lecture 3 - electric fieldtyson/p122-ece_lecture1b_extra.pdf · 2017-01-24 · visualization:...

Copyright R. Janow – Fall 2016 1

Physics 121 - Electricity and Magnetism Lecture 3 - Electric Field Y&F Chapter 21 Sec. 4 – 7

• Recap & Definition of Electric Field • Electric Field Lines • Charges in External Electric Fields • Field due to a Point Charge • Field Lines for Superpositions of Charges • Field of an Electric Dipole • Electric Dipole in an External Field: Torque

and Potential Energy • Method for Finding Field due to Charge

Distributions – Infinite Line of Charge – Arc of Charge – Ring of Charge – Disc of Charge and Infinite Sheet

• Motion of a charged paricle in an Electric Field - CRT example

Copyright R. Janow – Fall 2016

Recap: Electric charge

• Positive and negative flavors. Like charges repel, opposites attract • Charge is conserved and quantized. e = 1.6 x 10-19 Coulombs • Ordinary matter seeks electrical neutrality – screening • In conductors, charges are free to move around

• screening and induction • In insulators, charges are not free to move around

• but materials polarize Coulombs Law: forces at a distance enabled by a field

Basics:

q1

q2

r12

F12

F21

3rd law pair of forces

rr

qqkF 2

12

2112 =Force on

q1 due to q2 (magnitude)

Constant k=8.99x109 Nm2/coul2

repulsion shown

Superposition of Forces or Fields

.....4,13,12,1

n

2ii,11 on net FFF FF +

=

++== ∑


Fields

Scalar Field Examples:

• Temperature - T(r) • Pressure - P(r) • Gravitational Potential energy – U(r) • Electrostatic potential – V(r) • Electrostatic potential energy – U(r)

Vector Field Examples:

• Velocity - v(r) • Gravitational field/acceleration - g(r) • Electric field – E(r) • Magnetic field– B(r) • Gradients of scalar fields

Gravitational Field versus Electrostatic Field force/unit charge

)q

)r(F(Lim)r(E eq 000

>−=

q0 is a positive “test charge”

force/unit mass

)m

)r(F(Lim)r(g g

m 000

>−=

m0 is a “test mass” “Test” masses or charges map the direction and magnitudes of fields

Fields “explain” forces at a distance – space altered by source


Field due to a charge distribution

Test charge q0: • small and positive • does not affect the charge distribution that produces E. A charge distribution creates a field: • Map E field by moving q0 around and measuring the force F at each point • E(r) is a vector parallel to F(r) • E field exists whether or not the test charge is present • E varies in direction and magnitude x

y

z ARBITRARY

CHARGE DISTRIBUTION (PRODUCES E)

r

TEST CHARGE q0

)r(E

)q

)r(F(Lim)r(E

q 000

>−≡

0q)r(F

)r(E

=

most often… F = Force on test charge q0 at point r due to the charge distribution E = External electric field at point r = Force/unit charge SI Units: Newtons / Coulomb later: V/m

)r(Eq)r(F

0=


Electrostatic Field Examples

0qFE

=

• Magnitude: E=F/q0 • Direction: same as the force that acts on the

positive test charge • SI unit: N/C

Field Location Value

Inside copper wires in household circuits 10-2 N/C

Near a charged comb 103 N/C

Inside a TV picture tube (CRT) 105 N/C

Near the charged drum of a photocopier 105 N/C

Breakdown voltage across an air gap (arcing) 3×106 N/C

E-field at the electron’s orbit in a hydrogen atom 5×1011 N/C

E-field on the surface of a Uranium nucleus 3×1021 N/C


Electric Field due to a point charge Q

rrQq

F 20

41

0πε=

Coulombs Law test charge q0

Find the field E due to point charge Q as a function over all of space

rrr r

rQ

qF E 24

1

00

≡=≡ πε

⇒= EqF

0

• Magnitude E = KQ/r2 is constant on any spherical shell (spherical symmetry) • Visualize: E field lines are radially out for +|Q|, in for -|Q| • Flux through any closed (spherical) shell enclosing Q is the same: Φ = EA = Q.4πr2/4πε0r2 = Q/ε0 Radius cancels

The closed (Gaussian) surface intercepts all the field lines leaving Q

Q

r

E

E

E q0


Use superposition to calculate net electric field at each point due to a group of individual charges

n21net F...FFF

+++=

n21

0

n

0

2

0

1

0

totnet

E...EE qF...

qF

qF

qFE

+++=

+++==

rrq

41E

jij2

ij

i

0i at net ∑πε=

+ +

Do the sum above for every test point i

Example: for point charges at r1, r2…..


