Çankaya University
Department of Computer Engineering
2009 - 2010 Spring Semester
PHYS 122 - General Physics II
First Midterm Examination
1) Three charges are fixed at the corners of a square of side length L. We want the net force
on charge q to be zero. What is the charge we have to put on the fourth corner?
q
4q
4q
L
Answer:
Let’s call the fourth charge Q. The force between q and Q is:
→
−
F =
"
1
Qq
√
4πε0 (L 2)2
√
!#
√
2→
2→
−
−
i +
j
=0
2
2
The net force on q must be zero:
"
→
−
4q 2 →
4q 2 →
Qq
1
−
−
i
+
j + 2
F net =
2
2
4πε0 L
L
2L
√
4q 2
2Qq
+
=0
2
L
4L2
√
Q = −8 2q
√
!#
√
2→
2→
−
−
i +
j
=0
2
2
2) A total charge Q is distributed uniformly on a rod of length L. Find the electric field at
the point P in unit vector notation. (Do not evaluate the integrals)
P
b
+ + + + + + + + + + +
3L
4
+ + + +
L
4
Answer:
dq =
dE =
Q
dx
L
1
dq
2
3L
4πε0 b2 +
−x
4
dEx = dE sin θ,
dEy = dE cos θ
Z L
3L
−x
Q
4
Ex =
h
2 i3/2 dx
4πε0 L 0
3L
2
b + 4 −x
Z L
b
Q
Ey =
h
2 i3/2 dx
4πε0 L 0
3L
2
b + 4 −x
or alternatively
Q
Ex =
4πε0 L
Z
Q
Ey =
4πε0 L
Z
L
4
− 3L
4
x
[b2
L
4
− 3L
4
+ x2 ]3/2
b
[b2
→
−
→
−
→
−
E = Ex i + Ey j
+ x2 ]3/2
dx
dx
3) A non-conducting spherical shell has inner radius a and outer radius b. It has uniform
volume charge density ρ. Find the electric field for
a) r < a
b) a < r < b
c) b < r
ρ
b
O
a
Answer:
Using Gauss’ Law:
a) 4πr2 E =
qenc
ε0
E=0
b) 4πr2 E =
E=
2
− 43 πa3 ρ
ε0
(r3 − a3 )ρ
3r2 ε0
c) 4πr E =
E=
4
πr3
3
4
πb3
3
(b3 − a3 )ρ
3r2 ε0
− 34 πa3 ρ
ε0
4) Find the work we have to do to assemble the system shown in figure, bringing each charge
from infinity:
b
2b
q
2q
b
3q
4q
Answer:
1
U=
4πε0
U=
q 2q q 3q q 4q 2q 3q 2q 4q 3q 4q
+
+
+
+
+
b
3b
4b
2b
3b
b
65 q 2
3 4πε0 b
5) On a certain region of space, the electric potential is given by V (x, y, z) =
Find the electric field.
Answer:
−
→
−
∂V →
∂V →
∂V →
−
−
i −
j −
k
E =−
∂x
∂y
∂z
→
−
E =
x
yz 2 − p
x2 + y 2
!
→
−
i +
xz 2 − p
y
x2 + y 2
!
→
−
→
−
j + 2xyz k
p
x2 + y 2 −xyz 2 .
Çankaya University
Department of Computer Engineering
2009 - 2010 Spring Semester
PHYS 122 - General Physics II
Second Midterm Examination
1) a) A capacitor stores an energy of 1µ j when the potential difference between its plates is
3 V . Find its capacitance.
b) Find the charge on the 4 pF capacitor in the figure.
1 pF
4 pF
60 V
Answer:
a)
1
U = CV 2
2
C=
12µj
2U
=
2
V
(3V )2
C = 0.22µF = 2.2. × 10−7 F
b)
1
1
1
1
1
=
+
=
+
Ceq
C1 C2
1 pF
4 pF
Ceq = 0.8 pF
q = V C = 60V 0.8pF = 48 pC
q1 = q2 = 4.8 × 10−11 C
2) A cylindrical wire made of Nichrome has radius r = 1 mm and length L. Nichrome’s
resistivity is ρ = 5 × 10−7 Ω · m. When a potential difference of 100 V is applied, the
power dissipation is 2000 W . Find L.
Answer:
P =
(100V )2
V2
=
= 5Ω
R
2000W
R=
ρL
A
L=
π(10−3 m)2 5Ω
= 10π = 31.4 m
5 10−7 Ω m
⇒
L=
AR
ρ
3) In the following circuit,
a) Find all currents
b) Find Vab .
