Download PHYS 112 - General Physics for Engineering II FIRST MIDTERM

Çankaya University
Department of Mathematics and Computer Sciences
2010-2011 Spring Semester
PHYS 112 - General Physics for Engineering II
FIRST MIDTERM
1) Two fixed particles of charges q1 = 1.0µC and q2 = −3.0µC are 10 cm apart. Where
should a third charge be located so that no net electrostatic force acts on it?
Explain clearly your answer.
2) A uniformly charged plastic disk of radius R carries surface charge density σ. At what
distance z along the central axis is the magnitude of the electric field equal to one-half
the magnitude of the field at the center of the surface of the disk?
Explain clearly your answer.
3) a) A conducting sphere of radius 10 cm has an unknown charge. If the electric field 15
cm from the center of the sphere has the magnitude 3.0 × 103 N/C and is directed radially
inward, what is the net charge on the sphere?
b) Charge is distributed uniformly throughout the volume of an infinitely long
cylinder of radius R. The volume charge density is ρ. Calculate the electric field at a
distance r from the cylinder axis for r < R.
Explain clearly your answer.
4) A rod of length L lies on x−axis as shown in figure. Its linear charge density is λ. Find
the electric potential at the point (0, Y ) as an integral. Do not evaluate the integral.
Explain clearly your answer.
y
(0, Y )
λ
+++++++++
−
L
4
3L
4
x
5) In the following circuit, C1 = 1µF, C2 = 6µF, C3 = 4µF, C4 = 4µF and the potential
supplied by the battery is V = 60V .
a) Find the energy stored on C1
b) A dielectric material with dielectric constant κ = 6 is inserted between the plates of
C1 . Find the charge stored on C1 after insertion. (Battery is connected during and after
insertion)
C1
C2
C3
C4
V
Explain clearly your answer.
Answers
1) Consider a line that connects q1 and q2 which are d = 10cm apart. Clearly, the third
particle must be placed on this line to the left of q1 a distance x. At this point, the
electric field will be:
E=−
−
1 q1
q2
1
−
=0
2
4πε0 x
4πε0 (x + d)2
x2
q1
=
q2
(x + d)2
1
x
√ =
x + 10
3
⇒
⇒
x2
1
=
3
(x + d)2
10
x= √
cm = 13.66 cm
3−1
2) Electric field at the surface can be found by inserting z = 0 in the formula:
1
E(z) = E(0)
2
σ
z
σ
1− √
=
2ε0
4ε0
z 2 + R2
z
1
1− √
=
2
2
2
z +R
√
1
z
=
2
z 2 + R2
4z 2 = z 2 + R2
R
z=√
3
3) a) Outside the sphere, the charges act as if they are concentrated at the center.
E=
Q
4πε0 r2
⇒
Q = 4πε0 r2 E
Q = −3 × 103 N/C × 4π(0.15m)2 × 8.85 × 10−12 C 2 /N m2
Q = −7.5 × 10−9 C = −7.5nC
b)
qenc
ε0
πr2 Lρ
2πrLE =
ε0
rρ
E =
2ε0
Φ =
dq
√
4) dq = λ dx,
dV =
,
4πε0 x2 + Y 2
Z 3L/4
λ
dx
√
V =
4πε0 −L/4 x2 + Y 2
5) a) C13 = 5µF, C24 = 10µF
⇒
⇒
V1 = 30V
q1 = V1 C1 = 30V 6µF = 180µC
V =
5V1 = 10V2
U = 12 CV 2 = 21 1µF (40V )2 = 800µJ
b) C1 = 6µF
Z
dV
⇒
V1 = 40V
Çankaya University
Department of Mathematics and Computer Sciences
2010-2011 Spring Semester
PHYS 112 - General Physics for Engineering II
SECOND MIDTERM
1) a) (10 pts) A fuse in an electric circuit is a wire that is designed to melt, and thereby to
open the circuit, if the current exceeds a predetermined value. Suppose that the material
to be used in a fuse melts when the current density rises to 600 A/cm2 . What diameter
of cylindrical wire should be used to make a fuse that will limit the current to 1.2 A?
Explain clearly your answer.
b) (10 pts) When 230 V is applied across a wire that is 10 m long and has a 0.4 mm
radius, the current density is 6.3 × 104 A/m2 . Find the resistivity of the wire.
Explain clearly your answer.
2) Calculate the current through each ideal battery in Figure.
Explain clearly your answer.
5Ω
2Ω
6Ω
111V
20V
18V
4Ω
12Ω
3) a) (10 pts) A capacitor with capacitance C = 2µF and initial charge q0 is discharged
2
through a resistor of resistance R = 2Ω. After how much time will the charge be q0 ?
3
Explain clearly your answer.
b) (10 pts) A charged particle with mass m, charge q travels in a circular path in a
uniform magnetic field with magnitude B. Find the frequency of motion.
Explain clearly your answer.
4) a) (10 pts) A wire 2.5 m long carries a current of 8 A and makes an angle of 25◦ with
a uniform magnetic field of magnitude B = 0.7 T. Calculate the magnetic force on the
wire.
Explain clearly your answer.
b) (15 pts) A circular coil of 150 turns has a radius of 3 cm. Calculate the current
that results in a magnetic dipole moment of 5A.m2 .
Explain clearly your answer.
5) a) (15 pts) A square loop of wire of edge length a carries current i. Calculate the magnitude of the magnetic field at the center of the loop.
Explain clearly your answer.
I
→
− →
B · d−
s for the path shown in figure.
b) (10 pts) Evaluate
Explain clearly your answer.
