REVIEW:
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ELECTRIC FORCE, ELECTRIC FIELD,
ELECTRIC FIELD LINES, ELECTRIC FLUX, GAUSS’S LAW,
ELECTRIC POTENTIAL,
CONTINUOUS CHARGE DISTRIBUTIONS (ELECTRIC FIELD, ELECTRIC
POTENTIAL, GAUSS’S LAW),
ELECTRIC CURRENT,
MAGNETIC FORCE ON MOVING CHARGES AND WIRES,
BIO-SAVART-LAW,
FORCE BETWEEN PARALLEL CURRENT CARRYING WIRES
ELECTRIC FORCE:
Coulomb’s law:
• force on charge 1 due to charge 2 is
r
qq
F12 = k e 1 2 2 r̂12
r
Net force on a charge due to several other charges:
• VECTOR SUM of all forces on that charge due to other charges
• Called Principle of SUPERPOSITON
• Each charge exerts a force on charge 1
r
r
r
r
Resultant force is F1 = F21 + F31 + F41
• says net force on charge 1 equals sum of force on 1
from 2, force on 1 from 3, and force on 1 from 4
ELECTRIC FIELD
r
r
F
r
=
E
• If the force on q0 at a point is F , then electric field at that point is
q0
r
r
r
• If the electric field at a point is E , then the force on q0 at point is F = q0 E
r
q
ˆ
E
=
k
e 2 r
• Electric field at P due to a point charge is P
r
o Unit vector r̂ points from q → P
• Electric field points away from positive charge
• Electric field points toward negative charge
Superposition:
r
Total E at point P due to an arrangement of point charges is the VECTOR SUM
of the electric field contributions from all charges around P
• Total electric field at P is:
r
r
r
r
qi rˆi r
ET = k e ∑ 2 = E1 + E2 + E3 + E4
ri
i
o q i is the charge at i
o ri is the distance from q i → P
o r̂i is the unit vector from q i → P
o the sum is a VECTOR SUM
Did example with electric dipole
ELECTRIC FIELD LINES:
r
• E vector at a point in space is tangent to the EFL through that point
• “Density” of EFL is proportional to E (magnitude) in that region
o Larger E → closer packing of lines
• EFL start on positive charges and end on negative charges
• Number of EFL starting/ending on charge is proportional to its magnitude
• Electric field lines do not cross
Looked at motion of a particle in a uniform electric field:
ELECTRIC FLUX
General result for Electric Flux through element of area ∆Ai
r
r
∆Φ Ei = Ei ∆Ai cosθ i = Ei ⋅ ∆Ai
Total flux through a closed surface:
r r
Φ E = ∫ E ⋅ dA = ∫ En dA
over
closed
surface
GAUSS’S LAW (general statement):
r r qenclosed
Φ e = ∫ E ⋅ dA =
closed
surface
ε0
• Powerful way to calculate electric field if we can factor En out of integral
o Trick is to choose surface so that En is uniform over all or part of surface
No charge inside:
• net number of lines leaving = 0
• all lines go through
Positive charge inside:
• non-zero net number of lines leaving
• lines start on charge inside sphere
Used Gauss’s Law to calculate electric field around a point charge
ELECTRIC POTENTIAL
Difference in electric potential between points A and B is:
∆V = VB − V A =
∆U
q0
B
SO:
r r
∆V = VB − VA = − ∫ E ⋅ ds
A
o Potential difference between two points depends on electric field
o Note sign and order of integration limits
EQUIPOTENTIAL
o All points on plane perpendicular to uniform
r
E field have same electric potential
SIGN OF ∆V
o change in potential energy of q0 when moved
from A → B is
∆U = q0 ∆V = −q0 Ed
r
E
o Says that if A → B is in same direction as , then ∆V = VB − V A < 0
r
o If path is in the same direction as E , then potential difference ∆V is
negative
ELECTRIC POTENTIAL DUE TO A POINT CHARGE
Electric potential at distance r from a point charge q :
ke q
V=
r
•
assumes that electric potential at infinity is 0
IMPORTANT:
• Electric potential V is a scalar.
