Second Major-123
Monday, July 15, 2013
Coordinator: xyz
Zero Version
Page: 1
Q1.
Figure 1 shows three charges +q,–q and Q along with net force F on charge – q. If
 = 45 , the value of charge Q is:
Figure 1
A)
B)
C)
D)
E)
+ 4q
– 4q
– 3q
+ 3q
+ 2q
Ans:
⃗2
F
⃗1
F
Q must be (+).
⃗ will be in the second quadrant.
Otherwise, F
Since θ = 45° ∶ F1 = F2
k. q. Q
k. q. q
=
⇒ Q = +4q
4d2
d2
Q2.
Four identical metal spheres (A, B, C, D) have charges of qA = – 8.0 C, qB = – 2.0
C, qC = + 5.0 C, and qD = + 12.0 C. Then, three of the metal spheres are brought
together so that they touch each other simultaneously, and then they are separated.
Which of the three spheres were touched together, if the final charge on each of the
three spheres is + 3.0 C?
A)
B)
C)
D)
E)
A, C, D
A, B, C
B, C, D
A, B, D
none of the other answers
Ans:
Qf = 3 × 3.0 = +9.0 μC
∴ Qi = +9.0 μC ← charge conservation
A + B + C = −5.0 μC 
A + C + D = +9.0 μC 
B + C + D = +15 μC 
A + B + D = +2 μC 
Phys102
Coordinator: xyz
Q3.
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 2
Three charges ( – q, – q and Q ), with q = 2.5 C, are located at equal distance d from
the origin, as shown in Figure 2. If the resultant electric field at P1 due to the three
charges is zero, what are the magnitude and sign of charge Q?
𝐘
1 +Q
Figure 2
A)
B)
C)
D)
E)
+ 7.1 C
+ 1.7 C
– 7.1 C
– 1.7 C
+ 2.5 C
𝐝
2
𝐙
3
o
𝐝
−q
⃗2
E
𝐝
𝐝
𝛉
−q
𝐗
⃗E3
𝛉
Ans:
P1
Q must be (+) for the electric field to be zero at point P.
⃗E1
E1 = E2y + E3y = 2E2y (since E2 = E3 )
k. Q
k. q √2
=
2
.
⇒ Q = +2√2 q = +7.1 μC
4d2
2d2 2
Q4.
An electron enters a region of uniform electric field with an initial velocity of 4.2
km/s in the same direction as the electric field, which has a magnitude of 12 N/C.
What is the speed of the electron 2.0 ns after entering this region?
A) 0.0 km/s
B) 2.2 km/s
C) 1.3 km/s
D) 3.3 km/s
E) 1.7 km/s
⃗i
v
a⃗
⃗
E
⃗
E
⃗E
Ans:
a=
qE
eE
=
m
m
vf = vi − at
= 4.2 × 103 −
1.6 × 10−19 × 12 × 2.0 × 10−19
9.11 × 10−31
= 4.2 × 103 − 4.2 × 103 = 0.0 × 103 m/s = 0.0 km/s
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 3
Q5.
A torque of magnitude 0.1 N.m. has been applied to orient an electric dipole at a
particular angle with respect to a uniform electric field. When the dipole moment is
oriented along the field, the electric potential energy of the dipole is – 0.2 J. What was
the initial angle between the dipole moment and the electric field?
A)
B)
C)
D)
E)
30°
90°
45°
10°
0
Ans:
⃗ : θ = 0 ⇒ U = −p
⃗ = −p
⃗ = −p. E = −0.2 J
when p
⃗ // E
⃗ .E
⃗ .E
∴ pE = 0.2 J
τ = p. E. sinθ ⇒ sinθ =
τ
0.1
=
= 0.5 ⇒ θ = 30°
p. E 0.2
Q6.
Figure 3 shows three Gaussian surfaces A, B and C, with corresponding electric flux
A = – q/0 , B = + 3q/0 and C = –2q/0 through them, respectively. What is the
value of the charge q1?
Figure 3
A)
B)
C)
D)
E)
+2q
+q
−3q
+3q
−2q
Ans:
A = −
q
1
(q + q 3 ) ⇒ q1 + q 3 = −q ⟶ (1)
=
ε0
ε0 1
B = +
3q
1
(q + q 2 ) ⇒ q1 + q 2 = +3q ⟶ (2)
=
ε0
ε0 1
C = −
2q
1
(q + q 3 ) ⇒ q 2 + q 3 = −2q ⟶ (3)
=
ε0
ε0 2
From (1): q 3 = −q − q1
From (2): q 2 = +3q − q1
From (3) : + 3q − q1 − q − q1 = −2q
4q = 2q1
∴ q1 = +2q
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 4
Q7.
