Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
Formula Sheet for In-Class Exam #3
Reading and thoroughly familiarizing yourself with this formula sheet is an important part
of, but it is not a substitute for, proper exam preparation. The latter requires, among other
things, that you have re-worked all assigned homework problem sets (PS) and the in-class
quizzes, studied the posted PS solutions, and worked and studied the assigned conceptual
practice (CP) problems, as well as (optionally) some practice test (PT) problems, as posted
on the LON-CAPA homework and on the PHYS1112 examples and homework web pages.
You should consult the syllabus, and in particular review the Class Schedule on the last
syllabus page (posted on the PYS1112 course web site), to find out which topics you should
cover in preparing for this exam.
Capacitors, Capacitance, Electric Field Energy
(1) Definition of Capacitance: For two oppositely charged metallic objects a and b,
with −Q stored on a and Q stored on b, their electric potential difference V ≡ Vb − Va is
proportional to the charge Q. The capacitance of the two metallic objects is then defined
as:
Q
Q
C≡
,
hence
Q = CV
or
V =
V
C
(2) Voltage and Capacitance of a Planar Capacitor: For two oppositely charged,
parallel planar metallic plates, each of opposing surface area A, closely spaced with distance
d, the voltage V and capacitance C are
|V | = Ed =
|Q|d
;
κo A
C≡
|Q|
A
= κo ≡ κ Co
|V |
d
where κ is the dielectric constant of the dielectric (insulating) material between the plates
and κ = 1 for vacuum or air, and Co ≡ o A/d is the capacitance without dielectric.
(3) Electric Field Energy Storage in a Capacitor: The energy UE required to build up
a charge Q and a voltage V = Q/C in a capacitor is stored as electric field energy between
the capacitor plates and it is given by
UE =
1 2 1
Q = CV 2
2C
2
(4) Electric Field Energy Density: Energy per volume, uE , stored in an electric field is
given in terms of the field strength E
uE =
o 2
E
2
1
Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
Current, Resistance, Ohm’s Law, Electric Power Dissipation
(1) Definition of Current: For a charge ∆Q flowing through a wire or, more generally,
some cross-sectional area of a conducting object, over a time interval ∆t, the current I is
I≡
∆Q
∆t
(2) Ohm’s Law: The current I flowing through an ”ohmic” conductor (e.g., metallic wire)
and the applied voltage drop V ≡ Va − Vb across the conductor are proportional:
V = RI ∝ I
with R ≡
V
= constant ,
I
i.e., the conductor’s resistance R is independent of I or V . Here Va and Vb denote the
electric potential at current’s point of entry a and current’s point of exit b into/from the
conductor, respectively.
(3) Resistance and Resistivity: For current flow through an ohmic conductor, over a
constant cross-sectional area A and a length L, the conductor’s resistance R is proportional
to L and inversely proportional to A:
R=ρ
L
L
∝
A
A
where the resistivity ρ is independent of L or A or, more generally, independent of the size
and shape of the conductor: ρ depends on the material, the chemical composition and the
temperature of the conductor.
(4) General Electric Power Dissipation Law: In any electric device with current I
flowing from entry point a to exit point b, subject to a potential drop V ≡ Va − Vb , the
electric power dissipated, i.e., amount of electric energy per time consumed and heat per
time generated is
P = IV .
(5) Power Dissipation in Ohmic Resistors: For the special case of I flowing through
an ohmic resistor with voltage drop V = RI,
P = RI 2 =
1 2
V .
R
Electric Circuits
(1) Equivalent Capacitance: For any combination of capacitors C1 , C2 , ... having only one
point of entry a and one point of exit b of stored charge, the total voltage drop V ≡ Va − Vb
across the capacitor combination is proportional to the total stored charge Q, injected into
the capacitor combination at point a, with equal charge Q extracted at point b,
V =
1
Q∝Q
Ceq
with Ceq ≡
2
Q
= constant ,
V
Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
i.e., the capacitor combination’s equivalent capacitance Ceq is independent of Q or V .
(2) Capacitors C1 , C2 , ... in Series:
1
V
1
1
≡
=
+
+ ...
