Transcript Electric Field

Physics 121 - Electricity and Magnetism
Lecture 3 - Electric Field
Y&F Chapter 21 Sec. 4 – 7
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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
1
Copyright R. Janow – Fall 2016
Recap: Electric charge
Basics:
•
•
•
•
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
Constant
k=8.99x109 Nm2/coul2
3rd law pair
of forces
F12
q1
repulsion
shown
Force on
q1 due to q2
r12
q2
F21
(magnitude)
k q1q2

F12 
r̂
2
r12
Superposition of Forces or Fields
n 




Fnet on 1   F1,i  F1,2  F1,3  F1, 4 .....
i 2
Copyright R. Janow – Fall 2016
Fields
Scalar Field
Examples:
Vector Field
Examples:
• Temperature - T(r)
• Pressure - P(r)
• Gravitational Potential energy – U(r)
• Electrostatic potential – V(r)
• Electrostatic potential energy – U(r)
• Velocity - v(r)
• Gravitational field/acceleration - g(r)
• Electric field – E(r)
• Magnetic field– B(r)
• Gradients of scalar fields
Fields “explain” forces at a distance – space altered by source
Gravitational Field
force/unitmass
 
g(r )  Lim (
m0   0

Fg (r )
m0
m0 is a “test mass”
)
versus
Electrostatic Field
force/unit charge
 
 
Fe (r )
E(r )  Lim (
)
q0   0 q0
q0 is a positive “test charge”
“Test” masses or charges map the direction and magnitudes of fields
Copyright R. Janow – Fall 2016
Field due to a charge distribution
TEST
CHARGE q0
z
ARBITRARY
CHARGE
DISTRIBUTION
(PRODUCES E)
 
E(r )
Test charge q0:
•
•
A charge distribution creates a field:
•

r
y
•
•
•
x
 
E(r )  Lim (
q0   0
small and positive
does not affect the charge distribution
that produces E.
 
F (r )
q0
most often…
 
  F (r )
E(r ) 
q0
)
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
 
 
F(r )  q0E(r )
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
Copyright R. Janow – Fall 2016
later: V/m
Electrostatic Field Examples
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


F
E
q0
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•
Magnitude: E=F/q0
Direction: same as the force that acts on the
positive test charge
SI unit: N/C
Copyright R. Janow – Fall 2016
Electric Field
3-1 A positive test charge of +Q is at a point P. An external
electric field E points to the right. The force on +Q is
measured. The test charge is then replaced with another test
charge of -Q.
What happens to the external electric field at P and the force
on the test charge when the change in charge happens?
A. The field and force both reverse direction
B. The force reverses direction, the field is unaffected.
C. The force is unaffected, the field is reversed
D. Both the field and the force are unaffected
E. The changes cannot be determined from the
information given
Copyright R. Janow – Fall 2016
Electric Field due to a point charge Q
Coulombs Law
test charge q0

1
F 
Qq0
4 0
r
2
Find the field E due to point
charge Q as a function over
all of space
r̂
E
E
q0


F  q0E 
r


F
E 

q0
Q
1
4 0
Q r̂
r2

r
r̂ 
r
E
• 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:
F = EA = Q.4r2/40r2 = Q/0 Radius cancels
The closed (Gaussian) surface intercepts all the field lines leaving Q
Copyright R. Janow – Fall 2016
Use superposition to calculate net electric field at each
point due to a group of individual charges
Example: for point charges
at r1, r2…..
+
+

 

Fnet  F1  F2  ...  Fn





F
F
F
F
Enet  tot  1  2  ...  n
q0
q0 q0
q0



 E1  E2  ...  En

Enet at i 
1
qi
r̂

2 ij
4 0 j rij
Do the sum above
for every test point i
Copyright R. Janow – Fall 2016
Visualization: Electric field lines (Lines of force)
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•
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
Copyright R. Janow – Fall 2016
DETAIL NEAR A POINT CHARGE
STRONG
WEAK
NEAR A LARGE, UNIFORM SHEET OF + CHARGE
• No conductor - just an infinitely large charge sheet
• E approximately constant in the “near field” region (d << L)
L
+
+
+
+
+
+ d
q
+
+
+
+
TWO EQUAL
+
to
+
+
+
+
+
+
+
+
+
+
F
CHARGES (REPEL)

to

EQUAL
The field has
uniform intensity
& direction everywhere
except on sheet
+
AND
–
CHARGES ATTRACT
Copyright R. Janow – Fall 2016
Field lines for a spherical shell
or solid sphere of charge
Shell Theorem Conclusions
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
Copyright R. Janow – Fall 2016
Example: Find Enet at a point on the axis of a dipole
• Use superposition
• Symmetry  Enet parallel to z-axis
r  z  d/ 2
r-  z  d / 2
kq kq
Eat O  E  E  2  2
r r
• Limitation: z > d/2 or z < - d/2
1
1


