midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
E & M - Basic Physical Concepts
Current and resistance
Current: I = ddtQ = n q vd A
Ohm’s law: V = I R, E = ρJ
I , R = ρℓ
E = Vℓ , J = A
A
Electric force and electric field
Electric force between 2 point charges:
|q | |q |
|F | = k 1r2 2
k = 8.987551787 × 109 N m2 /C2
ǫ0 = 4 π1 k = 8.854187817 × 10−12 C2 /N m2
qp = −qe = 1.60217733 (49) × 10−19 C
mp = 1.672623 (10) × 10−27 kg
me = 9.1093897 (54) × 10−31 kg
~
~ =F
Electric field: E
2
Power: P = I V = VR = I 2 R
Thermal coefficient of ρ: α = ρ ∆ρ
0 ∆T
Motion of free electrons in an ideal conductor:
a τ = vd → qmE τ = nJq → ρ = n qm2 τ
|Q|
~2 + · · ·
~ =E
~1 + E
Point charge: |E| = k r2 , E
Field patterns: point charge, dipole, k plates, rod,
spheres, cylinders,. . .
Charge distributions:
Linear charge density: λ = ∆Q
∆x
Surface charge density: σsurf =
∆Q
Volume charge density: ρ = ∆V
∆Qsurf
∆A
Electric flux and Gauss’ law
~ · n̂∆A
Flux: ∆Φ = E ∆A⊥ = E
Gauss law: Outgoing Flux from S, ΦS = Qenclosed
ǫ0
Steps: to obtain electric field
~ pattern and construct S
–Inspect E
H
~ · dA
~ = Qencl , solve for E
~
–Find Φs = surf ace E
ǫ
0
Spherical: Φs = 4 π r2 E
Cylindrical: Φs = 2 π r ℓ E
Pill box: Φs = E ∆A, 1 side; = 2 E ∆A, 2 sides
σ
k
~ = 0, Esurf
Conductor: E
= 0, E ⊥ = surf
in
surf
Potential
ǫ0
Potential energy: ∆U = q ∆V 1 eV ≈ 1.6 × 10−19 J
Positive charge moves from high V to low V
Point charge: V = krQ V = V1 + V2 = . . .
1 q2
Energy of a charge-pair: U = k rq12
Potential difference: |∆V | = |E ∆sk |,
R
~ · ∆~s, V − V = − B E
~ · d~s
∆V = −E
B
A
A
¯
¯
d
V
∆V
E = − dr , Ex = − ∆x ¯
= − ∂V
∂x , etc.
f ix y,z
Capacitances
Q=CV
Series: V = CQ = CQ + CQ + CQ + · · ·, Q = Qi
eq
1
2
3
Parallel: Q = Ceq V = C1 V + C2 V + · · ·, V = Vi
ǫ A
Q
Parallel plate-capacitor: C = V
= EQd = 0d
2
RQ
Q
Energy: U = 0 V dq = 12 C , u = 12 ǫ0 E 2
2
1
2
Uκ = 21κ Q
C0 , uκ = 2 ǫ0 κ Eκ
Q
Q
Spherical capacitor: V = 4 π ǫ r1 − 4 π ǫ r2
0
0
~
Potential energy: U = −~
p·E
Dielectrics: C = κC0 ,
V =IR
Direct current circuits
q
Area charge density: σA = ∆Q
∆A
1
Series: V = I Req = I R1 + I R2 + I R3 + · · ·, I = Ii
V + V + V + · · ·, V = V
Parallel: I = RV = R
i
R2
R3
eq
1
Steps: in application of Kirchhoff’s Rules
–Label currents: i1 , i2 , i3 , . . .
