PHYSICS 2200 -- PRACTICE MIDTERM with Answers
1) Determine the equivalent capacitance of the combination shown when C = 24 µF.
a. 20 µF
b. 36 µF
c. 16 µF
d. 45 µF
e. 27 µF
Answer: C
2) In the figure, all the charges are point charges and the charge in the middle is Q = -3.1
nC. For what charge q1 will charge q2 be in static equilibrium?
A) 12 nC
B) 6.2 nC
C) 3.1 nC
D) 25 nC
Answer: A
3) The figure shows two tiny 5.0-g spheres suspended from two very thin
1.0-m-long threads. The spheres repel each other after being charged to +91
nC and hang at rest as shown. What is the angle θ? (k = 1/4πε0 = 8.99 × 109
N ∙ m2/C2)
A) 4.1°
B) 8.2°
C) 12°
D) 16°
Answer: A
4) In the figure Q = 5.8 nC and all other quantities are accurate to 2 significant figures.
What is the magnitude of the force on the charge Q? (k = 1/4πε0 = 8.99 ×
109 N ∙ m2/C2)
A) 1.8 × 10-3 N
B) 1.0 × 10-3 N
C) 9.0 × 10-4 N
D) 1.2 × 10-3 N
Answer: A
5) Two point charges of +20.0 μC and -8.00 μC are separated by a distance
of 20.0 cm. What is the magnitude of electric field due to these charges at a point midway
between them? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)
A) 25.2 × 106 N/C directed toward the negative charge
B) 25.2 × 106 N/C directed toward the positive charge
C) 25.2 × 105 N/C directed toward the negative charge
D) 25.2 × 105 N/C directed toward the positive charge
E) 25.2 × 104 N/C directed toward the negative charge
Answer: A
6) A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is
the electric flux through the sphere due to this charge? (ε0 = 8.85 × 10-12 C2/N ∙ m2)
A) 0.11 N ∙ m2/C
B) 8.9 N ∙ m2/C
C) 0.028π N ∙ m2/C
D) It cannot be determined without knowing the radius of the sphere.
Answer: A
7) An initially uncharged hollow metallic sphere with radius of 5 cm has a small object
with a charge of +10 µC carefully placed at the center of the sphere through a hole in the
latter’s surface. What charge resides inner surface of the sphere?
a. 4 000 µC
b. 10 µC
c. +10 µC
d. zero
Answer: B
8) A spherical volume of space has an electric field of intensity 100 N/C directed radially
outward from it surface of radius 0.600 m. What is the net charge enclosed within this
surface?
a. 6.8 nC
b. 4.0 nC
c. –6.8 nC
d. –4.0 nC
Answer: B
9) A point charge of +3.0 C is located at the origin of a coordinate system and a second
point charge of 6.0 C is at x = 1.0 m. At what point on the x axis is the electrical
potential zero?
a. 0.25 m
b. +0.25 m
c. +0.33 m
d. +0.75 m
Answer: C
10) If C1 = 25 µF, C2 = 20 µF, C3 = 10 µF, and V0 = 21 V, determine
C1
the energy stored by C2.
a. 0.72 mJ
b. 0.32 mJ
C2
V0
c. 0.40 mJ
d. 0.91 mJ
Answer: D
11) A pair of parallel plates, forming a capacitor, are connected to a battery. While the
capacitor is still connected to the battery maintaining a constant voltage, the plates are
C3
pulled apart to double their original distance. What is the ratio of the final energy stored
to the original energy stored?
a. 2
b. 1
c. 1/2
d. ¼
Answer: C
12) Two concentric spheres are shown in the figure. The inner sphere is a solid
nonconductor and carries a charge of +5.00 µC uniformly distributed over its outer
surface. The outer sphere is a conducting shell that carries a net charge of -8.00 µC. No
other charges are present. The radii shown in the figure have the values R1 = 10.0 cm, R2
= 20.0 cm, and R3 = 30.0 cm. (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)
(a) Find the total excess charge on the inner and outer surfaces of the conducting sphere.
(b) Find the magnitude and direction of the electric field at the following distances r from
the center of the inner sphere: (i) r = 9.5 cm, (ii) r = 15.0 cm, (iii) r = 27.0 cm, (iv) r =
35.0 cm.
Answer:
(a) -5.00 µC (inner surface), -3.00 µC (outer surface)
(b) (i) 0; (ii) 2.00 × 106 N/C, radially outward; (iii) 0; (iv) 2.20 × 105 N/C, radially
inward
13) The potential as a function of position x is shown in the graph in the figure. Which
statement about the electric field is true?
A) The electric field is zero at x = 0, its magnitude is at a maximum at x = 5 cm, and the
field is directed to the right there.
B) The electric field is zero at x = 5 cm, its magnitude is at a maximum at x = 0, and the
field is directed to the right there.
C) The electric field is zero at x = 0, its magnitude is at a maximum at x = 15 cm, and the
field is directed to the left there.
D) The electric field is zero at x = 10 cm, its magnitude is at a maximum at x = 5 cm, and
the field is directed to the left there.
Answer: A
14) Two capacitors of capacitance 6.00 μF and 8.00 μF are connected in parallel. The
combination is then connected in series with a 12.0-V voltage source and a 14.0-μF
capacitor, as shown in the figure.
(a) What is the equivalent capacitance of this combination?
(b) What is the charge on the 6.00-μF capacitor?
(c) What is the potential difference across the 6.00-μF capacitor?
Answer: (a) 7.00 µF (b) 36.0 µC (c) 6.00 V
15) If the potential in a region is given by V(x,y,z) = xy - 3z-2, then the y component of
