Description
What an unbelievable resource! I probably needed course on how to decipher my own handwriting, but not anymore...
Lecture 4
I. Electric Field Maps II. Electric Dipole Systems III.Fields of Finite Systems
Electric Field Line Electric field lines help us visualize the electric field and predict how a charged particle would respond to the field
Length of arrow represents À ? Strength of field
 strength
spucing represent
Studs
direction
+Q
Soup
Field Vector
Field lines
• direction of the field vector is tangential
to the electric field line.
BESTUS
• Field strength — We also discuss several other topics like What is a paradigm shift?
Density of lines We also discuss several other topics like What are the purposes of automatic thinking?
# lines/unit area perpendicular to field lines
Electric Field Lines Directed curves such that the .tangent at any point is in the direction of
the electric field at that point Don't forget about the age old question of How to measure the unemployment rate?
Direction of arrow
gives direction of , A
B field
En.
Electric Field Map We also discuss several other topics like Who is david buel jr?
EB Graphical representation of electric field by field lines such that the field of strength is proportional to the local density of lines Don't forget about the age old question of How to calculate the mean and median?
Weaker Field
Weaker Field
Density of lines: # lines per unit area normal to the lines tenser
than
Stronger Field
Stronger Field
Electric Field Lines
Study
For a point charge
E « 1/72 the density of lines
« 1/p2
StudySoup
Ye Pelletierberea
Does this 2D plot do justice to a 3D electrostatic field?
In 2D, the density of lines oc 1/r. In 3D, the density of lines of 1/r2. A useful "cartoon" for visualization Useful information on direction , symmetry, Don't forget about the age old question of What is the content of poor laws?
and qualitative sense of field strength
Drawing Field Maps (Rules)
• Draw charges
• Select # lines per Q.
• Draw stubs on charges.
• Lines start at + charges
end on  charges
• Draw circle at infinity and stubs
corresponding to net charge.
• Connect stubs: No kinks/crossing.
• Should reflect symmetry. Field map of two point charges +9 & 4
No kinks No crossing (only one tangent (direction) to a field line at each point). E they radiate rymmetrically
Soup
St
Stud SOU
Drawing Field Maps
net charge 20
• Lines diverge symmetrically from a +
point charge (converge to a  charge)
• Left right symmetry
• Lines leaving/entering the circle at infinity proportional to the net charge of the system
should be reflected Summetrcalin
net charge (+q)+(2q) = q
ta
29
Electric Field Simulator
Under Activity Guide on Blackboard for use in Activity 4, 9, and 10.
II. Electric Dipole Fields ( Mutt spoles) Monopole: System with net charge Dipole: System with
 zero net charge  Separated + and – charges Characterized by dipole moment vector Ď
T
ta
to
Magnitude: p=q;d Direction: O charge to
charge
In a uniform electric field, experiences no net force but may experience non
zero torque
Water molecule
Microwave oven
F. + F +eg) Ě + (g)Ě = 0
(+)
()
Shape of Dipole Field
circle of infiniti
.
*may !
SSOU
will be weakeroutsiell
V "circle of
infinity" (co)
http://phet.colorado.edu/sims/chargesand fields/chargesandfields en.html
Bond length: 0.096 nm
Polar molecule: Water
p=6x1030 cm
d=0.6x1010 m q* = 0.6e
Large dipole moment means Hydrogen Bonding As a result, it's the most unusual liquid: it is much denser than expected and as a solid it is much lighter than expected when compared with its liquid form. Anomalies: high freezing and melting point (due to this our planet is bathed in liquid water), large heat capacity, high thermal conductivity (high water content in organisms contribute to thermal regulation and prevent local temperature fluctuations), high latent heat of evaporation (resistance to dehydration and considerable evaporative cooling), excellent solvent due to its polarity, high dielectric constant, etc., etc.
http://www.isbu.ac.uk/water/anmlies.html
Microwave ovens: Alternating field exerts a torque on water molecules causing them to rotate. Rotating molecules push, pull, and collide with other molecules (through electrical forces), converting the energy of the electric field into the thermal energy (heat).
The electric field of continuous charge distribution
of
there!
• finite line of charge 2 will see several
• ring of charge
(aka study) disc of charge
• infinite sheet of charge
• infinite line of charge
• semicircle of charge
Field of charge distributions Charge Q uniformly distributed over a rod of length L. Find the electric field at a point p located a distance R on the perpendicular bisector of the rod.