o can just add contributions from different charges
• ALL points at distance r from a point charge q have the same potential
o Spherical surface around point charge is an equipotential surface
Electric Potential due to multiple point charges: SUPERPOSITION
• Electric potential at P is
q
V = ke ∑ i
i ri
o Not a vector sum. Contributions to V add
as scalars
r
Can get components of E from derivatives of V
Implies:
Ex = −
dV
dx
Ey = −
dV
dy
Ez = −
dV
dz
r
E
• Says:
is always perpendicular to equipotential surfaces
EQUIPOTENTIAL SURFACES:
• like contour maps
o “valleys” centred on negative charge
o “hills” centred on positive charge
• Example: Electric
dipole (contours
marked in volts)
o Electric field lines
start on positive
charge, end on
negative charge
and are
perpendicular to
equipotentials at
crossing
Cange in potential energy of charge + q 0
moving from A → B
• ∆U = q0 (VB − VA ) = q0 (24V − (− 16V ))
• So: ∆U = q0 × 40 V
• External force does Wext = q0 × 40 V
CONTINUOUS CHARGE DISTRIBUTIONS:
CHARGED OBJECTS WITH FINITE SIZE
Strategy:
• Break distribution into small charge elements
• Find contribution to electric potential or electric field from a given element
r
• Sum/integrate to find total V or E
ELECTRICAL POTENTIAL FROM A CONTINUOUS CHARGE DISTRIBUTION
• total V at P is:
V = lim
∆qi →0
∑ ke
i
∆qi
dq
= ke ∫
ri
r
over all
charge
dq
dV
k
=
dq
e
V
• Identify
r as the contribution to from the charge element
CHARGE DENSITY
• to convert sum/integral over charge elements into a sum/integral over spatial
variables (i.e. x, y, z)
o Linear charge density:
λ=
Q
L
Q
o Surface charge density: σ = A
Q
o Volume charge density: ρ = V
ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS
Approach:
• Break charge distribution into small elements (treat each as a point charge)
• Write vector sum of contributions from elements
• Take limit as elements become infinitesimally small → INTEGRAL
o r̂i is unit vector pointing from ∆qi toward P
r
r
o ∆Ei is the contribution to E due to ∆qi
r
∆q
E = lim ke ∑ 2 i rˆi
∆qi →0
ri
i
= ke ∫
dq
rˆ
r2
Two Ways to use GAUSS’S LAW:
qinside
Φ
=
• e
ε0
o relates flux to charge inside for surface of ANY shape
•
r r qinside
∫ E ⋅ dA =
over
closed
surface
ε0
r
o gives a way to calculate E for SPECIAL cases
r
E
is constant over part of
ƒ mostly useful if we normal component of
the surface and can be factored out of the integral
r
Using Gauss’s law to calculate Electric Field E
• Must know direction of electric field from symmetry of problem
o radial (spherical symmetry) for point charge
o radial (cylindrical symmetry) for a long line of
charge
o uniform for a large flat sheet of charge
• Must choose Gaussian surface that allows us to calculate
r r
Φ e = ∫ E ⋅ dA
over
closed
surface
r
o Must be able to factor E out of flux integral in region of space where we
want to find electric field
Two cases for which we can evaluate
r r
∫ E ⋅ dA for all or part of surface
surface
r
E
•
uniform and perpendicular to part or all of Gaussian surface
r r
o then flux is ∫ E ⋅ dA = En A for that part of the surface
r
• E parallel (tangent) to part of the Gaussian surface
CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM (19.11)
PROPERTIES: ISOLATED CONDUCTOR IN ELECTROSTATIC EQUILIBRIUM
st
1 :
r
E = 0 everywhere inside a conductor
r
E