P
P
i
–Node equations:
i =
P in
Pout
–Loop equations: “ (±E) + (∓iR)=0”
–Natural: “+” for loop-arrow entering − terminal
“−” for loop-arrow-parallel to current flow
RC circuit: if ddty + R1C y = 0, y = y0 exp(− RtC )
Charging: E − Vc − R i = 0, 1c ddtq + R ddti = ci + R ddti = 0
Discharge: 0 = Vc − R i = qc + R ddtq , ci + R ddti = 0
Magnetic field and magnetic force
µ0 = 4 π × 10−7 T m/A
µ a2 i
µ i
0
Wire: B = 2 π0 r
Axis of loop: B =
2 (a2 +x2 )3/2
~ → q ~v × B
~
Magnetic force: F~M = i ~ℓ × B
~ × B,
~
Loop-magnet ID: ~τ = i A
µ
~ = i A n̂
2
r
Circular motion: F = mrv = q v B, T = f1 = 2 π
v
~ + q ~v × B
~
Lorentz force: F~ = q E
~
Hall effect: V = FM d , U = −~
µ·B
H
q
~ and magnetism of matter
Sources of B
µ
~
µ
v ×r̂
0 q~
~ = 0 i ∆ℓ×r̂
Biot-Savart Law: ∆B
4 π r2 , B = 4 π r2
2
µ0 i ∆y
∆B = 4 π
sin θ, sin θ = ar , ∆y = r a∆θ
r2
H
~ · d~s = µ I
B
Ampere’s law: M =
L
0
encircled
Steps: to obtain magnetic field
~ pattern and construct loop L
–Inspect B
~
–Find M and Iencl , and solve for B.
d (E A)
ΦE = ǫ
Displ. current: Id = ǫ0 d dt
0
dt
Magnetism in atom:
Orbital motion: µ = i A = 2 em L
L = m v r = n h̄,
QA
= d dt
h̄ = 2hπ = 1.06 × 10−34 J s
h̄ = 9.27 × 10−24 J/T
µB = 2em
µspin = µB
Magnetism in matter:
0
B = B0 + BM = (1 + χ) B0 = (1 + χ) µ0 B
µ0 = κm H
Ferromagnetic: χ ≫ 1
Diamagnetic: −1 ≪ χ < 0
Paramagnetic: 0 < χ ≪ 1, M = C
TB
µorbit = n µB ,
Spin: S = h̄2 ,
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
2
.
Flow Diagram for iCliker test Mode
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from your liker, and memorize your box number.
quiz mode
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test mode
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bring your liker to lass.
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nd your box on the projeted sreen.
re-enter
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ode using your liker. DO NOT PRESS THE \E" (enter) BUTTON YET.
no
Version
Entry
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BUTTON.
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odes to navigate to whih question you want to answer. These
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Find and enter the 2-letter hoie ode for your answer from the \TWO-LETTER CODES" listed below. DO NOT PRESS \E" (enter) BUTTON YET.
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Entry
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on
esti
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ther
re-enter
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\E" (enter)
BUTTON.
no
TWO-LETTER CODES
Choie 1: AA
Choie 2: AB
Choie 3: AC
Choie 4: AD
Choie 5: BA
Choie 6: BB
Choie 7: BC
Choie 8: BD
Choie 9: CA
Choie 10: CB
previous question: CC
another question: CD
enter: E
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
= 1.75929 × 10−7 T .
Question 1, chap 29, sect 5.
part 1 of 1
10 points
The set up is shown in the figure, where
28 A is flowing in the wire segments, AB =
CD = 48 m, and the wire segment arc has a
radius 25 m subtending an angle of 90◦ .
The permeability of free space is
1.25664 × 10−6 T · m/A .
48 m
28 A
A
28
A
B
25
O
m
28 A
C
3
D
Question 2, chap 29, sect 3.
part 1 of 1
10 points
Four long, parallel conductors carry equal
currents. A cross-sectional view of the conductors is shown in the figure below. Each
side of the square has length of 0.9 m.
Note: The current direction is out of the
page at points indicated by the dots and into
the page at points indicated by the crosses.
The premeability of free space is
1.25664 × 10−6 N/A2 .
y
A
D
−6 A ×
+6 A
48 m
Find the magnitude of the magnetic field at
O due to the current segment ABCD.
Correct answer: 1.75929 × 10−7 .
−6 A ×
B
Explanation:
Let :
I = 28 A ,
µ0 = 1.25664 × 10−6 T · m/A ,
a = 25 m .
P
and
Along CD d~s and r̂ are antiparallel, so
again d~s × r̂ = 0 . Therefore that segment
of the current also creates no magnetic field
at O. Along BC d~s is perpendicular to r̂ so
|d~s ×r̂| = ds = a dθ. Also d~s ×r̂ is in the same
direction for all d~s along BC, while r = a, so
the magnitude of the magnetic field at O due
to ABCD is
Z π/2
π/2
µ0
µ0
a dθ
B=
I
=
I
θ
2
4π
a
4πa
0
0
µ0 I
=
8a
(1.25664 × 10−6 T · m/A) (28 A)
=
8 (25 m)
0.9 m
x
× −6 A
C
Which of the diagrams correctly denotes
the directions of the components of the magnetic field from each conductor at the point
P?