Ceq
Q
C1 C2
Ceq ≡
or
−1
Q 1
1
=
+
+ ...
V
C1 C2
and
Q = Q1 = Q2 = ... ;
V = V1 + V2 + ...
where Q is total charge injected at a and V ≡ Va − Vb is total voltage drop; Q1 , Q2 , ... are
the charges stored on a-side plates of C1 , C2 , ... respectively; and V1 = Q1 /C1 , V2 = Q2 /C2 ,
... are the voltage drops across C1 , C2 , ... respectively.
(3) Capacitors C1 , C2 , ... in Parallel:
Ceq ≡
Q
= C1 + C2 + ...
V
and
Q = Q1 + Q2 + ... ;
V = V1 = V2 = ...
where Q is total charge injected at a and V ≡ Va − Vb is total voltage drop; Q1 , Q2 , ... are
the charges stored on a-side plates of C1 , C2 , ... respectively; and V1 = Q1 /C1 , V2 = Q2 /C2 ,
... are the voltage drops across C1 , C2 , ... respectively.
(4) Equivalent Resistance: For any combination of ohmic resistors R1 , R2 , ... having
only one point of entry a and one point of exit b of current flow, the total voltage drop
V ≡ Va − Vb across the resistor combination is proportional to the total current I flowing
through the resistor combination
V = Req I ∝ I
with Req ≡
V
= constant ,
I
i.e., the resistor combination’s equivalent resistance Req is independent of I or V .
(5) Resistors R1 , R2 , ... in Series:
Req ≡
V
= R1 + R2 + ...
I
and
I = I1 = I2 = ... ;
V = V1 + V2 + ...
where I is total current; V ≡ Va − Vb is total voltage drop; I1 , I2 , ... are currents through
R1 , R2 , ... respectively; and V1 , V2 , ... are the voltage drops across V1 = R1 I1 , V2 = R2 I2 , ...
are the voltage drops across R1 , R2 , ... respectively.
(6) Resistors R1 , R2 , ... in Parallel:
1
I
1
1
≡
=
+
+ ...
Req
V
R1 R2
Req ≡
or
3
1
−1
V
1
=
+
+ ...
I
R1 R2
Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
and
I = I1 + I2 + ... ;
V = V1 = V2 = ...
where I is total current; V ≡ Va − Vb is total voltage drop; I1 , I2 , ... are currents through
R1 , R2 , ... respectively; and V1 = R1 I1 , V2 = R2 I2 , ... are the voltage drops across R1 , R2 ,
... respectively.
(7) Kirchhoff Rule Circuit Analysis: Any circuit can be analyzed, i.e., unknown voltages, currents and/or resistances, etc., can be calculated from known ones, by the following
steps:
(K1) Find all ”junctions” (≡where more than two wires meet); label them (e.g., a, b, ...).
(K2) Break up the circuit into ”branches” by cutting off all wires at each junction; assign
”current arrows” and current labels (e.g. I1 , I2 , ...) to each branch.
(K3) Mark ”check points” on all wires in circuit so that each circuit element (resistor,
battery, ...) is separated from every other element by at least one intervening check point;
add more check points for convenience, as needed; label all check points (e.g. f, g, ...); all
junctions also serve as check points.
(K4) Write down Kirchhoff’s ”Junction Rule” for enough junctions j:
X
X
I=
j,in
I
j,out
P
where j,in I is the sum of all currents with branches connected to j and arrows pointing
P
towards j and j,out I is the sum of all currents with branches connected to j and arrows
pointing away from j.
(K5) Write down Kirchhoff’s ”Loop Rule” for enough closed paths (”loops”) L in the circuit:
X
∆V = 0
L
P
where L ∆V is the sum of all voltage drops ∆V between adjacent check points, accumulated
as you walk around the closed loop from one check point to the next.
(K6) Express, as needed, each voltage drop ∆V ≡ Vx − Vy between two check points x and
y [used in Loop Rule (K5)], in terms of the voltage (EMF) E of intervening battery; in terms
of current I through intervening resistor R; or in terms of charge Q stored on intervening
capacitor C; etc. as follows:
For battery E between x and y,
Vx − Vy = +E if x conn. to + of battery;
Vx − Vy = −E if x conn. to − of battery .