R


less
Q


+

Begin with equation
dĒ = k 5idqp
Idap
It tells“ you how to work the problem:
• pick a dq of charge in the distribution
• draw in your diagram the dễ due to that da
• draw the components of dĒ
• for each component, check for
simplifications due to symmetry, then integrate over the charge distribution.
Field at P on the perpendicular bisector
*Always check for symmetry s in your illustrations
rdqp = V x2 + R2 = r
Tdge
q
dq
da

Starting equation: dĒ= K.dg . Paap 1. Define axes!
2. Pick a small charge patch da:
x = q dq= AdX
Uniform linear
charge density 3. Draw the dễ due to dq. 4. Now draw the components. 5. Can you see from symmetry that Ex = 0?
The electric field at a point p
P
p=V x2 + R2
e
fedap da
R
idge
dar da
6. Field strength due to dq dE = kaq 7. Unit vector from da to P:
From the green triangle with sides x and R, and hypotenuser
Ogre = ( x A) Field dỀ due to patch dq: dễ = k diq (2,5).
'
r'r
For each patch dq at x we have another patch dq at x so the two dEx will cancel one another. Only Ey survives.
The electric field at a point p
dq = \dx
RxdqX112
r= (x2+R2) 22
Ź
8. So we are left with
dey = K doo (B) = kR doo
9. Add the fields from all patches
уг Е, суу. We pulled all the constants outside The physics of the problem is done!
} dx
3
The electric field at a point p
x = 0
ER
I Xdq=adx
1/2 dx Ey = kR/
12 (x2 + R2)3/2
11/2
= kar [RevX +cam
L
L = kR
2R2 /(L2/4) + R2 2R2 V (L2/4) + R2 har I
RV(L2/4) + R2 Expectation: Remember derivatives and integrals of simple power and trig functions, as well as exponentials. The rest you can look up.
The electric field at a point p
RTL