BD
1.
P
BC
BB
BA
BA
2.
BC
P
BD
BB
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
4
A
−6 A ×
BB
3.
P
BC
BA
P
BD
BB
BA
−6 A ×
B
BC
4.
correct
P
BA
BB
BD
BB
Consider the magnetic field contributions
due to the currents in C and D only:
D
+6 A
BD
5.
BC
P
P
BC
BD
BA
× −6 A
C
BA
6.
P
BC
BD
BB
Consider the results of all four magnetic
field contributions superimposed on one another:
0.9 m
A
D
−6 A ×
+6 A
Explanation:
BC
P
Let :
I = 6 A and
ℓ = 0.9 m .
The directions of the magnetic field due
to each wire are given by the right hand rule,
where the thumb points in the direction of the
current and your fingers curl in the direction
of the magnetic field’s circular path.
Note: The magnetic field from each wire
circulates in a circle around that wire. At the
point P, you take the magnetic field direction
tangent to the circle formed by magnetic field
lines.
Consider the magnetic field contributions
due to the currents in A and B only:
BA
−6 A ×
B
BD
BB
× −6 A
C
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
A
−6 A ×
D
+6 A
so the current in the circuit is I =
the voltage across R2 is
× −6 A
C
At P, the direction of the resulting magnetic
~ net
1
B
field
= + √ (ı̂ − ̂) .
Bnet
2
Eighth of fourteen versions.
Since R2 and C are parallel, the potential
difference across each is the same. Hence the
charge on the capacitor is
Q = C V2
= (4 µF) (4.09091 V)
Question 3, chap 28, sect 7.
part 1 of 1
10 points
In the figure below the battery has an emf of
9 V and an internal resistance of 1 Ω . Assume
there is a steady current flowing in the circuit.
1Ω
5Ω
9V
5Ω
4 µF
= 16.3636 µC .
Question 4, chap 27, sect 4.
part 1 of 1
10 points
A 12 Ω resistor and a 6.0 Ω resistor are
connected in series with an emf source. The
potential difference across the 6.0 Ω resistor
is measured with a voltmeter to be 18 V.
Find the potential difference across the emf
source.
Correct answer: 54.
Explanation:
Find the charge on the 4 µF capacitor.
Correct answer: 16.3636.
Explanation:
Let : R1
R2
rin
V
C
= 5 Ω,
= 5 Ω,
= 1 Ω,
= 9 V , and
= 4 µF .
Let : R1 = 12 Ω ,
R2 = 6.0 Ω ,
∆V 2 = 18 V .
and
Basic Concepts:
Req = R1 + R2
∆V = IR
The equivalent resistance of the three resistors
in series is
Req = R1 + R2 + rin
= (5 Ω) + (5 Ω) + (1 Ω)
= 11 Ω ,
V
, and
Req
V2 = I R2
R2
=
V
Req
(5 Ω)
(9 V)
=
(11 Ω)
= 4.09091 V .
P
−6 A ×
B
5
I1 = I2 = I
Solution:
I=
18 V
∆V2
=
=3A
R2
6Ω
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
6
Req = 12 Ω + 6 Ω = 18 Ω
Let : q = 1.60218 × 10−19 C ,
m = 1.67262 × 10−27 kg ,
r = 5.8 × 1010 m , and
K = 31 MeV = 1 × 106 eV .
∆V = IReq
= (3 A)(18 Ω)
= 54 V .
Question 5, chap 27, sect 1.
part 1 of 1
10 points
A current of 87.4 mA exists in a metal wire.
How many electrons flow past a given crosssection of the wire in 10.7 min?
Correct answer: 3.50216 × 1020 .
From the kinetic energy K, we get the speed
of the proton is
r
2K
v=
.
m
The centripetal force is
qvB =
Explanation:
m v2
,
r
so the field is
Let :
I = 87.4 mA = 0.0874 A and
∆t = 10.7 min = 642 s .
The total charge is
q = N qe = I ∆t ,
so the total number of electrons is
I ∆t
qe
(0.0874 A) (642 s)
=
1.60218 × 10−19 C
N=
= 3.50216 × 1020 .
By convention, the current moves in the
opposite direction as the flow of electrons.
Question 6, chap 30, sect 1.
part 1 of 1
10 points
A cosmic-ray proton in interstellar space
has an energy of 31 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the sun (5.8 × 1010 m).
