Physics 260 – Exam 2 (Take-Home) – Hein
Skyline College, Spring 2020
NAME: ______________________
GENERAL INSTRUCTIONS:
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This is a take-home exam.
You may consult with others on this test, but you will turn in an individual solution. The work
you present needs to be your own, and may not be copied from others.
Your solution may be typed or handwritten.
FORMATTING YOUR ANSWERS:
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Justify your work sufficiently, with calculation or explanation, in order to receive credit.
Make your answers clear, legible and organized. If I can’t read it, I can’t give it credit.
I cannot grade crossed-out or erased work. Be sure before deleting work!
I cannot give full credit for a correct response if it is accompanied by another contradicting one.
However, “hedging your bets” will earn more credit than not having the correct response at all.
Do not forget to include units and express vectors appropriately for full credit.
GOOD LUCK!
In this exam, you will study various properties of electric dipoles. Electric dipoles are important to
describe molecules. Even without a net charge, the charge distribution can be such that they have a
dipole moment. This polar nature is, for example, what allows water molecules to bond to each other
and support a high surface tension.
Before attempting the problem, it is recommended that you read the following chapters in the text:
https://openstax.org/books/university-physics-volume-2/pages/5-7-electric-dipoles
https://openstax.org/books/university-physics-volume-2/pages/7-3-calculations-of-electric-potential
(TO BE FILLED BY INSTRUCTOR)
Problem
I
II
III
IV
Total
Points Obtained
Maximum Points
30
30
20
20
100
Question #1: Electric field due to a dipole (30 points total)
An electric dipole consists of a pair of point charges with equal
magnitude and opposite sign, +Q and –Q, and separated by a
distance 2a, as shown.
Part A (15 points): Show that at all points on the x-axis for
'(
which |𝑥| ≫ 𝑎, 𝐸 ≈
.
.
)*+, -
Part B (15 points): Show that at all points on the y-axis for which |𝑦| ≫ 𝑎, 𝐸 ≈
'(
*+, 0 .
.
Question #2: Electric Potential due to a dipole (30 points total)
Let’s consider the same configuration as Question #1.
Part A (10 points): Calculate the electric potential at any point on the x-axis.
Part B (10 points): Calculate the electric potential at any point on the y-axis.
Part C (10 points): Use your result in Question #2, Part B, to calculate the electric potential on
the y-axis, when |𝑥| ≫ 𝑎. Use this result to confirm the answer to Question #1, Part B.
Question #3: Torque on an electric dipole (20 points total)
The dipole is placed in an electric field, as shown.
Part A (5 points): Draw the electric forces acting on
each of the charges on the figure.
The net force on the electric dipole is zero, but there is
a torque that tends to rotate the dipole clockwise.
Part B (15 points): We define the electric dipole
moment 𝑝. It has a magnitude 𝑝 = 2𝑞𝑎 and its
direction is along the dipole axis from the negative
charge to the positive charge. Show that the net torque
with respect to the center of the dipole is 𝜏 = 𝑝×𝐸.
Note: One method for increasing capacitance is to insert a dielectric between the conductors
that reduces the voltage because of its effect on the electric field. When the molecules of a
dielectric are placed in the electric field 𝐸, their negatively charged electrons separate slightly
from their positively charged cores, and dielectric polarization occurs. The molecules acquire
an electric dipole moment. We define the polarization vector 𝑃 = 𝑁𝑝, where N is the number of
dipoles per unit volume in the material. This polarization is proportional to 𝐸.
Question #4: RC circuits (20 points total)
Consider the RC circuits shown below. Find the magnitude and direction for the currents at the
instant the switch S is closed.
Formula Sheet
Electric Potential due to a continuous charge distribution
Z
dq
V =k
(12)
r
q1 q2
U =k
(13)
r12
Physics 260/Spring 2020
Electricity
Coulomb’s Law
Fe = k
Electric Field
|q1 ||q2 |
r2
~ = −∇V
~
E
(14)
~ = − ∂V ı̂ − ∂V ̂
E
∂x
∂y
(15)
(1)
~
~ = Fe
E
q0
(2) Capacitance
q
V
Capacitance for parallel plate capacitor
C=
Electric Field due to a point charge
~ = kq r̂
E
r2
(3)
C=
ε0 A
d
Electric Field due to a continuous charge distribution
Capacitors in parallel
Z
dq
~
E=k
r̂
(4)
Ceq = C1 + C2 + ...
r2
Electric Field due to an infinite plane of charge
E=
σ
2ε0
Electric Field just outside of a conductor
Electric Flux
I
φE =
1
1
1
=
+
+ ...
Ceq
C1
C2
U=
(7)
Q2
1
1
CV 2 =
= QV
2
2C
2
C = κC0
(8)
p~ = |q|d~
Z
~
~ ds
E.
(10)
q
r
(22)
Electric Current
(11)
(23)
Potential Energy of Electric Dipole
~
U = −~
p.E
Electric Potential due to a Point Charge
V =k
(21)
(9) Torque on Electric Dipole
~
~τ = p~ × E
∆U
=−
q0
(20)
Electric Dipole Moment
Electric Potential
∆V =
(19)
Capacitor with dielectric
Gauss’s Law
qenc
φE =
ε0
Electric Potential Energy
Z
~
~ ds
∆U = −q0 E.