Eat O  kq 

2
2
 (z  d / 2) (z  d / 2) 
z
E+
O
Er+
Eat O
r-


z
 2kqd  2
2
2
(
z

d
/
4
)


DIPOLE MOMENT

q

d
points from – to +
q

p  qd
Exact
For z >> d : point “O” is “far”
from center of dipole
+q
d
and
 
1
z3
since
d
 1
z
z=0
-q
 Eat O  
Exercise: Do these formulas
describe E at the point
midway between the charges
Ans:
E = -4p/20d3
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•
•
•
qd 1
p 1


2 0 z 3
2 0 z 3
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?
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
.B
.C
.A
Copyright R. Janow – Fall 2016
A Dipole in a Uniform EXTERNAL Electric Field
Feels torque - Stores potential energy (See Sec 21.7)
ASSSUME RIGID DIPOLE


p  qd
Dipole
Moment
Vector
  
  pxE
Torque = Force x moment arm
= - 2 q E x (d/2) sin(q)
= - p E sin(q)
(CW, into paper as shown)
Potential Energy U = -W
U    dq  pE sin(q)dq
  pE cos(q)
 
UE  p  E
• |torque| = 0 at q = 0 or q = 
• |torque| = pE at q = +/- /2
• RESTORING TORQUE: (q)  (q)
OSCILLATOR
• U = 0
for q = +/- /2
• U = - pE for q = 0
minimum
• U = + pE for q = 
maximum
Copyright R. Janow – Fall 2016
3-3: In the sketch, a dipole is free to rotate in a
uniform external electric field. Which configuration
has the smallest potential energy?
A
C
B
E
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
This process is just
superposition

EP 
1
q
1
lim  2 i r̂i 
4 0 q 0 i ri
4 0

dq
r
2
r̂
1. Find an expression for dq, the “point charge”
within a differentially “small” chunk of the
distribution
 dl for a linear distributi on
dq   dA for a surface distributi on
 dV for a volume distributi on



2. Represent field contributions at P due to a point
charge dq located anyhwere in the distribution.
Use symmetry where possible.

E 
1 q
r̂
2
4 0 r

dE 
dq
4  0r
2
r̂
3. Add up (integrate) the contributions dE over the
whole distribution, varying the displacement
and direction as needed.
Use symmetry where possible.

EP 

 dE
dist
(line, surface, or volume integral)
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.
  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.
dq  dx
1 dq
1 dx
dE 

4 0 x 2 4 0 x 2
•
Add up contributions to the field
from all locations of dq along the
rod (x  [a, L + a]).
L a
E

a
 dx


4 0 x 2 4 0
 E
L a

a
Q
4 0a(L  a)
dx


x 2 4 0
L a
 1
 x 
a
• Interpret
• L =>
• L <<
• L >>

1 Q 1
1 
 

4 0 L  a L  a 
Limiting cases:
0 rod becomes point charge
a same, L/a << 1
a a/L << 1,
Copyright R. Janow – Fall 2016
Electric field at center of an ARC of charge
L
• Uniform linear charge density  = Q/L
dq
dq = ds = Rdq
• P on symmetry axis at center of arc
 Net E is along y axis  need Ey only

kdq
k dq cos(q)
dEP  2 r̂  dEP, y  
ĵ
2
R
R
• Angle q is between –q0 and +q0
R
q
q0
dEp
q0
P
• Integrate:

 q0
q
1 

1
EP,y 
ĵ
R  cos(q)dq  
ĵ sin(q) |q0
 q0
4 0 R 2
4 0 R
0
note :
 cos(q)dq
 sin(q)


 EP,y   2k sin(q0 ) ĵ
R
• For a semi-circle, q0 = /2


EP,y   2k
ĵ
R
and
sin( q)   sin(q)
In the plane of the arc
• For a full circle, q0 = 

EP,y  0
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
L
dq = dx
x
q0
q0
y
r
q

dEP ( q)
P

dEP ( q)
• uniform linear charge density:  = 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

kdq
dEP  2 r̂
r
 dEP, y  
k dq cos( q)
r
2
ĵ
• “1 + tan2(q)” cancels in numerator and denominator

k  cos(q)dq
dE y,P  
ĵ
y
• Integrate from –q0 to +q0

k   q0
k
q
E y,P  
ĵ  cos(q)dq  
ĵ sin(q)  q0
0
y  q0
y
x  y tan(q)
r 2  x2  y2
r 2  y2 [1  tan2 (q)]
Find dx in terms of
SEE Y&F Example
21.10
q:
dx
d [tan(q)]
d sin(q)
y
y
dq
dq
dq cos(q)
2
 y [1  tan (q)]
dx  y [1  tan (q)] dq
2
dq  dx  y [1  tan2 (q)] dq

2k 
EP  
ĵ sin(q0 )
y
Finite length wire
• For y << L (wire looks infinite) q0  /2