The proton has a charge of 1.60218 ×
10−19 C and a mass of 1.67262 × 10−27 kg.
What is the magnetic field in that region of
space?
Correct answer: 1.38704 × 10−11 .
Explanation:
r
m 2K
mv
=
B=
qr
qr
m
1 √
2mK .
=
qr
=
10−19
1
C) (5.8 × 1010 m)
(1.60218 ×
q
× 2 (1.67262 × 10−27 kg) (1 × 106 eV)
= 1.38704 × 10−11 T .
Question 7, chap 29, sect 1.
part 1 of 1
10 points
An electron is in a uniform magnetic field
B that is directed into the plane of the page,
as shown.
B
B
e−
B
v
B
When the electron is moving in the plane
of the page in the direction indicated by the
arrow, the force on the electron is directed
1. toward the left
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
2. toward the top of the page.
3. toward the bottom of the page. correct
4. toward the right
5. into the page.
When the switch is closed, the charge on the
capacitor will rise exponentially:
q = Q (1 − e−t/τ )
h
i
= C E 1 − e−t/(R C)
h
i
s
− (1 MΩ)9 (4.9
µF)
= (4.9 µF) (28 V) 1 − e
= 0.000115339 C .
6. out of the page.
Explanation:
The force on the electron is
~ = q ~v × B
~ = −e ~v × B.
~
F
The direction of the force is thus
b = −b
b,
F
v×B
Question 9, chap 27, sect 2.
part 1 of 1
10 points
The figure represents two possible ways to
connect two lighbulbs X and Y to a battery.
Bulb X has less resistance than bulb Y .
X
pointing toward the bottom of the page , usb and reversing
ing right hand rule for b
v × B,
the direction due to the negative charge on
the electron.
Question 8, chap 28, sect 7.
part 1 of 1
10 points
Consider a series RC circuit. The capacitor
is initially uncharged when the switch is open.
1 MΩ
28 V
7
Y
A
X
4.9 µF
S
Find the charge on the capacitor 9 s after
the switch is closed.
Correct answer: 0.000115339.
Explanation:
Let : R = 1 MΩ ,
C = 4.9 µF ,
E = 28 V , and
T = 9 s.
Y
B
Which bulb has the most current running
through it?
1. Bulb X in B
2. Bulb Y in A
3. Bulb Y in B
4. Bulb X in A correct
Explanation:
In A, a parallel circuit, the voltage across
X is the same as Y , but X has less resistance,
so bulb X has more current running through
it.
In B, a series circuit, the current is the
same through both bulbs.
Power (W)
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
5.
Power (W)
Power (W)
Power (W)
E2
= I2 R ,
R
where the last two are simply derived from the
first equation together with the application of
the Ohm’s law.
Since the resistor is connected to a constant
voltage source E = constant
5
4
3
2
1
0
P =
E2
constant
=
,
R
R
tells us that the power is inversely
propor
1
tional to the resistance P ∝
.
R
cor-
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
rect
Power (W)
P =EI =
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
4.
Power (W)
Explanation:
The power dissipated in the resistor has
several expressions
5
4
3
2
1
0
5
4
3
2
1
0
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
3.
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
7.
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
2.
6.
Power (W)
Power (W)
1.
5
4
3
2
1
0
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
Question 10, chap 28, sect 5.
part 1 of 1
10 points
A variable resistor is connected across a
constant voltage source.
Which of the following graphs represents
the power P dissipated by the resistor as a
function of its resistance R?
8
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8 910
Resistance (Ω)
Question 11, chap 26, sect 2.
part 1 of 2
10 points
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
Consider the group of capacitors shown in
the figure.
C1 , C23 , and C4 are connected in series with
equivalent capacitance
8.84 µF
8.62 µF
a
b
5.35 µF
5.88 µF
c
d
9
Cba =
=
10.4 V
1
1
1
+
+
C1 C23 C4
−1
1
1
1
+
+
8.62 µF 14.19 µF 5.88 µF
−1
= 2.80466 µF .
Find the equivalent capacitance between
points a and d.
Correct answer: 2.80466.
Explanation:
Let : C1
C2
C3
C4
EB
= 8.62 µF ,
= 8.84 µF ,
= 5.35 µF ,
= 5.88 µF ,
= 10.4 V .
Question 12, chap 26, sect 2.
part 2 of 2
10 points
Determine the charge on the 8.84 µF capacitor at the top center part of the circuit.