Visualization: Electric field lines (Lines of force)

• Map direction of an electric field line by moving a positive test charge around.

• The tangent to a field line at a point shows the field direction there.

• The density of lines crossing a unit area perpendicular to the lines measures the strength of the field. Where lines are dense the field is strong.

• Lines begin on positive charges (or infinity and end on negative charges (or infinity).

• Field lines cannot cross other field lines


DETAIL NEAR A POINT CHARGE

• No conductor - just an infinitely large charge sheet • E approximately constant in the “near field” region (d << L)

The field has uniform intensity

& direction everywhere except on sheet

NEAR A LARGE, UNIFORM SHEET OF + CHARGE

TWO EQUAL + CHARGES (REPEL) EQUAL + AND – CHARGES ATTRACT

WEAK STRONG

∞+ to+ + + + + + + + + +

∞− to

+ + + + + + + + + +

L d q

F


Field lines for a spherical shell or solid sphere of charge

Outside point: Same field as point charge

Inside spherical distribution at distance r from center: • E = 0 for hollow shell; • E = kQinside/r2 for solid sphere

Shell Theorem Conclusions


Example: Find Enet at a point on the axis of a dipole • Use superposition • Symmetry Enet parallel to z-axis

DIPOLE MOMENT

points from – to +

dq p

≡ d

q−

q+

22 /dz r /dz r - and +≡−≡+

Exercise: Do these formulas describe E at the point midway between the charges Ans: E = -4p/2πε0d3

z = 0

- q

+q

O

z E+

E-

r-

r+

d

22−+

−+ −=−=rkq

rkqEEE O at

+

−−

= 22O at )2/dz(1

)2/dz(1 kqE

)/dz(

z kqdE O at

−= 222 4

2 Exact

[ ] 113 <<≈

zd

zsince

z

p z

qd E O at 30

30

12

12 πε

+=πε

+≈∴

• Fields cancel as d 0 so E 0 • E falls off as 1/z3 not 1/z2 • E is negative when z is negative • Does “far field” E look like point charge?

For z >> d : point “O” is “far” from center of dipole

• Limitation: z > d/2 or z < - d/2


Electric Field

3-2: Put the magnitudes of the electric field values at points A, B, and C shown in the figure in decreasing order. A) EC>EB>EA

B) EB>EC>EA C) EA>EC>EB D) EB>EA>EC E) EA>EB>EC

.C

.A

.B


A Dipole in a Uniform EXTERNAL Electric Field Feels torque - Stores potential energy (See Sec 21.7)

ASSSUME RIGID DIPOLE

dqp

≡

Torque = Force x moment arm = - 2 q E x (d/2) sin(θ) = - p E sin(θ) (CW, into paper as shown)

Exp

=τ

• |torque| = 0 at θ = 0 or θ = π • |torque| = pE at θ = +/- π/2 • RESTORING TORQUE: τ(−θ) = τ(+θ)

OSCILLATOR

EpUE

⋅−=

Potential Energy U = -W

)cos(pE d)sin(pEdU

θ−=θθ+=θτ−= ∫∫ • U = 0 for θ = +/- π/2

• U = - pE for θ = 0 minimum • U = + pE for θ = π maximum

Dipole Moment Vector


3-3: In the sketch, a dipole is free to rotate in a uniform external electric field. Which configuration has the smallest potential energy?

E

A B C

D E

3-4: Which configuration has the largest potential energy?


Method for finding the electric field at point P - - given a known continuous charge distribution

1. Find an expression for dq, the “point charge” within a differentially “small” chunk of the distribution

ρσλ

= dV

dA dl

dq ondistributi volume a for

ondistributi surface a forondistributi linear a for

2. Represent field contributions at P due to a point charge dq located anyhwere in the distribution. Use symmetry where possible.

rr4

dq Ed 20πε

=

rrqE 2

041 ∆πε

=∆

3. Add up (integrate) the contributions dE over the whole distribution, varying the displacement and direction as needed.

