Download TEST 2 Giving or receiving aid in any examination is cause for

TEST 2
Giving or receiving aid in any examination is cause for dismissal from the
university. Perform the necessary calculation in the spaces provided. If additional
space is required, use the backs of the questions sheets. All work must be shown in
order to receive FULL credit. Work must be legible and comprehensible, and
answers should be clearly indicated.
1. A parallel plate capacitor is constructed by filling the space between two square
plates with blocks of three dielectric materials, as shown in figure below. You
may assume the space that l>> d.
(a)
C1 =
κ 1 ∈0 A 2
d
⎛ 1
1⎞
⎜⎝ C + C ⎟⎠
2
3
−1
=
; C2 =
κ 2 ∈0 A 2
d2
; C3 =
κ 3 ∈0 A 2
d2
∈ A⎛ κ κ ⎞
C 2C 3
= 0 ⎜ 2 3 ⎟
C2 + C3
d ⎝κ2 +κ3 ⎠
⎛ 1
1⎞
C = C1 + ⎜
+
⎝ C 2 C 3 ⎟⎠
−1
=
κ κ ⎞
∈0 A ⎛ κ 1
+ 2 3 ⎟
⎜
d ⎝ 2 κ2 +κ 3 ⎠
a) Find expression for a capacitance of the device in terms of the area A and d, k1
and k2 and k3.
b) Calculate the capacitance using the values A=1.00 cm2, d=2.00 mm, k1=4.9,
k2=5.60 and k3=2.10
Using the given values one can find: C total = 1.76 × 10−12 F = 1.76 pF .
2. Find the equivalent capacitance between points a and b in the combination of
capacitors shown in figure below
FIG. 1
1 ⎞
⎛ 1
PC s = ⎜⎝ 5.00 + 7.00⎟⎠
−1
= 2.92 μF
2C = 2.92 + 4.00 + 6.00 = 12.9 μF
p
6
.
2
9
FIG. 2
3. Consider the circuit shown in figure below, C1 = 6.00 µF and C2 = 3.00 µF and
∆V=20.0 V. Capacitor C1 is first charged by closing switch S1. Switch S1 is then opened
and the charged capacitor is connected to the uncharged capacitor C2= 3.00 µF by the
closing the switch S2.
a) Calculate the initial charge on C1 = 6.00 µF
b) Calculate the final charge on C2 = 3.00 µF
c) Calculate the final charge on C1 = 6.00 µF
Q
so
ΔV
and Q = 120 μ C
C=
6.00 × 10−6 =
Q
20.0
Q 1 = 120 μ C − Q 2
and
or
120 − Q 2 Q 2
=
C1
C2
120 − Q 2
Q
= 2
6.00
3.00
ΔV =
Q
C
:
FIG. 3
( 3.00) (120 − Q 2 ) = ( 6.00) Q 2
360
= 40.0 μ C
9.00
Q 1 = 120 μ C − 40.0 μ C = 80.0 μ C
Q2 =
4. Consider a series RC circuit, shown in figure below, for which R=1.00MΩ, C= 5.00
µF and ε=30.0 V.
Find
a) The time constant of the circuit
b) The maximum charge on a capacitor after the switch is closed
c) Find the current in resistor 10 s after the switch was closed
(
)(
)
(a)
RC = 1.00 × 106 Ω 5.00 × 10−6 F = 5.00 s
(b)
Q = C ε = 5.00 × 10−6 C ( 30.0 V ) = 150 μ C
(c)
(
I( t) =
)
⎡
−10.0
⎛ 30.0 ⎞
⎢
e− t RC = ⎜
exp
⎟
⎝ 1.00 × 106 ⎠
⎢ 1.00 × 106 5.00 × 10−6
R
⎣
ε
(
)(
)
⎤
⎥ = 4.06 μ A
⎥
⎦
FIG. 4
5. What is escape speed of an electron launched from the surface of a 1.0 cm diameter
plastic sphere that has been charged to 10 nC
Energy is conserved. The electron ends up so far away from the plastic sphere that we
can consider its potential energy to be zero.
The charge of the plastic sphere must be positive. The minimum speed to escape is the
speed that allows the electron to reach rf = ∞ when vf = 0 m/s. The conservation of energy
equation Kf + Uf = Ki + Ui is
⎛ 1 q⎞
0 J + 0 J = 21 mvi2 + qVi ⇒ 0 J = 21 mvi2 + ( −e ) ⎜
⎟
⎝ 4πε 0 R ⎠
⇒ vi =
2e 1 q
=
m 4πε 0 R
(
2 1.60 × 10 −19 C
9.11 × 10
−31
kg
)
( 9.0 × 10
9
⎛ 10 × 10 −9 C ⎞
7
N m 2 /C2 ⎜
⎟ = 7.95 × 10 m/s
−2
⎝ 0.5 × 10 m ⎠
)
6. The four identical 1.0 g spheres with charge q=10x10-9C are released simultaneously
from rest at the vertices of a square of side L = 1 cm and allowed to move away from
each other.
a) How fast is each particle moving when their distance from the center is doubled?