Correct answer: 18.1712.
Explanation:
The voltage drops across C1 and C4 are
then
and
V1 =
Q1
29.1685 µC
=
= 3.38381 V
C1
8.62 µF
V4 =
29.1685 µC
Q4
=
= 4.96062 V .
C4
5.88 µF
C2
a
C1
b
C3
c
and
C4
d
EB
The remaining voltage is
For capacitors in series,
X 1
1
=
Cseries
C
X i
Vseries =
Vi ,
and the individual charges are the same.
For parallel capacitors,
X
Cparallel =
Ci
X
Qparallel =
Qi ,
and the individual voltages are the same.
In the given circuit C2 and C3 are connected
parallel with equivalent capacitance
C23 = C2 + C3
= 8.84 µF + 5.35 µF
= 14.19 µF .
Vremain = Vtotal − V1 − V4
= 10.4 V − 3.38381 V − 4.96062 V
= 2.05556 V .
This remaining voltage is the same across the
parallel capacitors, so
Q2 = C2 Vremain
= (8.84 µF) (2.05556 V)
= 18.1712 µC .
Question 13, chap 26, sect 1.
part 1 of 1
10 points
A parallel-plate capacitor is charged by connecting it to a battery.
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
1. The charge and the electric potential decrease.
2. The charge decreases and the electric potential increases.
κ = 8.9 fills the lower half of the capacitor.
Neglect edge effects.
dielectric 4.98
constant
top
17 cm
If the battery is disconnected and the separation between the plates is increased, what
will happen to the charge on the capacitor
and the electric potential across it?
10
dielectric
constant 8.9
bottom
0.99 mm
Calculate the capacitance C of the device.
Correct answer: 1793.79.
3. The charge remains fixed and the electric
potential increases. correct
Explanation:
4. The charge decreases and the electric potential remains fixed.
Let : ǫ0 = 8.85419 × 10−12 F/m ,
L = 17 cm = 0.17 m ,
d = 0.99 mm = 0.00099 m ,
κt = 4.98 , and
κb = 8.9 .
5. The charge and the electric potential remain fixed.
6. The charge increases and the electric potential decreases.
7. The charge remains fixed and the electric
potential decreases.
The original capacitor may be considered
to be two capacitors in parallel, as in the
diagram.
8. The charge and the electric potential increase.
Ct
9. The charge increases and the electric potential remains fixed.
Explanation:
Charge is conserved, so it must remain constant since it is stuck on the plates. With the
battery disconnected, Q is fixed.
C=ǫ
A
d
A larger d makes the fraction smaller, so C is
Q
smaller. Thus the new potential V ′ = ′ is
C
larger.
Question 14, chap 26, sect 3.
part 1 of 1
10 points
A capacitor is constructed from two square
metal plates. A dielectric κ = 4.98 fills the
upper half of the capacitor and a dielectric
Cb
Note: Both in the original arrangement and
in the equivalent circuit, the potential across
the dielectrics κt and κb is equal.
Note: Also, one side of the equivalent cir1
cuit capacitors (Ct and Cb ) is reduced by ,
2
1
so the areas are reduced by .
2
1
Thus the area for Ct and Cb is L2 . Finally,
2
since the capacitors are in parallel, and
C = κ C0 = κ ǫ0
so
C = Ct + Cb
A
,
d
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
κt ǫ0 L2 κb ǫ0 L2
+
2d
2d
κt + κb ǫ0 L2
=
·
2
d
4.98 + 8.9 1012 pF
·
=
2
1F
8.85419 × 10−12 F/m (0.17 m)2
×
0.00099 m
= 1793.79 pF .
=
Question 15, chap 29, sect 4.
part 1 of 1
10 points
What current is required in the windings
of a long solenoid that has 2850 turns uniformly distributed over a length of 0.416 m
in order to produce a magnetic field of
magnitude 0.000308 T at the center of the
solenoid? The permeability of free space is
4π × 10−7 T · m/A .
Correct answer: 35.7758.
Explanation:
Let : N = 2850 ,
L = 0.416 m ,
B = 0.000308 T , and
µ0 = 4π × 10−7 T · m/A .
The magnetic field inside a long solenoid is
N
B = µ0 n I = µ0
I.
L
A 14.4 V battery is connected to a 168 Ω
resistor.
Neglecting the internal resistance of the
battery, calculate the power dissipated in the
resistor.