For resistor R with current I between x and y,
Vx − Vy = +RI if I−arrow from x to y;
Vx − Vy = −RI if I−arrow from y to x .
4
Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
For capacitor C between x and y, with charge Q stored on x-connected plate and −Q stored
on y-connected plate,
1
Vx − Vy = Q .
C
For wire only, without any other circuit element, between x and y,
Vx − Vy = 0 .
(K7) Solve system of coupled Junction Rule equations (K4) and Loop Rule equations (K5)
for all unknown quantities, with voltage drops ∆V in (K5) expressed in terms of unknown
currents, battery voltages, resistancesetc., using (K6) as needed.
(K8) To find the electric potential difference Vp − Vq between any two check points p and
q in the circuit (incl. non-adjacent check points), walk through the circuit from q to p,
along any open path L connecting q and p, and add up the potential drops ∆V across each
element, as you step from one check point to the next:
X
Vp − Vq =
∆V .
L, q→p
Use (K6) to express voltage drops ∆V in terms of currents, resistances, battery voltages,
etc., as needed.
Forces from Magnetic Fields
(1) Lorentz Force Law: the magnetic force F~ on particle of charge q moving with velocity
~ is
~v at angle θ to magnetic field B
~ ;
F~ = q~v × B
~ ;
F~ ⊥ ~v , B
F = |q|vB sin θ (with 0 ≤ θ ≤ 180o );
~
with F~ -dir. given by right-hand (RH) turn of q~v into B.
(2) Motion of a Charged Particle in Uniform Magnetic Field: a particle of charge
~ in a
q and mass m moving with speed v ≡ |~v |, subject only to magnetic force F~ = q~v × B
~ has a circular trajectory of radius r, period T and frequency f of
uniform magnetic field B,
revolution, given by:
mv
1
m
r=
; T = = 2π
|q|B
f
|q|B
(3) Magnetic Force on Straight Wire Segment: the magnetic force F~ on I-wire seg~ pointing along wire, at angle θ to uniform magnetic field B,
~
ment, with length vector L
~
~ =length of wire, is given by
current I > 0 flowing in L-dir.,
L ≡ |L|
~ ×B
~ ;
F~ = I L
~ B
~ ;
F~ ⊥ L,
F = ILB sin θ (with 0 ≤ θ ≤ 180o );
~ into B.
~
with F~ -dir. given by right-hand (RH) turn of I L
5
Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
(4) Magnetic Torque on Current Loop: the magnetic torque ~τ on a planar I-wire loop
~ at angle θ to uniform magnetic field B,
~ A
~ perpendicular to plane of
with loop area vector A
~ (i.e.,
loop, with current I > 0 flowing around loop in right-handed (RH) dir. relative to A
~ must be chosen so that, if RH thumb points in A-dir.,
~
A
then RH 4 fingers point in I-dir.
~
around loop), and A ≡ |A| =area enclosed by I-wire loop, is given by
~×B
~ ;
~τ = I A
~ B
~ ;
~τ ⊥ A,
τ = IAB sin θ (with 0 ≤ θ ≤ 180o );
~ into B.
~ This results holds for any shape of I-loop (not
with ~τ -dir. given by RH turn of I A
just rectangular!), as long as the loop is planar. Note that ~τ acts to rotate the loop, with
~τ -dir. as rotation axis, in RH dir. (i.e., if RH thumb points in ~τ -dir., then RH 4 fingers
point in τ -induced rotation dir.).