Use symmetry where possible.

integral) volume or surface, (line,dist

P Ed E ∫=

r rdq r

rqlimE

ii

i

iqP ∑ ∫πε

⇒∆

πε=

→∆ 20

200 41

41This process is just

superposition


Example: Find electric field on the axis of a charged rod

• Rod has length L, uniform positive charge per unit length λ, total charge Q. λ = Q/L.

• Calculate electric field at point P on the axis of the rod a distance a from one end. Field points along x-axis.

20

20 4

14

1xdx

xdqdE

dxdqλ

πε=

πε=

λ=

+−

πε=

−

πελ

=πελ

=πελ

=+++

∫∫ aL1

a1

LQ

41

x1

4xdx

4xdx

4E

0

aL

a 0

aL

a2

0

aL

a2

0

• Interpret Limiting cases: • L => 0 rod becomes point charge • L << a same, L/a << 1 • L >> a a/L << 1,

• Add up contributions to the field from all locations of dq along the rod (x ε [a, L + a]).

)aL(a

QE +πε

=∴04


Electric field at center of an ARC of charge

• Uniform linear charge density λ = Q/L dq = λds = λRdθ

jR

)cos( dq kdE rRkdq Ed 2yP,2P

θ−=→=

+θ0 −θ0

θ

R

dq

P dEp

L

• P on symmetry axis at center of arc Net E is along y axis need Ey only

• Angle θ is between –θ0 and +θ0

• For a semi-circle, θ0 = π/2 • For a full circle, θ0 = π

j R

2k E y,Pλ

−= 0=y,PE

0

0

|)sin( R

j4

d)cos( RR

j4

1 E0

20

y,Pθθ

θ

θθ

πελ

θθλ

πε+−

+

−−=

−= ∫

10

0

• Integrate:

)sin()sin( and )sin( d)cos( :note θ−=θ−θ=θθ∫

j )sin( R 2k E 0y,P θ

λ−=∴

In the plane of the arc


Electric field due to a straight LINE of charge Point P on symmetry axis, a distance y off the line

• uniform linear charge density: λ = Q/L • point “P” is at y on symmetry axis • by symmetry, total E is along y-axis x-components of dE pairs cancel • solve for line segment, then let y << L

jy

d)cos( k Ed P,yθθλ

−=

• “1 + tan2(θ)” cancels in numerator and denominator

)tan(yx θ=

)](tan[1 y

)cos()sin(

ddy

d)][tan( dy

ddx

2 θ+=θθ

θ=

θθ

=θ

θθ+= d )](tan[1 ydx 2

22 yx r +=2

Find dx in terms of θ:

θθ+λ=λ≡ d )](tan[1 y dxdq 2

)](tan[1 y r 22 θ+=2

j ˆ r

) cos( dq k dE r ˆ r

kdq E d 2 y P, 2 P θ

− = → =

SEE Y&F Example 21.10

x

y

P

L dq = λdx

r

θ

+θ0 −θ0

)(Ed P θ+

)(Ed P θ−

• Integrate from –θ0 to +θ0

0

0

0

0

θ+θ−

θ+

θ−θ

λ−=θθ

λ−= ∫ )sin(j

y kd)cos(j

y k E P,y

jy

k E Pλ

−=2 Falls off as 1/y

• For y << L (wire looks infinite) θ0 π/2

)sin( jy

k E 0P θλ

−=2

Along –y direction

Finite length wire


Electric field due to a RING of charge at point P on the symmetry (z) axis

• Uniform linear charge density along circumference: λ = Q/2πR • dq = λds = charge on arc segment of length ds = Rdφ • P on symmetry axis xy components of E cancel • Net E field is along z only, normal to plane of ring

φλ=λ≡ dR ds dq z/r )cos( =θ 22 zR r 2 +=

kr

)cos( dq kdE rr

kdq Ed 2Pz,2Pθ

=→=

kr

Rzd k Ed 3z,Pφλ

=

• Integrate on azimuthal angle φ from 0 to 2π

∫π

φ+λ

=2

0232 d k]zR[ Rz k E /2z,P

integral = 2π

disk on charge total QR ≡λπ2

k]zR[

zkQ E /2z,P 232+=

Ez 0 as z 0 (see result for arc)