Correct answer: 1.23429.
Explanation:
Let : V = 14.4 V and
R = 168 Ω .
Power is
P =IV =
Question 16, chap 28, sect 5.
part 1 of 1
10 points
V2
(14.4 V)2
=
= 1.23429 W .
R
(168 Ω)
Question 17, chap 26, sect 4.
part 1 of 1
10 points
Two devices with capacitances of 24 µF and
3.9 µF are each charged with separate 103 V
power supplies.
Calculate the total energy stored in the two
capacitors.
Correct answer: 0.147996.
Explanation:
Let : C1 = 24 µF = 2.4 × 10−5 F ,
C2 = 3.9 µF = 3.9 × 10−6 F ,
∆V = 103 V .
and
The potential energy stored is
U=
Thus the required current is
BL
I=
µ0 N
(0.000308 T) (0.416 m)
=
(4π × 10−7 T · m/A) (2850)
1000 mA
×
1A
= 35.7758 mA .
11
1
C (∆V )2 ,
2
so
Utot = U1 + U2
1
1
= C1 (∆V )2 + C2 (∆V )2
2
2
1
= (C1 + C2 ) (∆V )2
2
1 (2.4 × 10−5 F) + (3.9 × 10−6 F)
=
2
× (103 V)2
= 0.147996 J .
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
R
R
Let : R = 7.6 Ω ,
2 R = 15.2 Ω ,
E = 31 V .
Explanation:
ǫ0 V
σ
(8.85419 × 10−12 C2 /N · m2 ) (222 V)
=
(0.00014 C/m2 )
= 1.40402 × 10−5 m
Question 19, chap 28, sect 4.
part 1 of 1
10 points
7.6 Ω
.2
15
31 V
Ω
Consider the resistor network shown.
7.6 Ω
7.6 Ω
At what rate is thermal energy being generated in the 15.2 Ω resistor in the center of
the circuit?
Correct answer: 15.8059.
Explanation:
and
Basic Concepts:
Req = R1 + R2 + R3 + · · ·
1
1
1
1
=
+
+
+···
Req
R1
R2
R3
P = I2 R .
s=
= 14.0402 µm .
R
E
When a potential difference of 222 V is
applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density
of 14 nC/cm2 .
The permittivity of a vacuum is 8.85419 ×
10−12 C2 /N · m2 .
What is the spacing between the plates?
Correct answer: 14.0402.
2R
Question 18, chap 26, sect 1.
part 1 of 1
10 points
Let : σ = 14 nC/cm2 = 0.00014 C/m2 ,
V = 222 V , and
ǫ0 = 8.85419 × 10−12 C2 /N · m2 .
12
Solution: The resistors R on the right and
on the bottom are in series, and this combination is parallel with the 2 R resistor on
the diagonal. The equivalent resistance of the
above three resistors is
1
1
1
=
+
′
R
2R 2R
so that
R′ = R .
The equivalent resistance of the entire circuit
is
Req = R + R = 2 R .
From Ohm’s Law, a current
Itot =
E
2R
flows through this equivalent circuit.
implies that a current
I=
This
E
Itot
=
2
4R
flows through the 2 R resistor on the diagonal.
midterm 02 – JYOTHINDRAN, VISHNU – Due: Oct 18 2007, 11:00 pm
Hence the power dissipated by the 2 R resistor is
P = I 2 (2 R)
2
E
=
2R
4R
E2
=
8R
(31 V)2
=
8 (7.6 Ω)
= 15.8059 W .
Note: You may also arrive at this answer by
considering the voltage across the 2 R resistor.
Question 20, chap 27, sect 2.
part 1 of 1
10 points
A 1.01 V potential difference is maintained
across a 2.3 m length of tungsten wire that
has a cross-sectional area of 0.61 mm2 and the
resistivity of the tungsten is 5.6 × 10−8 Ω · m.
What is the current in the wire?
Correct answer: 4.78339.
Explanation:
Let :
V = 1.01 V ,
ℓ = 2.3 m ,
A = 0.61 mm2 = 6.1 × 10−7 m2 ,
ρ = 5.6 × 10−8 Ω · m .
The resistance is
R=
V
ρℓ
=
,
I
A
so the current is
VA
ρℓ
(1.01 V) (6.1 × 10−7 m2 )
=
(5.6 × 10−8 Ω · m) (2.3 m)
I=
= 4.78339 A .
and
13