III and IV
B) I and II
C) I, III, and IV
D) II only.
E) II and IV
Solution: Gauss’s law tells you that the flux through a closed surface depends
only on the total charge enclosed. The size and shape of the surface and location
of the charge inside do not matter. So any action other than changing the charge
enclosed will leave the flux unchanged. You can increase or decrease the size of
the sphere or move the charge around inside.
, 4. Three charges are arranged at the vertices of an equilateral triangle with sides of
length 0.25 m as shown. The charges have values q1 = 3.69 µC, q2 = −5.85 µC,
and q3 = 3.31 µC. What is the vertical component of the electrostatic force on
q3 ? Take the positive direction to be toward the top of the page.
q3
d d
q1 d q2
A) 0.51 N
B) −0.89 N
C) 3.93 N
D) 0.73 N
E) 2.27 N
Solution: See the figure below. The relative lengths and signs of the force
arrows are not necessarily accurate.
o
30
q3
d d
q1 d q2
The vertical component of the force on q3 is:
kq3
Fy = (q1 + q2 ) cos 30◦ = −0.8903219917460056 N.
d2
If you insert the correct signs for the charges into the equation above, the algebra
will give you the correct magnitude and direction.