A charged conducting spherical shell has an inner radius of 6.0 cm and an
outer radius of 10 cm. A point charge is placed at the center of the shell such
that the resulting surface charge densities on the inner and outer surfaces of
the shell are –100 nC/m2 and +100 nC/m2, respectively. What is the electric
field at a distance of 12 cm from the center of the shell?
A)
B)
C)
D)
E)
7.9 × 103 N/C , outward
7.9 × 103 N/C , inward
9.7 × 103 N/C , outward
9.7 × 103 N/C , inward
5.3 × 103 N/C , outward
Ans:
r = 12 cm is outside the shell
∴ ⃗⃗⃗
E is supplied by q out ⇒ outward
kq out
k
= 2 . (σout . Aout )
2
r
r
k
= 2 . (σout . 4πR out 2 ) = 7.9 × 103 N/C
r
E=
Q8.
An electron experiences a force of magnitude F when it is 2 cm away from a
very long, charged wire that has a uniform linear charge density +. If the
linear charge density is increased to +2, at what distance from the wire will
the electron experience a force of the same magnitude F?
A)
B)
C)
D)
E)
4 cm
1 cm
3 cm
2 cm
6 cm
Ans:
E=
2kλ
r
F = qE =
∴r =
2kqλ
r
2kqλ
F
If λ is doubled but F is the same, then r is doubled.
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 5
Q9.
Figure 4 shows cross sections through two large parallel non-conducting
sheets with surface charge densities σ1 = – 1.8 µC/m2 and σ2 = + 1.2 µC/m2.
What is the electric field at point P (in units of 104 N/C)?
A) – 3.4 î
B) – 6.8 î
C) + 3.4 î
D) + 6.8 î
Figure 4
⃗1
E
⃗2
E
E) + 1.7 î
Ans:
σ1
(−î)
2ε0
σ
⃗ 2 = 2 (−î)
E
2ε0
∴ ⃗Enet = ⃗E1 + ⃗E2
⃗E1 =
σ2 − σ1
= (
) î
2ε0
1.2 × 10−6 − 1.8 × 10−6
= (
) î
2 × 8.85 × 10−12
= −3.4 × 104 î (N/C)
Q10.
A uniformly charged solid insulating sphere has a radius of 5.0 cm. If the
magnitude of the electric field due to this sphere at r = 8.0 cm is 2.0 × 105
N/C, what is the magnitude of the field at r = 3.0 cm? [r is the distance from
the center of the sphere]
A)
B)
C)
D)
E)
3.1 × 105 N/C
1.8 × 105 N/C
9.0 × 104 N/C
2.7 × 105 N/C
7.2 × 105 N/C
Ans:
R = 5.0 cm, r0 = 8.0 cm, ri = 3.0 cm
E0 =
kq
⇒ kq = E0 . r0 2
r0 2
Ei =
kq
E0 . r0 2 . ri
2.0 × 105 × 64 × 3.0
r
=
=
= 3.1 × 105 N/C
R3 i
R3
125
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 6
Q11.
In Figure 5, seven charged particles are fixed in place to form a square with
an edge length of 5.0 cm. How much work must we do to bring a particle of
charge + 5e initially at rest from an infinite distance to the center of the
square?
A)
B)
C)
D)
E)
+ 1.4 × 10-25 J
– 1.4 × 10-25 J
– 2.8 × 10-25 J
0
+ 2.8 × 10-25 J
Ans:
Figure 5
d
The only charge that contributes the
potential at the center of the square is
the +3e charge at the center of the
bottom line.
(k)(+3e)
kq
+3ke
VC =
=
=
r
d
d
+3ke
W = q. Vc = (+5e). (
)
d
+3ke
W = q. Vc = (+5e). (
)
d
15 × 9 × 109 × (1.6 × 10−19 )2
= +
0.025
= + 1.4 × 10−25 J
Q12.
In a certain region of space, the electric field is given by E  0.40 x î (N/C).
If the electric potential at the origin is + 5.0 V, what is the electric potential at
the point (3.0, 0, 0) m?
A)
B)
C)
D)
E)
+ 3.2 V
+ 1.8 V
– 1.8 V
+ 6.8 V
– 6.2 V
Ans:
3
ΔV = − ∫ ⃗E . ds = − ∫ 0.4 xdx = −0.2[x 2 ]30 = −1.8 V
0
V3 − V0 = −1.8 ⇒ V3 = V0 − 1.8 = 5.0 − 1.8 = +3.2 V
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 7
Q13.
Two point charges lie along the x axis. One charge ( q1 ), located at the origin, has a
magnitude of + 2q. The other charge ( q2 ) is located at x = + 5 cm. If the electric
potential, due to the two charges, at x = + 4 cm is equal to zero, what are the
magnitude and sign q2?