For R >>
for rssi
=P ( VIL TA + )
RL0
= kod
Field of a charge Q!
R
C R 01=Q)
For R«< L
3  Jim SÀ (VILLA+ R)
la 5 = vya kā var han tro)
charge Q!
Ē = lim ka
R<L R
1=
= Kilit
1 R
Field of an infinite line charge!
=hl
27
De
dE = kahe
Х
r
de
#field will be pointing
in xdirection
dĒ = kaina 0, ,, 0) "
Consider the da'(=dq) on the opposite side (negative yaxis) da' gives rise to dĒ' at P
On adding dễ and dỂ', their ycomponents cancel pairwise!
Study S
Same for the zcomponents (not shown) for other dq elements. So after adding them all Ey = Ez = 0.
Lecture 5: Gauss' Law From the concept of electric field flux to the calculation of electric fields of complex charge distributions. Why? Because (F= qĒ 1. Idea of Gauss' Law
2. Mathematics of Gauss' Law 3. Using Gauss' Law to calculate field
4. Gauss' Law for Spherical Systems
Electric Flux
Ô = surface normal
A cos 0
Electric flux through an area perpendicular
a planar surface A to field lines
*d of product
de = Ē Ā= E Á cos 0 * field strength À = AÑ 0 = angle between ñ and È
E « # lines per unit area perpendicular to
field lines
Electric flux through a surface A is the number of Winther
electric field line crossing it.
If the electric field is not uniform or the surface is not flat, divide the surface into infinitesimal surface elements dA and add the flux through each
û
doe = Fodă
on
= ZondA
da
* = 5, 8nda
E and may vary with dA
The flux of interest is when the surface is a closed surface ( encloses a volume )
de = d Ē.ndA
JS
closed surface = Gaussian surface
V
Study Soup
Sphere and cylinders are closed surfaces For closed surfaces, în points from the inside to the outside . We count* lines exiting as positive and lines entering as negative Flux through a closed surface equals the net # of lines exiting the surface.
*There are 10 kinds of people in this world: those who can count, and those who can't.
Gauss' Law
The electric flux through a closed surface, S, is proportional to the net charge enclosed by s
K245E,
enclosed
=41k Qenc
SOU
85rom a Gene Hele Pene
*Given an electric charge distribution , we
can find the flux through a closed surface.
*Given flux through a closed surface,
find the total charge enclosed by that surface.
rather than
* Can we find the Efield itself
just the flux?
YES! For highly symmetric distributions.
What is the flux through the surfaces A, B, C, and D?
OB = q/80
0A = +q/80
0 = 0 !! 1 *Anytime there is
a dipole, Flux=0 Flux is positive if net enclosed charge is positive : & negative if net charge is negative.
+q
9
3. Using Gauss' law to find Electric Field
• Avoid Integration
• Find a Gaussian surfaces on which Ē  N
"Closed surface" ĒL Ñ Eñeo Everywhere on that surface
Ē l ñ & h = En n = < tuxt?
I where tu Su Cu, v)
and tu= Si (u, v) 1. Reason what the shape of field must be.
Symmetry 2. Choose s over which Ē. N is constant
The flux is then
A Ē. ñdA = En fi, dA = En s
preme
Surface area
→
En =
Useful symmetries:
• Spherical
• plane
• cylindrical + translational along axis
5. Gauss' law : spherically symmetric systems
Study Sc
• Charge distribution which depends only on r
• By symmetry, the field points radially
and strength depends only on the distance from the origin "r".
nor is constant
Sou
outward unit Ē (7) = E(r) i radial vector
• Choose Gaussian Surface to be a sphere
& radius r : outward surface normal ñ = ño
• The flux through Gaussian
surface is
$ Jsphere
E(r)f. fdA = Ecr) QdA
sphere
41012
Study
0= E(r) 40 42 4
• From Gauss' Law
$ = E(r)46r2 = Qenc
Qens
E(r) = enter
Qenc
and Ē (7) = 47€, 52
Gaussian
surface
By computing the charge i Qencer's by the Gaussian surface (sphere of radius r), we can calculate electric field
415€
The field of a thin spherical shell of radius a with uniform surface charge density o
"...P
(distribution of
Charge is the i same at all points)
Ototas  541502 TQ total = 4 lines
* " has to be a unit vector
Region I: r<a
Inside
Ēr (a) = = 0 Region II: r>a Outside
Ēr (69 = berdering Sortera
enc
orĒj(r) = 4#a” ŕ
4 H Eor2
See Fig. above
The electric field of a spherical volume charge of radius a and density P (constante

Cif (t))
spherical symmetry ?

6o.
Region I r<a: Inside Gaussian sphere encloses only part of charge up to radius r<a
@enc = P 4тт°
3
3
Ēr(r) = 60
The electric field of a spherical volume charge of radius a and uniform density p
Region II: r>a Outside Gaussian sphere encloses the entire charge up to radius a
Qene =pVenc = P410'3 Where toutes
_bic that's
where the
7e Stud
enc
3
Charge is
Er(m) (41ao/3) – pasa
DI)
300m2
47€.p2' Qtotal f 41€.p2
3€.p2 O total
E(r)
all
As you move inside, charge gets smaller

d smaller >
Stoup
inside
outside
The three spheres have radii 1, 0.79, 0.5 corresponding to 24, 12, 3 lines crossing them