Magnetic Fields from Electric Currents
(1) Magnetic Field from a Very Short Straight Wire (Biot-Savart Law): magnetic
~ generated at observation point P by a short I-wire segment, with
field contribution ∆B
vector ∆~s pointing along wire and ∆s ≡ |∆~s| =length of wire segment, with distance vector
~r pointing from wire segment to P at angle θ to ∆~s, with current I > 0 flowing in ∆~s-dir.,
~ =
∆B
µo
I ∆~s × ~r ;
4πr3
~ ⊥ ∆~s, ~r ;
∆B
~ = µo I ∆s sin θ
∆B ≡ |∆B|
4πr2
~ at oberva(2) Magnetic Field from Infinitely Long Straight Wire: magnetic field B,
tion point P , at distance r measured perpendicular to the line of the wire carrying current
I>0
~ = µo I ,
B ≡ |B|
2π r
~
B-dir.
is given by concentric circular field line (FL) passing through P and looping around
I in right-handed (RH) dir. (i.e., if RH thumb points in I-dir., then RH 4 fingers point in
FL dir.).
(3) Magnetic Force Between Two Long Straight Parallel Wires: Two long straight
parallel wires, both of length L, spaced a distance d apart with d L, and carrying currents
I1 and I2 , will attract or repel each other with force F on each wire, given by
F ≡ |F~ | =
µo I1 I2
L ,
2π d
Force between wires is attractive if I1 and I2 flow in the same direction; else it is repulsive.
~ at observation point
(4) Magnetic Field from Finite Straight Wire: magnetic field B
P , at distance r measured perpendicular to the line of the wire carrying current I > 0
~ =
B ≡ |B|
µo I cos α + cos β
,
2π r
2
6
Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
where α ≡ 6 (P, a, b) and β ≡ 6 (P, b, a) are the angles enclosed between the wire and lines
~
drawn from observation point P to the endpoints a and b of the wire, respectively. B-dir.
is given by concentric circular field line (FL) passing through P and looping around I in
right-handed (RH) dir. (i.e., if RH thumb points in I-dir., then RH 4 fingers point in FL
dir.).
~ at of center of
(5) Magnetic Field at Center of Circular Wire Loop: magnetic field B
wire loop of radius R, with N turns (i.e., wire is wrapped around loop N times), each turn
carrying same current I > 0
~ = µo N I .
B ≡ |B|
2 R
~
B-dir.
is perpendicular to the plane of the I-loop and in in right-handed (RH) dir. rel. to I
~
(i.e., if RH 4 fingers point in I-dir., then RH thumb points in B-dir.).
~ inside of a cylindrical
(6) Magnetic Field Inside Long Thin Solenoid: magnetic field B
(or, more generally, prismatic) solenoid of height L with N turns of wire carrying current I
wrapped around the mantle:
~ = µo N I.
B ≡ |B|
L
~
B-dir. inside is along the cylinder (prism) axis (of length L) in right-handed (RH) dir. rel.
to I [i.e., if RH 4 fingers point in I-dir., looping around the cylincer (prism) axis, then RH
~
thumb points in B-dir.].
~ is being generated
(7) Superposition Principle of Magnetic Field: If a magnetic field B
~ at any observation point P is the
by multiple current-carrying objects (I1 , I2 , ...), then B
~ 1, B
~ 2 , ... that would be
vector sum (resultant vector) of the magnetic field contributions B
generated by each of the current-carrying objects in isolation at that point P :
~ =B
~1 + B
~ 2 + ...
B
~ generated by electric currents flowing in closed
(8) Ampere’s Law: For a magnetic field B,
~
loops or circuits, without time-dependent accumulation of charges anywhere, the B-field
circulation C(L), around a closed, directed path L, is
C(L) = µo I(L) .
Here, I(L) the total current passing through a surface area bounded by L. A current
contribution Ii enclosed by L counts as a positive contribution to I(L) if L loops around Ii
in a right-handed (RH) direction (i.e., if RH 4 fingers point in L-dir., then RH thumb points
~ around a directed,
in Ii -dir.); else, Ii is a negative contribution to I(L). The circulation of B
closed path L is defined as
X
C(L) ≡
Bk ∆s
L
where L is broken up into small length vectors ∆~s pointing around L, ∆s ≡ |∆~s| and Bk is
~
the B-field
vector component parallel to ∆~s at the location of ∆~s.