(18)
Energy stored in a capacitor
(6)
~ A
~
E.d
(17)
Capacitors in series
(5)
σ
E=
ε0
(16)
I=
dq
dt
(24)
(25)
Current Density
J=
Magnetism
Magnetic Force on a moving charge
I
= nqvd
A
~
J~ = σ E
(26)
~
F~B = q~v × B
(41)
(27)
FB = |q|vB sin θ
(42)
Resistance
Magnetic Force on a current carrying wire
V
R=
I
ρL
R=
A
Temperature dependence
(28)
~ ×B
~
F~B = I L
(29) Magnetic Dipole Moment
~
µ
~ = IA
ρ = ρ0 (1 + α(T − T0 ))
(44)
(30) Torque on Current Loop
~
~τ = µ
~ ×B
Resistors in series
Req = R1 + R2 + ...
(45)
(31) Magnetic Dipole Potential Energy
~
U = −~
µ.B
Resistors in parallel
1
1
1
=
+
+ ...
Req
R1
R2
Electric Power
V2
R
P = IV = I 2 R =
d~l × r̂
r2
(47)
~ s = µ0 Ienc
B.d~
(48)
Kirchhoff’s Junction and Loop Rules:
(34) Magnetic Flux
Σclosed ∆V = 0
(35)
Z
φB = N
RC Circuit - Charging Capacitor
t
q(t) = Q(1 − e− RC )
ε − t
e RC
R
RC Circuit - Discharging Capacitor
I(t) =
t
− RC
q(t) = Qe
Q − t
e RC
RC
~ A
~
B.d
Gauss’s Law of Magnetism
I
~ A
~=0
B.d
(37)
(49)
(36)
(50)
Faraday’s Law
(38)
ε=−
dφB
=
dt
I
~ s
E.d~
(51)
(39) Magnetic Field due to a long, straight wire
Time Constant
τ = RC
Z
(33) Ampere’s Law
I
ΣIin = ΣIout
(46)
(32) Biot-Savart Law
~ = µ0 I
B
4π
I(t) = −
(43)
(40)
B=
µ0 I
2πr
(52)
Magnetic Field of solenoid
LC angular frequency
B = µ0 In
n=
N
L
(53)
ω=√
1
LC
(54) LC energy
U = UC + UL = constant
Inductance and Inductors
Self Inductance
N φB
L=
I
Inductance of a solenoid
L
= µ0 n2 A
l
dI
dt
Vmax
Imax
Irms = √ ; Vrms = √
2
2
(56) Inductive Reactance
XL = ωL
(69)
1
ωC
(70)
XC =
(58)
Impedance
Z=
(59)
Current Decay in RL Circuit
L
τ=
R
Energy stored in an inductor
(72)
(60)
Pave = Irms Vrms cos φ
(73)
2
Pave = Irms
R
(74)
(61) RLC Circuit
Vrms
Z
(75)
1
LC
(76)
N2
V1
N1
(77)
I1 V1 = I2 V2
(78)
Irms =
1 2
LI
2
(62)
Resonance frequency
ω0 = √
Mutual inductance
ε2 = −M
(71)
XL − XC
)
R
φ = tan−1 (
RL time constant
dI1
dt
(63) Transformers
V2 =
LC Circuit
Q(t) = Qmax cos(ωt + φ)
I(t) =
p
R2 + (XL − XC )2
Average Power
t
ε
I = e− τ
R
U=
(68)
(57) Capacitive Reactance
RL Circuit
dI
L
+ RI = ε
dt
Current in RL Circuit
t
ε
I = (1 − e− τ )
R
(67)
(55) AC Circuits
Self-induced emf
εL = −L
(66)
dQ
= −ωQmax sin(ωt + φ)
dt
(64)
(65)
Surface charge density σ
Math Stuff
Z
Q=
~B
~ = |A||
~ B|
~ cos θ = Ax Bx + Ay By + Az Bz
A.
~ × B|
~ = |A||
~ B|
~ sin θ
|A
σdA
(93)
ρdV
(94)
(79)
Volume charge density ρ
(80)
Z
Q=
~×B
~ = (Ay Bz − Az By )ı̂ + (Az Bx − Ax Bz )̂ (81)
A
+(Ax By − Ay Bx )k̂ (82)
Constants
Quadratic Formula
x=
−b ±
√
b2 − 4ac
2a
k=
(83)
1
= 8.99 × 109 N m2 /C 2
4πε0
Permittivity of free space
Surface Area of Sphere
ε0 = 8.85 × 10−12 C 2 /N m2
4πr
2
(95)
(96)
(84)
Permeability of free space
Volume of Sphere
4 3
πr
3
Integrals
Z
Z
µ0 = 4π × 10−7 T m/A
(85)
Elementary Charge
e = 1.602 × 10−19 C
dx
1
=
ln(k1 x + k2 ) + C
k1 x + k2
k1
p
dx
√
= ln( x2 + a2 + x) + C
x2 + a2
Z
p
xdx
√
= x2 + a 2 + C
2
2
x +a
Z
dx
x
= √
+C
(x2 + a2 )3/2
a2 x2 + a2
Z
−1
xdx
=√
+C
2
2
3/2
(x + a )
x2 + a2
Z
x3 dx
x2 + 2a2
√
=
+C
(x2 + a2 )3/2
x2 + a2
me = 9.11 × 10−31 kg
(87)
(89)
(90)
(91)
(92)
(99)
Mass of a Proton
(88)
Z
λds
(98)
(86) Mass of an Electron
Charge Densities
Linear charge density λ
Q=
(97)
mp = 1.67 × 10−27 kg
(100)
c = 3.00 × 108 m/s
(101)
Speed of Light
...

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