4Ω
8Ω
b
1Ω
3Ω
2Ω
7V
a
80V
60V
Answer:
i1 + i2
i2
i1
80 − 7(i1 + i2 ) − i1 − 7 = 0
7 + i1 − 10i2 + 60 = 0
8i1 + 7i2 = 73
−i1 + 10i2 = 67
The solution of the system of equations give:
i2 = 7 A,
i1 = 3 A,
i1 + i2 = 10 A
Vab = +7 + 3 − 56 = −46V
4) A charged particle with mass m, charge q, kinetic energy K enters a magnetic field and
then follows a circle of radius r as seen in the figure. Find the magnitude and direction
→
−
of the magnetic field B .
r
→
−
B
m, q
Answer:
1
K = mv 2
2
r=
⇒
v=
p
2K/m
mv
qB
p
m 2K/m
mv
B=
=
qr
qr
√
2Km
B=
qr
→
−
Using the right hand rule, we see that B is into the page.
5) A current of I flows around a square of side length a. Find the magnitude of the magnetic
field at the center of the square.
Answer:
I
a
C
a
2
Using Biot-Savart law for one side of the square, we obtain the magnetic field at C as:
Z
π/4
B=
−π/4
µ0 I cos θ dθ
4π(a/2)
π/4
µ0 I
B=
sin θ
2πa
−π/4
√
µ0 I 2
B=
2πa
If we add the magnetic fields of all four sides:
√ µ0 I
Btotal = 2 2
πa
Çankaya University
Department of Computer Engineering
2009 - 2010 Spring Semester
PHYS 122 - General Physics II
Final Examination
1) Find all currents in the following circuit:
8Ω
8Ω
4Ω
17Ω
40V
5Ω
15V
10Ω
10V
47V
6Ω
Answer: Let’s choose the currents as follows:
i1
i2
i1 + i2
Now, using Kirchoff’s loop rule, we obtain:
+10 − 10i2 + 40 − 8i2 − 8i2 + 17i1 − 15 = 0
+15 − 17i1 − 4(i1 + i2 ) − 5(i1 + i2 ) + 47 − 6(i1 + i2 ) = 0
−17i1 + 26i2 = 35
32i1 + 15i2 = 62
The solution of this system of equations give:
i1 = 1A,
i2 = 2A,
i1 + i2 = 3A
2) a) The time constant of an RC circuit is τ = RC = 0.05 s. Find the time necessary for
initially uncharged capacitor to be charged to 99% of its final charge.
Answer:
q = CE(1 − e−t/RC )
Final charge is CE therefore q = 0.99 CE.
0.99 = 1 − e−t/τ
e−t/τ = 0.01
−t/τ = ln 0.01
t = 0.05s. × 4.6 = 0.23s.
b) A particle of mass m, charge q travels in a circular path of radius R in a uniform
magnetic field B. Find its period of revolution.
Answer:
r=
mv
qB
T =
2πr
v
T =
2πm
qB
3) A variable magnetic field of magnitude B = B0 (1 + t3 ) is perpendicular to a rectangular
loop. The loop has dimensions a × b and resistance R. Find the magnitude and direction
of the induced current.
R
→
−
B
a
Answer:
ΦB = BA = Bab = B0 (1 + t3 )ab
dΦB
= −B0 3t2 ab
dt
3B0 t2 ab
E
=−
I=
R
R
Positive direction is counter-clockwise.
Therefore I is in clockwise direction.
E =−
b
4) a) In a series RLC circuit, L = 90 mH, C = 1.5 µF and f = 0.8 kHz. If the phase
constant is 55◦ , what is the resistance of the coil?
Answer:
ω = 2πf = 5027s−1
XL = ωL = 452 Ω
1
XC =
= 132 Ω
ωC
X L − XC
= tan 55 = 1.428
R
XL − X C
R=
= 224 Ω
tan φ
b) An oscillating LC circuit is made of a 4 nF capacitor and a 9 mH inductor. The
maximum current is 3 A. What is the maximum charge on the capacitor?
2
1 2
1 qmax
LImax =
2
2 C
√
qmax = LC I
√
qmax = 9 · 10−3 · 4 · 10−9 3
qmax = 18 µC
5) a) The displacement current through a rectangular loop of area 3.5 m2 is 8 A. At what
rate is the electric field through the loop changing?
id = ε0
d(EA)
dE
dΦE
= ε0
= ε0 A
dt
dt
dt
dE
id
8
=
=
dt
ε0 A
8.85 · 10−12 · 3.5
dE
= 2.58 × 1011 V /m.s
dt
b) The maximum electric field of an electromagnetic wave is 5 V /m. What is the
intensity?
I=
2
Erms
E2
= m
cµ0
2cµ0
52
I=
2 · 3 · 108 · 4π · 10−7
I = 0.033 W/m2
Prepared by: Dr. Emre Sermutlu
(2010)