• Must be true or else charges would move until = 0
2nd: Any NET CHARGE on conductor must be on surface
• can prove with Gauss’s Law
ANY NET CHARGE ON THE CONDUCTOR MUST BE
ON THE SURFACE
3rd: Electric field JUST OUTSIDE a CHARGED CONDUCTOR:
• MUST be perpendicular to the surface. If not, charge flows
along surface
σ
• MUST have magnitude E n = ε
0
4th: Surface charge density is HIGHEST where radius of surface curvature is
smallest
r
• Means highest E at most “pointed” regions of surface
ELECTRIC POTENTIAL AT THE SURFACE OF AND INSIDE CHARGED
CONDUCTORS (text section 20.6)
r
• E ⊥ surface means that the surface is an equipotential
r
• E = 0 inside says potential V inside conductor must be
same as potential V at surface
o entire conductor is an equipotential
ELECTRIC CURRENT – FLOWING CHARGE (Quick Review from Chap 21)
CURRENT: rate at which charge crosses a specified
surface
∆Q
=
I
Average current is ave ∆t
• ∆Q is amount of charge across surface in time ∆t
DIRECTION OF CURRENT:
• SAME as direction of POSITIVE charges crossing
surface
• OPPOSITE direction of NEGATIVE charges
crossing surface
CONDUCTIVITY: current density J = n q vd = σ E
• vd is drift velocity (proportional to electric field)
• n is density of charge carriers with charge q
+
+
-
Current (I)
l
1
RESISTIVITY: ρ = σ
A
IMPORTANT: resistivity and conductivity
are properties of the material
a
b
RESISTANCE: relates current through particular object to pot. diff. across it
• Look at wire with cross-sectional area A
o Potential difference between point a and point b is Vb − Va = ∆V
A ∆V
=
I
o
ρl
Ohm’s Law:
ρl
=
R
∆V = IR
Define resistance
A so that
J =σ E =
E
ρ
Resistance unit is Ohm (Ω): 1Ω = 1 V/A
Resistivity unit is ohm-metre (Ω·m)
Conductivity unit is (Ω·m)-1
I (current)
MAGNETIC FORCE & MAGNETIC FIELD (Chapter 22)
• Magnetic Field Lines – RULES
N
S
o Lines start and stop on poles of a magnet OR lines form closed loops
around current-carrying wires
• No such thing as a magnetic monopole
• For magnet – lines start at North pole and end at South pole
r
o MAGNITUDE of magnetic field | B | indicated by DENSITY of lines
o DIRECTION of magnetic field at a point is tangent to field lines at point
• Direction of magnetic field at some location is the direction that a
compass needle would point at that location.
r
MAGNETIC FORCE FB on a moving charge
r
r r
FB = q v × B
v
Cyclotron motion:
Charge with velocity perpendicular to magnetic field moves on
circular path (cyclotron orbit) in plane perpendicular to magnetic field
Radius of orbit is
r=
r
mv
qB
• angular frequency for motion (cyclotron frequency)
• Period of circular motion
• Period independent of speed:
T=
v qB
ω= =
r
m
m
circumference 2 π r
=
= 2π
v
v
qB
+
F
Devices that use magnetic force on a moving charge (22.4)
r
r
r
r
r r
In space with E and B , force on moving charge is F = qE + qv × B (Lorentz Force)
VELOCITY SELECTOR
• Contains region of space with uniform electric and magnetic fields
o perpendicular to each other
o perpendicular to path of charged particle
Barrier with hole
E (down)
+ + + + + + +
• In selector region:
r
r r
o FB = q v × B is “up” by RHR
r
r
o FE = q E is “down”
• Net force on charge is zero IF
v
+q
qv B = q E
E
v
=
• So particles with speed
B are undeflected
- - - - - - B (into screen)
MASS SPECTROMETER
