Physics 1502: Lecture 2
Today’s Agenda
• Announcements:
– Lectures posted on: www.phys.uconn.edu/~rcote/
– HW assignments, solutions etc.
• Homework #1:
– On Masterphysics this Friday
• Homeworks posted on Masteringphysics
– You need to register (included in cost of book)
– Go to masteringphysics.com and register
– Course ID: MPCOTE33308
• Labs: Begin in two weeks
Today’s Topic :
• End of Chapter 20
–
–
–
–
–
Define Electric Field in terms of force on "test charge"
Electric Field Lines
Example Calculations
Continuous charge distributions => integrate
Moving charges: Use Newton’s law
• Demonstration of Mastering Physics
Coulomb's Law
q1
q2
r
F21
r
F12
q1q2
1
F12=
r
4pe0 r2
SI Units:
• r in meters
• q in Coulombs
• F in Newtons

1 = 8.987 109 N m2/C2
4pe0
Charles Coulomb (1736-1806)
Electric Fields
The force, F, on any charge q due to some collection of
charges is always proportional to q:
Introducing the Electric Field:
a quantity, which is independent of that charge q,
and depends only upon its position relative to the
collection of charges.
A FIELD is something that can be defined anywhere in space
it can be a scalar field (e.g., a Temperature Field)
it can be a vector field (as we have for the Electric Field)
Lecture 2, ACT 1
• Two charges, Q1 and Q2 , fixed along the x-axis as
shown, produce an electric field E at a point (x,y) =
(0,d) which is directed along the negative y-axis.
d
– Which of the following statements is true?
y
E
Q1
(a) Both charges Q1 and Q2 must be positive.
(b) Both charges Q1 and Q2 must be negative.
(c) The charges Q1 and Q2 must have opposite signs.
Q2
x
How Can We Visualize the E Field?
• Vector Maps:
arrow length indicates
vector magnitude
+
O
• Graphs:
Ex, Ey, Ez as a function of (x, y, z)
Er, Eq, EF as a function of (r, q, F)
Ex
x
+ chg
Example
E
y
r
• Consider a point charge fixed at the origin of
a co-ordinate system as shown.
– The following graphs represent the
functional dependence of the Electric
Field.
y=
Q
f
i(x)
0.67
Er
Er
0
0
0
9
r
x=
0
f
2p
x
• As the distance from the charge increases, the field falls off
as 1/r2.
• At fixed r, the radial component of the field is a constant,
independent of f!!
x
Lecture 2, ACT 2
y
r
• Consider a point charge fixed at the origin of
a co-ordinate system as shown.
– Which of the following graphs best
represents the functional dependence of
the Electric Field at the point (r,f)?
Ex
Fixed
r>0
0
Q
2p
0
x
Ex
Ex
f
f
f
2p
0
f
2p
Another Way to Visualize E ...
• The Old Way:
Vector Maps
• A New Way:
Electric Field Lines
O
+
O
+
O
+ chg
- chg
• Lines leave positive charges and return to negative charges
• Number of lines leaving/entering charge = amount of charge
• Tangent of line = direction of E
• Density of lines = magnitude of E
y
Electric Dipole
+Q
What is the Electric Field
generated by this charge
arrangement?
a
q
E
a
r
x
E
-Q
Calculate for a pt along x-axis: (x,0)
Ex = ??
Symmetry
Ey = ??
Electric Dipole: Field Lines
• Lines leave positive charge
and return to negative charge
What can we observe about E?
• Ex(x,0) = 0
• Ex(0,y) = 0
• Field largest in space between
the two charges
• We derived:
... for r >> a,
Field Lines from 2 Like Charges
• Note the field lines from 2
like charges are quite
different from the field lines
of 2 opposite charges (the
electric dipole)
• There is a zero halfway
between charges
• r>>a: looks like field of
point charge (+2q) at
origin.
Lecture 2, ACT 3
y
2a
+Q
a
a
x
a
-Q
• Consider a dipole aligned with the y-axis as shown.
– Which of the following statements about Ex(2a,a) is
q1
true?
q2
(a) Ex(2a,a) < 0
(b) Ex(2a,a) = 0
(c) Ex(2a,a) > 0
E
y
Electric Dipole
+Q
a
a
-Q
x
What is the Electric Field
generated by this charge
arrangement?
Now calculate for a pt along y-axis: (0,y)
Ey = ??
Ex = ??
Coulomb Force
Radial
Ey (0, y ) =
E x (0, y) = 0
Ey (0, y ) =
Q  1
1