× 2I
· 3I
· I
· 3I
× I
Answers
i
i
1) a) J = = 2
A
πr
r
r
i
i
r=
⇒ R=2
= 0.05cm = 0.5mm
πJ
πJ
ρL
= JρL
b) V = iR = i
A
V
ρ=
= 3.65 × 10−4 Ω.m
JL
2) Let i1 be the current through 18 V battery and i2 through the 20 V battery. Then, a
current of i1 + i2 passes through the 111 V battery.
111 − 5(i1 + i2 ) − 6i1 − 18 − 4(i1 + i2 ) = 0
−2i2 − 20 − 12i2 + 18 + 6i1 = 0
⇒
i1 = 5 A,
i2 = 2 A,
i1 + i2 = 7 A
2
2
t
q0 = q0 e−t/RC ⇒ ln = −
3
3
RC
3
−6
t = RC ln = 1.62µs = 1.62 × 10 s
2
mv
v
qB
b) r =
, f=
=
qB
2πr
2πm
3) a)
4) a) F = BiL sin θ = 5.92N
b) µ = N iA
⇒
i=
5A.m2
= 11.8A
150π(0.03m)2
√
√
µ0 I 2
2 2µ0 I
5) a) B =
·4=
2πa
πa
I
−
→
− →
b)
B × ds = 3µ0 I
Çankaya University
Department of Mathematics and Computer Sciences
2010-2011 Spring Semester
PHYS 112 - General Physics for Engineering II
FINAL
1) A parallel-plate capacitor has a capacitance of 100 pF, a plate area of 100cm2 and a mica
dielectric (κ = 5.4) filling the space between the plates. At 50 V potential difference,
calculate
a) (6 Pts.) the electric field magnitude E in the mica
b) (7 Pts.) the magnitude of the free charge on the plates
c) (7 Pts.) the magnitude of the induced surface charge on the mica.
Explain clearly your answer.
2) (20 Pts.) In the given circuit, determine the value of R that maximizes the energy
dissipation rate (power) at R.
Explain clearly your answer.
24V
3Ω
R
5Ω
3) a) (13 pts.) An AC generator is to be connected in series with an inductor of L = 2 mH
and a capacitance C. You are to produce C by using capacitors of capacitance C1 = 4 µF
and C2 = 6 µF either singly or together. What resonant frequencies can the circuit have,
depending of how you use C1 and C2 ?
b) (7 pts.) Average power in an AC circuit is 60W. The maximum value of current
is 15 A and the maximum value of emf is 20V. Find the phase constant.
Explain clearly your answer.
4) A circular loop of radius 2 cm is placed in a uniform magnetic field as seen in figure. The
magnetic field is changed uniformly from 0.2 T to 0.8 T in a time interval of 1 s, beginning
at t = 0.
a) (5 pts.) Find the magnitude of magnetic field t = 0, t = 0.5 and t = 1 s.
b) (10 pts.) What emf is induced in the loop at t = 0.5 s?
c) (10 pts.) Find the direction of induced current in the loop.
Explain clearly your answer.
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
−
×→
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
5) a) (10 Pts.) At what rate must the potential difference between the plates of a parallelplate capacitor with a 2 µF capacitance be changed to produce a displacement current of
1.5 A?
b) (8 Pts.) Calculate the intensity of a plane traveling electromagnetic wave if
Bm is 2.0 × 10−4 T.
c) (7 Pts.) 3G cellphones use electromagnetic waves of frequency 2100 MHz. What
is the corresponding wavelength?
Explain clearly your answer.
Answers
1)
V
d
κε0 A
A
⇒ d=
C = κε0
D
C
VC
E=
= 1 × 104 V/m
κε0 A
a) E =
b) q = CV = 5 × 10−9 C
c) qi = q − ε0 AE = 4.1 × 10−9 C=4.1 nC
2) 24 − 3i − 5i − Ri = 0
⇒
i=
P =i R
⇒
242
P =
R
(8 + R)2
dP
=0
dR
⇒
R=8Ω
2
3) a) f1 =
f2 =
f3 =
f4 =
24
8+R
1
√
= 1.78 × 103 Hz
2π LC1
1
√
= 1.45 × 103 Hz
2π LC2
1
2π
p
2π
p
LC1 C2 /(C1 + C2 )
1
L(C1 + C2 )
= 1.13 × 103 Hz
b) Pavg = Erms Irms cos φ
cos φ =
4
5
= 2.3 × 103 Hz
4) a) B = 0.2 + 0.6t
⇒ B(0) = 0.2T, B(0.5) = 0.5T, B(1) = 0.8T.
dΦB
d
dB
= − (BA) = − A
dt
dt
dt
−2
2
= − π(2 × 10 m) 0.6T /s = −7.54 × 10−4 V
b) E = −
c) The magnetic field is into the page and increasing. By Lenz’s law, induced current
must decrease it. So the direction of the induced current is counter clock-wise.
dΦE
dt
dEA
= E0
dt
E0 A dV
=
d dt
dV
= C
dt
id
1.5 A
dV
=
=
= 7.5 × 105 V /s
dt
C
2 × 10−6 F
5) a) id = E0
b) I =
=
=
=
=
c) λ =
1
E2
cµ0 rms
2
1 Em
cµ0 2
2
c Bm
µ0 2
3 × 108 m/s (2 × 10−4 T )2
4π × 10−7 T.m/a
2
6
2
4.777 × 10 W/m
3 × 108 m/s
c
=
= 0.14m
f
2100 × 106 s−1