Physics 2049 Exam 1 Solutions
1. The figure shows five pairs of plates; A, B, and D are charged plastic plates and C is an electrically
neutral copper plate. The electrostatic forces between the pairs of plates are shown for three of the
pairs. Which of the following statements is correct?
A
C
D
C
B
B
A
A
D
D
Answer: C and D attract, B and D attract.
Solution: From the first and third pairs of plates we see that B and D must have opposite charges,
so they attract. The charge in a neutral copper plate will be polarized in such a way that if D is,
say, positively charged, negative charge will be drawn to the face of C closest to D, resulting in a net
attraction. See Checkpoint 1, p. 508 of Halliday, Resnick, and Walker.
2. A charge Q is placed on a line halfway between two charges, each of charge q. If this system is to be
in equilibrium, what is Q?
q
Q
d
q
d
Answer: −q/4
Solution: Consider the net force on the charge q on the right; we have
Fx =
kq 2
kqQ
+ 2 .
(2d)2
d
(1)
In equilibrium, Fx = 0; solving for Q, we find that Q = −q/4. You would get the same result for the
charge q on the left (why?).
1
3. Charge q1 = 3.0 × 10−6 C is located at (x, y) = (0, 10 cm), charge q2 = 4.0 × 10−6 C is located at
(x, y) = (10 cm, 0), and charge q3 = −1.0 × 10−6 C is at the origin, as shown in the figure. What is
the magnitude of the force on q3 ? (k = 9.0 × 109 N m2 /C2 )
y
q1
q3
q2
x
Answer: 4.5 N
Solution: The x-component of the force on q3 is
kq2 q3
d2
(2)
kq1 q3
.
d2
(3)
Fx = −
with d = 0.10 m. Likewise,
Fy = −
The magnitude of the force is
F =
q
Fx2 + Fy2 = 4.5 N
(4)
Remember that the magnitude is the length of the vector, which is always a positive number!
4. The figure shows two electrons e on the x axis. Also shown are three points A, B, and C. Which of
the following statements about the direction of the electric field at these three points is correct?
e
A
e
B
C
Answer: A right, B left, C right
Solution: Remember that electrons are negatively charged, and that the field lines begin on positive
charges and end on negative charges.
2
5. A charge Q = 27 × 10−9 C is distributed uniformly on a string which is stretched between x = 0 and
x = 3.0 m. What is the magnitude of the electric field at x = 4.0 m?
y
What’s the field here?
0
3.0
4.0
x
Answer: 61 N/C
Solution: Let L = 3.0 m be the length of the string, and let d = 1.0 m be the distance between the
right end of the string and the point P at x = 4.0 m. Divide the string into segments of length dx,
each of which has a charge dQ = (Q/L)dx. The electric at point P due to one of these segments at a
distance x to the right of the origin is
dEx =
kdQ
k(Q/L)dx
=
.
2
(L + d − x)
(L + d − x)2
(5)
To find the net electric field at P , we integrate this over the length of the string:
Z L
1
1
k(Q/L)dx
−
= k(Q/L)
.
Ex =
2
d d+L
0 (L + d − x)
(6)
Substituting the numbers given, we find Ex = 61 N/C at point P .
6. Two semicircular plastic rods of radius R are arranged as shown in the figure; the left rod has a
uniformly distributed charge +q, and the right rod has a uniformly distributed charge −q. What is
the electric field at the center, point P ?
y
+q
-q
P
x
R
3
~ = (4kq/πR2 )î
Answer: E
Solution: See Sample Problem 23-3 on p. 529 of HRW, and Problem 23-20. Let’s start with the left
rod. Consider a segment on the left rod of length ds = R dθ and charge dQ = (q/πR)ds = (q/πR)Rdθ,
which is at an angle θ with respect to the x-axis. The magnitude of the electric field at the origin from
~ = k|dQ|/R2 ; taking x and y components and integrating, we have
this segment is |dE|
Z
Ex
π/2
=
−π/2
=
2kq
,
πR2
Z
Ey
π/2
=
−π/2
=
k(q/πR)R
cos θ dθ
R2
(7)
k(q/πR)R
sin θ dθ
R2
0.
(8)
This is just the contribution from the left rod. The right rod makes a contribution which is equal in
magnitude but opposite in sign, so that the net field is Ex = 4kq/πR2 and Ey = 0. In vector notation
~ = (4kq/πR2 )î, with î the unit vector in the x-direction.
this would be E
7. A ball of mass m = 0.75 grams is suspended from a thread of negligible mass. The ball carries a charge
q = +32 × 10−6 C, and is placed in a uniform electric field of magnitude E = 120 N/C which points
2
to the left. What is the angle θ which the thread makes with the vertical? (g = 9.8 m/s )
θ
g
q
uniform E field
Answer: 28◦
Solution: This is basically the same as Webwork 2, Problem 3. Draw a free-body diagram of the ball;
let T~ be the tension in the thread. In the x-direction we have
Fx = 0 = −qE + T sin θ,
(9)
and in the y-direction we have
Fy = 0 = −mg + T cos θ.
(10)
◦
Eliminating T and solving for θ, we find tan θ = qE/mg; putting in the numbers, we find θ = 28 .
4
8. A positive point charge Q is placed at the center of a spherical shell of radius R; the flux of the electric
field through the surface of the shell is Φ0 . We now triple the amount of charge, and reduce the radius
of the shell to half of its original value. What is the new flux through the surface of the shell?
Answer: 3Φ0
Solution: From Gauss’ Law we know that the net flux through a closed surface only depends upon
the net charge enclosed by that surface; tripling the amount of charge inside the shell triples the flux
through its surface, irrespective of the change in radius.
9. A long, very thin plastic rod has a constant charge per unit length λ. It is coaxial with a long, hollow
conducting cylinder with an inner radius R and an outer radius 3R; the conductor is electrically neutral.
What is the magnitude of the electric field at the a distance 2R from the plastic rod?
3R
λ
conductor
R
Cross section of rod and conductor
Answer: E = 0
Solution: A point 2R from the plastic rod is inside the conducting cylinder, and we know that in
equilibrium the electric field inside conductors is zero. See Section 24-6 in HRW.
10. A point charge of charge +Q is placed at the center of a solid spherical conducting shell of inner radius
R and outer radius 2R. The shell is in static equilibrium and has a net charge +Q. What is the total
charge on the inner surface (at r = R) of the shell?
2R
R
Net charge +Q on conductor
+Q
Answer: −Q
Solution: The electric field inside the conductor (between R and 2R) is zero. Draw a Gaussian surface
with a radius r inside the conductor (R < r < 2R); since the electric field is zero at this surface, the
net flux through the Gaussian surface is also zero. From Gauss’ Law we know that this means that
the net charge inside the surface is zero, so there must be a charge −Q on the inner surface of the
conductor (i.e., the point charge at the center has polarized the charge in the conductor). The net
charge on the outer surface of the conductor is then 2Q.
5