7
Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
Mechanics Memories: Velocity, Acceleration, Force, Energy, Power
(1) Velocity
~v =
∆~r
∆t
if constant; else ~v = lim
∆~r
∆t→0 ∆t
~a =
∆~v
∆t
if constant; else ~a = lim
(2) Acceleration
∆~v
∆t→0 ∆t
(3) Constant-Acceleration Linear Motion: for ∆~r ≡ ~rf − ~ri and ∆~v ≡ ~vf − ~vi
∆~r =
1
(~vi + ~vf ) t ;
2
∆~r = ~vi t +
1
~a t2 ;
2
∆~v = ~a t .
(4) Constant-Speed Circular Motion: for motion at constant speed v ≡ |~v | around a
circular trajectory of radius r.
The velocity vector ~v is always tangential to trajectory and perpendicular to acceleration
vector ~a: ~v ⊥ ~a.
The acceleration vector ~a always points towards the center of the circular trajectory.
Period T and frequency f of revolution, angular velocity ω, and orbital speed v:
T =
1
2πr
2π
=
=
f
v
ω
2π
v
=
T
r
ω = 2πf =
v = ωr = 2πf r =
2πr
T
Circular centripetal acceleration:
v2
= ω2r
r
Orbital angle ∆φ and arc of circumference ∆s covered during time interval ∆t:
a=
∆φ = ω ∆t =
v ∆t
∆s
=
r
r
∆s = v ∆t = ω r ∆t = ∆φ r
(5) Newton’s 2nd Law:
m~a = F~
(6) Kinetic Knergy (KE), Work, Work-KE-Theorem: K=kinetic energy of object of
mass m moving at speed v; W =work done by force F~ on an object moving/moved with
displacement ∆~r, with ∆~r pointing at an angle θ from F~ and 0o ≤ θ ≤ 180o ; ∆K = Kf −Ki =
change of kinetic energy due to work done by total force F~ :
1
K = m v2 ,
2
W = F ∆r cos θ ,
∆K = W .
(7) Energy Conservation Law for ∆K ≡ Kf − Ki and ∆U ≡ Uf − Ui :
Ki + Ui = Kf + Uf
or
8
∆K + ∆U = 0
Physics 1112
Spring 2010
University of Georgia
Instructor: HBSchüttler
(8) Mechanical Power: P =rate of work done by force F~ on an object moving at speed
~v , with ~v pointing at an angle θ from F~ and 0o ≤ θ ≤ 180o :
P = F v cos θ .
Algebra and Trigonometry
az 2 + bz + c = 0
sin θ =
opp
,
hyp
⇒
cos θ =
z=
adj
,
hyp
−b ±
√
b2 − 4ac
2a
tan θ =
opp
sin θ
=
adj
cos θ
sin2 θ + cos2 θ = 1
For very small angles θ (with |θ| 90o ):
sin θ ∼
= tan θ ∼
= θ (in radians)
Numerical Data
Acceleration of gravity (on Earth):
Speed of light in vacuum:
Biot-Savart’s constant:
Permittivity of vacuum:
Permeability of Vacuum:
Electron mass:
Proton mass:
c = 3.00 × 108 m/s
k = 8.99 × 109 Nm2 /C2
Coulomb’s constant:
Elementary charge:
g = 9.81m/s2
km ≡
µo
4π
= 1 × 10−7 Tm/A (exact)
o ≡ 1/(4πk) = 8.85 × 10−12 C2 /Nm2
µo ≡ 4πkm = 4π × 10−7 Tm/A (exact)
e = 1.60 × 10−19 C
me = 9.11 × 10−31 kg
mp = 1.67 × 10−27 kg
Other numerical inputs will be provided with each problem statement.
SI numerical prefixes:
y = yocto =10−24 , z = zepto =10−21 , a = atto =10−18 , f = femto =10−15 , p = pico =10−12 ,
n = nano =10−9 , µ= micro =10−6 , m = milli =10−3 , c = centi =10−2 , d = deci =10−1 ,
da = deca =10+1 , h = hecto =10+2 , k = kilo =10+3 , M = Mega =10+6 , G = Giga =10+9 ,
T = Tera =10+12 , P = Peta =10+15 , E = Exa =10+18 , Z = Zetta =10+21 , Y = Yotta =10+24 .
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