the electric field in that region is
A) x + y - 6z-3.
B) -y.
C) -x.
D) x + y.
Answer: C
16) Three equal charges of magnitude +8.00 µC are placed at the corners of an
equilateral triangle of side 6.00 cm. What is the electric field potential energy of the
system of these charges?
A) 90.0 J
B) 0 J
C) 289 mJ
D) 28.9 J
Answer: D
17) Which of the following will increase the capacitance of a parallel plate capacitor?
A) a decrease in the plate area and an increase in the plate separation
B) a decrease in the potential difference between the plates
C) an increase in the potential difference between the plates
D) an increase in the plate area and a decrease in the plate separation
Answer: D
18) When a dielectric material is introduced between the plates of a parallel plate
capacitor the capacitance increases by a factor of 4. What is the dielectric constant of the
material introduced between the plates?
A) 0.4
B) 1/4
C) 4
D) None of the other choices is correct.
Answer: C
19) The equipotential surfaces between the two point charges are shown in Figure with
the value of potential marked for each line. If the
distance between points C and D is also 2.5 cm, what is
the intensity of the electric field between these two
points?
A) 2 V/m
B) 4 V/m
C) 6 V/m
D) 8 V/m
Answer: D
20) Four equal point charges of magnitude 6.00 µC and of
varying signs are placed at the corners of a square 2.00 m on
each side, as shown in Figure. What is the electric potential at
the center of the square?
A) 0 V
B) 76.4 kV
C) 152.7 kV
D) 61.0 kV
Answer: A
21) A capacitor consists of a set of two parallel plates of area A
separated by a distance d. This capacitor is connected to a
battery that maintains a constant potential difference between the plates. A slab of a
dielectric material is inserted in the region between the plates so as to completely fill it.
What changes would you observe?
A) Only the charge in the capacitor would change.
B) Only the capacitance would change.
C) Both the charge in the capacitor and its capacitance would change.
D) Nothing would change.
Answer: C
22) Suppose you have two point charges of opposite sign. As you move them farther and
farther apart, the potential energy of this system relative to infinity
A) increases.
B) decreases.
C) stays the same.
Answer: A
23) Four point charges of equal magnitude and signs are arranged
on three of the corners and at the center of the square of side d as
shown in Figure above. Which of the arrows shown represents the
net force acting on the center charge?
A) A
B) B
C) C
D) D
Answer: B
Pre Test Problems with Solutions
Chapter 21
1. A charge of 6.00 mC is placed at each corner of a square 0.100 m on a side. Determine
the magnitude and direction of the force on each charge.
We find the magnitudes of the individual forces on the
charge at the upper right corner:
F1 = F2 = kQQ/L2 = kQ2/L2
y
F1
= (9.0109 N · m2/C2)(6.0010–3 C)2/(0.100 m)2
F3
7
L
= 3.2410 N.
+
+
F3 = kQQ/(L√2)2 = kQ2/2L2
Q F2
Q
9
2
2
–3
2
2
= (9.010 N · m /C )(6.0010 C) /2(0.100 m)
= 1.62107 N.
L
The directions of the forces are determined from the signs
x
of the charges and are indicated on the diagram. For the
forces on the upper-right charge, we see that the net force
will be along the diagonal. For the net force, we have
Q
Q
+
F = F1 cos 45° + F2 cos 45° + F3
+
= 2(3.24107 N) cos 45° + 1.62107 N
= 6.20107 N along the diagonal, or away from the center of the square.
From the symmetry, each of the other forces will have the same magnitude and a direction away
from the center: The net force on each charge is 6.20107 N away from the center of the square.
Note that the sum for the three charges is zero.
2. Given the two charges Q1= Q and Q1=-Q on a distance d=1 m from each other shown in
Figure below, find at what position x is electric field is equal zero? Is the field zero at
any other points, not on the x axes?
Because the charges have opposite signs, the location where the electric
field is zero must be outside the two charges, as shown.
x
P
The fields from the two charges must balance:
–
x
+
Q1
kQ1/(d + x)2 = kQ2/x2;
Q
E
E
2
2
1
d
Q/(d + x)2 = Q/2x2, or d + x = ± x√2.
which gives x = d/(√2 – 1), – d/(√2 + 1), or x = d(1 + √2), d(1 – √2),
Because d(1 – √2) is between the charges, the location is
d(1 + √2) from the negative charge, and d(2 + √2) from the positive charge.
Other than at infinity, there is no place, not on the x-axis, where the vectors sum to zero.
Chapter 22
3. Figure shows a solid metal sphere at the center of a hollow metal sphere. What is a total
charge on (a) the exterior of the inner sphere, (b) the inside surface of the hollow
sphere, and (c) the exterior surface of the hollow sphere?
The excess charge on a conductor resides on the outer surface. The field inside, outside, and within the hollow
metal sphere has spherical symmetry.
The figure shows spherical Gaussian surfaces with radii r  a, a  r  b, and r  b. These surfaces match the
symmetry of the charge distribution.
Solve: (a) For r  a, Gauss’s law is
 e   E  dA 
Qin
0