• 1st: ionize molecules and fragments
• 2nd: use velocity selector to pick out
fragments with vs = Es / Bs
Detector
r
• 3 : inject ions with speed vs into uniform B
with magnitude B0
m vs
=
r
o Radius of path is
q B0
rd
r
velocity selector
Bs, Es
+ + + + + + +
m rBs B0
• SO: q = E
s
vs
v
+q
+q
- - - - - - -
B0
CYCLOTRON – device to accelerate charged particles to high energy
• Potential difference across
gap changes each time
particle reaches gap
Bup
o “kick” raises K by q∆V
r
• for charge q in uniform B
mv
r
=
o found
qB
o period of orbit is: T =
+q
circumf . 2 π r 2 π m
=
=
v
v
qB
• NOTICE: period (i.e. timing of “kick”) does NOT depend on speed
alternating
voltage
source
MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR
r
r
r
F
=
q
v
×
B
• Force on each charge in wire is B
d
A
Bin
l
• For cross-section area A and n carriers per unit volume,
r
r r
force on wire segment of length l is FB = q vd × B n A l
(
)
FB
I
• Current in wire is I = n q v d A so force on length l of wire is
r r
r
FB = I l × B
r
o Vector l points in direction of current; magnitude is length of segment
r
• Force on segment d s of arbitrary shaped wire is
r
r r
dFB = I ds × B
ds
Magnetic Dipole Moment
µ = IA
I
• For current I circulating around loop of area A
r
o µ = IA is the magnetic dipole moment of the loop
o Magnitude: µ = IA
r
o Direction: r
• vector µ perpendicular to the plane of the loop
• direction by Right-hand rule:
• fingers curl in direction of current
r
• thumb shows direction of µ
• Units of magnetic dipole moment: A·m2
• For a coil of n loops, µ = nIA
A
Can express torque as vector product of magnetic dipole moment and field
µ = IA
r
τ = µ×B
r
r
F2
θ
• Magnitude of torque is τ = I A B sin θ
r r
• Direction of µ × B is into screen/page
• True for any current loop in a magnetic field!
B
2
τ
4
F4
r µ 0 I dsr × rˆ
BIOT-SAVART LAW: dB = 4π r 2
r
• For drawing, direction of d s × rˆ is out of screen/page
P
r
r
θ
r
r
o So dB at P due to ds points out of screen/page
ds
r
d
s
× rˆ = ds sin θ
• Magnitude
r
r
d
B
d
s
o For a given r, contributions
from
are maximum for points on
r
plane perpendicular to ds
r
r
r
d
s
d
B
d
s
makes NO contribution to
at points along direction
o Current in
I
MAGNETIC FIELD AROUND A LONG (INFINITE) WIRE
• Result of using Biot-Savart Law:
r
o Magnetic field lines circle wire → no component of B parallel to wire
B
I
Iout
r
B
• Magnitude of
inversely proportional to perpendicular a distance from wire
r µ0 I
B=
2π a
(IMPORTANT RESULT)
• Direction of magnetic field lines:
r
B
o Another Right-Hand Rule: thumb along I ; fingers curl in direction of
µ0 I
=
B
USE
2π a TO FIND MAGNETIC FORCE BETWEEN PARALLEL WIRES
I1
a
B2
F1
I2
I1
B2
a
F1
I2
µ0 I 2
=
B
• Field at I1 due to I2 is 2 2π a
F1 µ0 I1 I 2
• Force on I1 per unit length is l = 2π a toward I2 (for currents in same dir.)
F2 µ 0 I1 I 2
• Force on I2 per unit length is l = 2π a toward I1 (for currents in same dir.)
o By Newton’s 3rd law.
• For parallel conductors, current in same direction:
F µ 0 I1 I 2
o Wires ATTRACT with force per unit length l = 2π a
• For parallel conductors, current in opposite direction:
F µ 0 I1 I 2
o Wires REPEL with force per unit length l = 2π a
Provides definition of AMPERE:
−7
• For 2 wires, 1 m apart, I 1 = I 2 = 1 A → force per unit length is 2 × 10 N/m