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 +
=
++== ∑
Copyright R. Janow – Fall 2016
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
Copyright R. Janow – Fall 2016
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=
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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
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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
Copyright R. Janow – Fall 2016
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…..
Copyright R. Janow – Fall 2016
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
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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
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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
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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
Copyright R. Janow – Fall 2016
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
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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
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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?
Copyright R. Janow – Fall 2016
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
Copyright R. Janow – Fall 2016
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
Copyright R. Janow – Fall 2016
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
Copyright R. Janow – Fall 2016
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
Copyright R. Janow – Fall 2016
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
Copyright R. Janow – Fall 2016
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
Copyright R. Janow – Fall 2016
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
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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
Copyright R. Janow – Fall 2016
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)
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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