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UA / Physics / PHYS 2074 / The direction of the field vector is tangential to what?

# The direction of the field vector is tangential to what? Description

##### Description: These notes cover the material we went through in lecture 4 on February 1st and lecture 5 on February 3rd.
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Lecture 4

## The direction of the field vector is tangential to what? 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

## What is 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

## What is an electric field line? 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)

Bond length: 0.096 nm

Polar molecule: Water

p=6x10-30 cm

d=0.6x10-10 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:

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

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 >>

=P ( VIL TA + )

R-L0

= 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 x-direction

dĒ = kaina 0, -,, 0) "

Consider the da'(=dq) on the opposite side (negative y-axis) da' gives rise to dĒ' at P

On adding dễ and dỂ', their y-components cancel pairwise!

Study S

Same for the z-components (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, 8-nda

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

85-rom 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 E-field 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

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