Physics 260 – Exam 2 (Take-Home) – Hein
Skyline College, Spring 2020
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:
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.
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:
(TO BE FILLED BY INSTRUCTOR)
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.
Electric Potential due to a continuous charge distribution
Physics 260/Spring 2020
Fe = k
|q1 ||q2 |
~ = −∇V
~ = − ∂V ı̂ − ∂V ̂
~ = Fe
Capacitance for parallel plate capacitor
Electric Field due to a point charge
~ = kq r̂
Electric Field due to a continuous charge distribution
Capacitors in parallel
Ceq = C1 + C2 + ...
Electric Field due to an infinite plane of charge
Electric Field just outside of a conductor
CV 2 =
C = κC0
p~ = |q|d~
Potential Energy of Electric Dipole
U = −~
Electric Potential due to a Point Charge
(9) Torque on Electric Dipole
~τ = p~ × E
Electric Dipole Moment
Capacitor with dielectric
Electric Potential Energy
∆U = −q0 E.
Energy stored in a capacitor
Capacitors in series
Magnetic Force on a moving charge
J~ = σ E
F~B = q~v × B
FB = |q|vB sin θ
Magnetic Force on a current carrying wire
F~B = I L
(29) Magnetic Dipole Moment
~ = IA
ρ = ρ0 (1 + α(T − T0 ))
(30) Torque on Current Loop
~τ = µ
Resistors in series
Req = R1 + R2 + ...
(31) Magnetic Dipole Potential Energy
U = −~
Resistors in parallel
P = IV = I 2 R =
d~l × r̂
~ s = µ0 Ienc
Kirchhoff’s Junction and Loop Rules:
(34) Magnetic Flux
Σclosed ∆V = 0
φB = N
RC Circuit - Charging Capacitor
q(t) = Q(1 − e− RC )
ε − t
RC Circuit - Discharging Capacitor
q(t) = Qe
Q − t
Gauss’s Law of Magnetism
(39) Magnetic Field due to a long, straight wire
τ = RC
(33) Ampere’s Law
ΣIin = ΣIout
(32) Biot-Savart Law
~ = µ0 I
I(t) = −
Magnetic Field of solenoid
LC angular frequency
B = µ0 In
(54) LC energy
U = UC + UL = constant
Inductance and Inductors
Inductance of a solenoid
= µ0 n2 A
Irms = √ ; Vrms = √
(56) Inductive Reactance
XL = ωL
Current Decay in RL Circuit
Energy stored in an inductor
Pave = Irms Vrms cos φ
Pave = Irms
(61) RLC Circuit
I1 V1 = I2 V2
ω0 = √
ε2 = −M
XL − XC
φ = tan−1 (
RL time constant
Q(t) = Qmax cos(ωt + φ)
R2 + (XL − XC )2
I = e− τ
(57) Capacitive Reactance
+ RI = ε
Current in RL Circuit
I = (1 − e− τ )
(55) AC Circuits
εL = −L
= −ωQmax sin(ωt + φ)
Surface charge density σ
~ = |A||
~ cos θ = Ax Bx + Ay By + Az Bz
~ × B|
~ = |A||
~ sin θ
Volume charge density ρ
~ = (Ay Bz − Az By )ı̂ + (Az Bx − Ax Bz )̂ (81)
+(Ax By − Ay Bx )k̂ (82)
b2 − 4ac
= 8.99 × 109 N m2 /C 2
Permittivity of free space
Surface Area of Sphere
ε0 = 8.85 × 10−12 C 2 /N m2
Permeability of free space
Volume of Sphere
µ0 = 4π × 10−7 T m/A
e = 1.602 × 10−19 C
ln(k1 x + k2 ) + C
k1 x + k2
= ln( x2 + a2 + x) + C
x2 + a2
= x2 + a 2 + C
(x2 + a2 )3/2
a2 x2 + a2
(x + a )
x2 + a2
x2 + 2a2
(x2 + a2 )3/2
x2 + a2
me = 9.11 × 10−31 kg
Mass of a Proton
(86) Mass of an Electron
Linear charge density λ
mp = 1.67 × 10−27 kg
c = 3.00 × 108 m/s
Speed of Light
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