Each charge moves off on its diagonal line. All charges have equal speeds.
∑ ( K + U )i = ∑ ( K + U ) f
0+
2
2
4keq2 2keq2
⎛1
⎞ 4k q 2k q
+
= 4⎜ m v2 ⎟ + e + e
⎝2
⎠
2L
L
2L
2 2L
1 ⎞ keq2
⎛
2
+
= 2m v2
⎜⎝
⎟
2⎠ L
1 ⎞ keq2
⎛
v = ⎜ 1+
⎝
8 ⎟⎠ m L
b) What is the speed of each sphere when they are very far apart?
The total potential energy of the four spheres before they are allowed to move is
Ui = (U12 + U23 + U34 + U41) + (U13 + U24)
All the charges are identical,
potential energy is
(
r12 = r23 = r34 = r41 = L = 1 cm, and r13 = r24 = 2 cm .
)(
)
(
)(
The initial
)
⎛ 1 10 × 10 −9 C 10 × 10 −9 C ⎞
⎛
10 × 10 −9 C 10 × 10 −9 C ⎞
⎟ + 2⎜ 1
⎟ = 48.73 × 10 −5 J
Ui = 4 ⎜
⎜ 4πε 0
⎟
⎜ 4πε 0
⎟
0.01 m
0.01414 m
⎝
⎠
⎝
⎠
Since all charges are at rest, Ki = 0 J. As the spheres are allowed to move away from one
another and they are far apart, Uf = 0 J. The final kinetic energy is
Kf = 4
(
1
2
) (
)
mvf2 = 2 1.0 × 10 −3 kg vf2
From the energy conservation equation Uf + Kf = Ui + Ki,
(
)
2 1.0 × 10 −3 kg vf2 = 48.73 × 10 −5 J ⇒ vf = 0.494 m/s
7. If the current carried by conductor is doubled, what happens to the
(a)
Charge carrier density? n is unaffected
(b)
Current density? J =
I
∝I
A
So current doubles .
(c)
Electron drift velocity? J = nevd
So vd also doubles .
(d)
The average time interval between the collisions? τ =
unchanged
mσ
nq2
is
as long as σ does not change due to a temperature change
in the conductor.
1.
Charge of uniform density (3.5 nC/m) is distributed along the circular arc
shown. Determine the electric potential (relative to zero at infinity) at point P.
+
+
+
+
+
+
+
+
+
+
+
a.
b.
c.
d.
e.
R
60°
60
P
R
61 V
42 V
52 V
33 V
22 V
d
2.
The circuit below contains three 100 watt light bulbs. The emf E = 110 V. Which
light bulb(s) is(are) the brightest?
A
+
E
–
B
a.
b.
c.
d.
e.
a.
C
A
B
C
B and C
All three are equally bright.
3.
The circuit below contains 5 light bulbs. The emf is 110 V. Which light bulb(s)
is(are) brightest?
A
C
110 V
E
D
B
a.
b.
c.
d.
e.
A: The one closest to the positive terminal of the battery.
A and C: The bulbs closest to the positive terminal of the battery.
A and B: Because they are closest to the terminals of the battery.
C and D: Because they receive current from A and B and from E.
E: Because the potential difference across E is that of the battery.
e.
4.
If I = 2.0 A in the circuit segment shown below, what is the potential difference
VB – VA?
A
10 Ω
I
+
+
–
20 V
a.
b.
c.
d.
e.
–
B
10 V
+10 V
–20 V
–10 V
+20 V
+30 V
b.
5.
At what rate is thermal energy generated in the 30-Ω resistor?
10 Ω
+
30 V
30 Ω
–
5.0 Ω
a.
b.
c.
d.
e.
d.
20 W
27 W
60 W
13 W
30 W
5.0 Ω
6.
Which of the following represents the equipotential lines of a dipole?
(c)
(b)
(a)
(d)
(e)
e.
7.
If a = 30 cm, b = 20 cm, q = +2.0 nC, and Q = –3.0 nC in the figure, what is the
potential difference, V A − V B ?
a
A
q
a.
b.
c.
d.
e.
b
a
B
Q
+60 V
+72 V
+84 V
+96 V
+48 V
a
8. Extra: Determine the equivalent capacitance of the combination shown in figure below.
(Suggestion: consider the symmetry involved)
a.
b.
c.
d.
e.
c.
5C
6C
4
C
3
3
C
4
2C