A)
B)
C)
D)
E)
 q/2
 q/4
 2q
+ q/2
+ 2q
Ans:
kq1 kq 2
+
=0
4
1
q1
2q
q
⇒ q2 = − = −
= −
4
4
2
V4 =
Q14.
In a certain situation, the electric potential varies along an x axis as shown in Figure
7. For which range of x is the magnitude of the electric field the largest?
A)
B)
C)
D)
E)
from x = 2 to x = 4 
from x = 4 to x = 6 
from x = 0 to x = 2 
from x = 5 to x = 6 
from x = 0 to x = 1 
Figure 7
Ans:
E=
dV
= slope
dx
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 8
Q15.
Two particles have the same charge of +1 µC. Initially, they are held at rest, separated
by a distance d = 1 cm. One of the particles is released and moves away from the
other fixed particle. When the moving particle is a distance of 3d from the other
particle, what is its kinetic energy?
A)
B)
C)
D)
E)
Ans:
0.6 J
0.2 J
0.3 J
0.1 J
0.0 J
0
K i + Ui = K f + U f
K f = Ui − Uf
kq2 kq2 2kq2
2 9 × 109
=
−
=
= ×
× 10−12 = 0.6 J
d
3d
3d
3
0.01
Q16.
A conducting sphere of radius 16 cm has a net charge of 2.0×10-8 C. If the electric
potential is zero at infinity, at what distance from the sphere’s surface has the electric
potential decreased by 500 V from its value on the surface?
A)
B)
C)
D)
E)
13 cm
29 cm
36 cm
11 cm
22 cm
Ans:
Vs =
kq
9 × 109 × 2 × 10−8
=
= 1125 V
R
0.16
Vx = Vs − 500 = 625 V
Vx =
kq
kq 9 × 109 × 2 × 10−8
⇒r=
=
= 0.288 m = 28.8 cm
r
Vx
625
⇒ x = r − R = 28.8 − 16 = 12.8 cm ⇒ 13 cm
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 9
Q17.
The potential difference between the plates of a parallel plate capacitor is 35 V, and
the electric field between the plates has a magnitude of 750 V/m. If the plate area is
400 cm2, what is the capacitance of this capacitor?
A)
B)
C)
D)
E)
7.6 × 10-12 F
7.6 × 10-14 F
7.6 × 10-11 F
7.6 × 10-10 F
None of the other choices is correct
Ans:
V = E. d ⇒ d =
V
35
=
= 0.047 m
E
750
ε0 A
8.85 × 10−12 × 400 × 10−4
C=
=
= 7.6 × 10−12 F
d
0.047
Q18.
Four capacitors, with capacitances C1 = 3.0 F, C2 = 2.0 F, C3 = 5.0 F, and C4 = 4.0
F, are connected in a circuit to a 10 V battery, as shown in Figure 6. How much
energy is stored by the combination?
A)
B)
C)
D)
E)
97 µJ
64 µJ
53 µJ
59 µJ
25 µJ
Figure 6
Ans:
Series: C23 =
C2 C3
10
=
μF
C2 +C3
7
10
38
+4=
μF = 5.4 μF
7
7
C234 × C1 5.4 × 3.0
=
=
= 19.3 μF
C234 +C1
5.4 + 3.0
Parallel: C234 = C23 + C4 =
Ceq
1
C . V2
2 eq
1
= × 1.93 × 10−6 × 100
2
U=
= 9.7 × 10−5 J = 97μJ
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0
Phys102
Coordinator: xyz
Second Major-123
Monday, July 15, 2013
Zero Version
Page: 10
Q19.
When the potential difference between the plates of a parallel-plate capacitor is V, the
energy density in the capacitor is u. If the potential difference is doubled, which of the
following changes would keep the energy density equal to its previous value u?
A)
B)
C)
D)
E)
1
ε
V2
doubling the spacing between the plates  u = 2 ε0 E 2 = 20 . d2
doubling the area of the plates 
reducing the area of the plates by half 
reducing the spacing between the plates 
The energy density is unaffected by a change in the potential difference 
Ans:
A
Q20.
An isolated parallel-plate capacitor stores an energy of 3.4 J. How much work is
required by an external agent to insert a dielectric of dielectric constant κ = 2.8
between the plates of the capacitor?
A)
B)
C)
D)
E)
– 2.2 J
+ 2.2 J
– 1.2 J
+ 1.2 J
– 3.5 J
Ans:
qi2
2C
qi2
Ui
Uf =
=
2kC U
Ui =
1
1
W = ΔU = Uf − Ui = ( − 1) Ui = (
− 1) (3.4) = −2.2 J
k
2.8
King Fahd University of Petroleum and Minerals
Physics Department
c-20-n-20-s-0-e-1-fg-1-fo-0