Exercise: Where is Ez a maximum? Set dEz/dz = 0 Ans: z = R/sqrt(2)

• Limit: For P “far away” use z >> R

z

kQ E z,P 2→ Ring looks like a point charge if point P is very far away!

dφ φ= dR ds

SEE Y&F Example 21.9


Electric field due to a DISK of charge for point P on z (symmetry) axis

• Uniform surface charge density on disc in x-y plane σ = Q/πR2 • Disc is a set of rings, each of them dr wide in radius • P on symmetry axis net E field only along z • dq = charge on arc segment rdφ with radial extent dr

φσσφ d dr r dA dq d dr r dA =≡=

z/s )cos( =θ 22 zr s 2 +=

[ ]k

zr

d dr r z k )cos(sdqk Ed 3/2 2z 202 4

1

+==

φσπε

θ

See Y&F Ex 21.11 dEz

φ

R

x

z θ

P

r

dA=rdφdr

s

• Integrate twice: first on azimuthal angle φ from 0 to 2π which yields a factor of 2π then on ring radius r from 0 to R

k ]zr[ dr r z

42 E

R

/20

z

∫ +=

0232πε

πσ

[ ] k Rz

z 1 2

E 1/2 20disk

+−

εσ

=2

+−

=+ 212232

1/2/2 ]zr[

drd

]zr[ r

Note Anti-derivative


Electric field due to a DISK of charge, continued

[ ] k Rz

z 1 2

E 1/2 20disk

+−

εσ

=2

“near field” is constant – disk approximates an infinite sheet of charge

Near-Field: z<< R: P is close to the disk. Disk looks like infinite sheet.

k 2

E 0

disk εσ

≈

Exact Solution:

[ ] 001/2

0disk 2

k Rz 1

2 k

)R/z(1 R

z 1 2

E :1z/R forεσ

εσ

εσ

≈

−≈

+−=<<

2

Far-Field: R<< z: P is far from to the disk. Disk looks like a point charge.

[ ]2

1/2 0

disk R / Q Recall k )z/R(1 z

z 1 2

E :1 R/z for πσεσ

=

+−=<<

2

1s for quickly converges... !/s)n(n !/ ns)s(

Expansion Seriesn

<<+−++=+ 21111 2

kzR

4 k

zR

21 1 1

2 E

00disk 2

2

2

2

εσ

εσ

=

+−≈∴

[ ]...

zR

21 1

R/z)( 1

1 :eapproximat 1/22+

−≈

+

2

kz1

4Q E 2

0disk πε

≈ Point charge

formula


Infinite (i.e.”large”) uniformly charged sheet

Non-conductor, fixed surface charge density σ

L d

Infinite sheet d<<L “near field” uniform field

sheet charged conducting-non infinite, for E02ε

σ=

Method: solve non-conducting disc of charge for point on z-axis then approximate z << R


Motion of a Charged Particle in a Uniform Electric Field

EqF

=

amF net

=

•Stationary charges produce E field at location of charge q •Acceleration a is parallel or anti-parallel to E. •Acceleration is F/m not F/q = E •Acceleration is the same everywhere in uniform field

Example: Early CRT tube with electron gun and electrostatic deflector

ELECTROSTATIC DEFLECTOR

PLATES electrons are negative so acceleration a and electric force F are in the direction opposite the electric field E.

heated cathode (- pole) “boils off” electrons from

the metal (thermionic emission)

FACE OF CRT TUBE ELECTROSTATIC

ACCELERATOR PLATES

(electron gun controls intensity)


Motion of a Charged Particle in a Uniform Electric Field

EqF

= amF net

=

electrons are negative so acceleration a and electric force F are in the direction opposite the electric field E.

x

∆y

L

vx

∆y is the DEFLECTION of the electron as it crosses the field Acceleration has only a constant y component. vx is constant, ax=0

meEa y −=

t

Lv x ∆=

Measure deflection, find ∆t via kinematics. Evaluate vy & vx

( )

vv v ta v

t meE t ay

/ 2y

2xyy

y21

2212

21

+=∆=

∆−=∆=∆

vx yields time of flight ∆t

Kinematics: ballistic trajectory

Use to find e/m

lecture 3 - electric fieldtyson/p122-ece_lecture1b_extra.pdf · 2017-01-24 · visualization:...

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