Q
0
Notice that the electric field is everywhere perpendicular to the spherical surface. Because of the spherical
symmetry of the charge, the electric field magnitude E is the same at all points on the Gaussian surface. Thus,


 e  EAsphere  E 4 r 2 
Q
0
 E
Q
4 0r
2
E
1
Q
rˆ
4 0 r 2
where we made use of the fact that E is directed radially outward. The field depends only on the enclosed charge,
not on the charge on the outer sphere.
For a  r  b, Gauss’s law is
Q
e   E  dA  EAsphere  E 4 r 2  in


0
Here Qin  0 C. It is not Q, because the charge in the cavity polarizes the metal sphere in such a way that E = 0
in the metal. Thus a charge Q moves to the inner surface. Because the hollow sphere has a net charge of 2Q,
the exterior surface now has a charge of 3Q. Thus, the electric field E  0 N/C.
For r  b,
Qin  Qexterior  Qinterior  Qcavity  3Q  (Q)  (Q)  3Q
Gauss’s law applied to the Gaussian surface at r  b yields:


e   E  dA  EAsphere  E 4 r 2 
Qin
0

3Q
0
 E
1  3Q 
1  3Q 
E
rˆ
4 0  r 2 
4 0  r 2 
(b) As determined in part (a), the inside surface of the hollow sphere has a charge of Q, and the exterior surface
of the hollow sphere has a charge of 3Q.
Assess: The hollow sphere still has the same charge 2Q as given in the problem, although the sphere is
polarized.
Chapter 23
4. Four point charges are located at the corners of a square that is 8.0 cm on a side. The
charges, going in rotation around square, are Q, 2Q, -3Q and 2Q, where Q=4.5 µC (see
Figure). What is total electric potential energy stored in the system, relative to U=0 at
infinite separation?
For the potential energy of the four charges we have
U = (1/4Å0)(Q1Q2/r12 + Q1Q3/r13 + Q1Q4/r14 + Q2Q3/r23 +
Q2Q4/r24 + Q3Q4/r34)
= (Q2/4Å0)[(1)(2)/b + (1)(– 3)/b√2 + (1)(2)/b + (2)(–3)/b +
(2)(2)/b√2 + (– 3)(2)/b]
9
2
2
–6
2
= (9.010 N · m /C )(4.810 C) 
[2 – 3/√2 + 2 – 6 + 4/√2 – 6]/(0.080 m)
= – 19 J.
Q +
b
+ 2Q
b
+
2Q
5. Two identical 7.5 µC point charges are initially at rest and spaced 5.5 cm from each
other. If they are released at the same instant moment from rest, how fast will they be
moving when they are very far away from each other? Assume that they have identical
masses, 1.0 mg.
We find the electric potential energy of the system by considering one of the charges to be at the
potential created by the other charge. This will be zero when they are far away. Because the masses
are equal, the speeds will be equal. From energy conservation we have
K + U = 0;
½mv2 + ½mv2 – 0 + Q(0 – V) = 0, or
2(½mv2) = mv2 = Q(kQ/r) = kQ2/r;
(1.010–6 kg)v2 = (9.0109 N · m2/C2)(7.510–6 C)2/(0.055 m),
which gives v = 3.0103 m/s.
Chapter 24
6. A 3.0µF and 4.0 µF capacitor are connected on series and this combination is connected
in parallel with a 2.00 µF capacitor (see Figure below). (a) What is the net capacitance?
(b) If 26.0 V is applied across the whole network, calculate the voltage across each
capacitor.
–
– 3Q
(a) From the circuit, we see that C2 and C3 are in series
and find their equivalent capacitance from
1/C4 = (1/C2) + (1/C3);
1/C4 = (1/3.00 F)+ (1/4.00 F), which gives C4 = 1.71 F.
From the new circuit, we see that C1 and C4 are in parallel,
with an equivalent capacitance
Ceq = C1 + C4 = 2.00 F + 1.71 F = 3.71 F.
(b) Because Vab is across C1 , we have
Vab = V1 = 26.0 V.
For the series combination we have
Q2 = Q3 = Q4 = C4Vab = (1.71 F)(26.0 V) = 44.6 C.
We find the other voltages from
V2 = Q2/C2 = (44.6 C)/(3.00 F) =14.9 V;
V3 = Q3/C3 = (44.6 C)/(4.00 F) =11.1 V.
As a check we see that V2 + V3 = Vab .
C2
C3
c
b
a
C1
V ab
C4
a
b
C1
V ab