2k 
Falls off as 1/y
EP  
ĵ
y Copyright
Along
–y direction
R. Janow
– Fall 2016
Electric field due to a RING of charge
at point P on the symmetry (z) axis
•
•
•
•
SEE Y&F Example
21.9
Uniform linear charge density along circumference:  = Q/2R
dq = ds = charge on arc segment of length ds = Rdf
P on symmetry axis  xy components of E cancel
Net E field is along z only, normal to plane of ring

kdq
dEP  2 r̂
r
 dEz,P 
dq   ds   R df
k dq cos(q)
r
2
k̂
cos(q)  z/r
r 2  R2  z 2

k Rzdf
dEP,z 
k̂
3
r
• Integrate on azimuthal angle f from 0 to 2

EP,z 
k Rz
 2
k
df
2
2 3 / 2 0
[R z ]
2R  Q total charge on disk

EP,z 
kQz
k̂
2
2 3/2
[R z ]
integral = 2
Ez  0 as z  0
(see result for arc)
ds  R df df
• Limit: For P “far away” use z >> R
EP,z 
kQ
z2
Ring looks like a point charge
if point P is very far away!
Exercise: Where is Ez a maximum?
Set dEz/dz = 0
Ans:R. Janow
z = R/sqrt(2)
Copyright
– Fall 2016
Electric field due to a DISK of charge
for point P on z (symmetry) axis
See Y&F Ex 21.11
dEz
• Uniform surface charge density on disc in x-y plane
 = Q/R2
• 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 rdf with radial extent dr
dA  r dr df
dq   dA   r dr df
cos(q)  z/s
P
q
s
z
R
s2  r 2  z 2
f

k dq
1  z r dr df
dEz 
cos(q) k̂ 
k̂
2
3/2
4 0 2
s
r  z2

dA=rdfdr

r
x
• Integrate twice: first on azimuthal angle f from 0 to 2 which yields a factor of 2
then on ring radius r from 0 to R
R


2 
r dr
Ez 
z
k
4 0  [r 2  z2 ]3 / 2
0
Note Antiderivative
r
[r 2  z 2 ]3 / 2

d 
1

 2

dr  [r  z 2 ]1 / 2 

 
1 
Edisk 
2 0 



 k̂
2
2 1/2 
z R

z

Copyright R. Janow – Fall 2016
Electric field due to a DISK of charge, continued
Exact Solution:

 
1 
Edisk 
2 0 



 k̂
2
2 1/2 
z R

z

Near-Field: z<< R: P is close to the disk. Disk looks like infinite sheet.



 
z
 k̂  
for z/R  1 : Edisk 
1
20 
2 0
2 1/2 
R
1

(
z
/
R
)




Edisk 

k̂
2 0

z


1

k̂


R 
20
“near field” is constant – disk
approximates an infinite sheet of charge
Far-Field: R<< z: P is far from to the disk. Disk looks like a point charge.



 
z
 k̂
for R/z  1 : Edisk 
1
1/2
2 0 

z 1  (R / z)2



Series Expansion
(1  s)n  1  ns / 1!  n(n  1)s2 / 2!  ...
converges quickly for s  1

Recall   Q / R 2
1R 
approximate :

1

 
2 z 
2 1/2
1  (R/z)

 
1 R2 
 R2
 Edisk 
k̂
1  1 
 k̂ 
20 
2 z 2 
4 0 z 2


Edisk 
1

Q 1
k̂
2
4 0 z
2
 ...
Point charge
formula
Copyright R. Janow – Fall 2016
Infinite (i.e.”large”) uniformly charged sheet
Non-conductor, fixed surface charge density 
Infinite sheet  d<<L  “near field”  uniform field

E
2 0
for infinite, non - conducting charged sheet
d
L
Method: solve non-conducting disc of charge
Copyright
for point on z-axis then approximate z <<
R R. Janow – Fall 2016
Motion of a Charged Particle in a Uniform Electric Field


F  qE


Fnet  ma
•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
ACCELERATOR
PLATES
(electron gun controls
intensity)
FACE OF
CRT TUBE
heated cathode (- pole)
“boils off” electrons from
the metal (thermionic
emission)
ELECTROSTATIC
DEFLECTOR
PLATES
electrons are negative so
acceleration a and electric
force F are in the direction
opposite
the
electric
Copyright
R. Janow
– Fallfield
2016 E.
Motion of a Charged Particle in a Uniform Electric Field


F  qE


Fnet  ma
Kinematics: ballistic trajectory
y is the DEFLECTION of the
electron as it crosses the field
Acceleration has only a constant y
component. vx is constant, ax=0
L
vx
ay  
x
y
eE
m
vx yields time of flight t
L
vx 
t
electrons are negative so
acceleration a and electric
force F are in the direction
opposite the electric field E.
Use to find e/m
Measure deflection, find t via
kinematics. Evaluate vy & vx
eE 2
y 
t  
t
m
v y  a y t
v  v 2x  v 2y
1
a
2 y
2
1
2
(
)
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Copyright R. Janow – Fall 2016