2
2

4 pe0  y - a
y
a
( ) ( + )
Q
4pe0
4ay
2 2
 a 
y4 1 - 2 
 y 




y
Electric Dipole
+Q
r
a
a
x
-Q
For pts along x-axis:
Case of special interest:
(antennas, molecules)
r>>a
For pts along y-axis:
E x (r, 0) = 0
E y (r, 0) = -2
E x (0, r ) = 0
1
Qa
4 pe 0 (r 2 + a 2 )3/2
Ey (0, r ) =

For r >>a,
For r >>a,
E y (r ,0 )  -2
1 Qa
4pe0 r 3
Q
4pe0
4ar
2 2
 a 
r 4 1 - 2 
 r 
Ey (0, r )  +4
1 Qa
4pe0 r3
y
Electric Dipole Summary
+Q
r
Case of special interest:
a
(antennas, molecules)
x
a
r>>a
-Q
• Along x-axis
• Along y-axis
E x (r, 0) = 0
E y (r ,0 )  -2
E x (0, r ) = 0
1 Qa
4 pe0 r3

1 Qa
Ey (0, r )  +4
4pe0 r3

• Along arbitrary angle Q
E  Qa
1
E 3
r
dipole moment
with
Electric Fields from
Continuous Charge Distributions
See Examples in text
• Principles (Coulomb's Law + Law of Superposition) remain
the same.
Only change:
S


P
Dq
r
r
DE

Dq
DE = k e 2 rˆ
r

Dqi
dq
ˆ
E = k e lim  2 ri = k e  2 rˆ
Dq0
ri
i
V r
Charge Densities
• How do we represent the charge “Q” on an extended object?
total charge
Q
small pieces
of charge
dq
• Line of charge:
l
= charge per
unit length
• Surface of charge:
s
= charge per
unit area
dq = l dx
dq = s dA
• Volume of charge:
r
= charge per
unit volume
dq = r dV
Example: Infinite line of charge
r
E(r) = ?
++++++++++++++++++++++++++
How do we approach this calculation?
r
E(r) = ?
In words:
++++++++++++++++++++++++++
“add up the electric field contribution from each bit of charge,
using superposition of the results to get the final field”
In practice:
• Use Coulomb’s Law to find the E field per segment of charge
• Plan to integrate along the line…
• x: from -to
+
OR
q: from
-p/2to +p/2
Q
+++++++++++++++
• Any symmetries ? This may help for easy cancellations.
Infinite Line of Charge
y
dE
Charge density = l
Q
r
r'
++++++++++++++++
dx
We need to add up the E field
contributions from all segments
dx along the line.
x
Infinite Line of Charge
We use Coulomb’s Law to find dE:
dq
dE =
4 pe0 r2
dE
y
1
Q
r'
r
But what is dq in terms of dx?
++++++++++++++++ x
dx
dq = l dx
 what is r’ in terms of r?
And

r=
Therefore,

But x and q are not independent!
r
cosq
dE =
dE =
ldx
4 pe 0 (r / cos q )2
1
1
l cos2 qdx
4 pe 0
r2
x = r tanq
dx = r sec2q dq
dE =
1
ldq
4 pe0 r


Infinite Line of Charge
• Components:
dEx = -
dEy = +
1
ldq
4 pe0 r
1
ldq
4 pe0 r
Ey
sin q
dE
q
y
Ex
q
r
cosq
++++++++++++++++ x
dx
• Integrate:
+p/2
1 ldq
Ex = dEx = - 
sin q
4
r
- p / 2 pe0
Ey = dEy =
r'
+ p/ 2
1 ldq
cosq
 4
- p / 2 pe0 r
Infinite Line of Charge
y
• Solution:
dE
+p / 2
sinq dq = 0
-p / 2
Ex = 0
+p / 2
Ey =
cosq d q = 2
Q
2l
4 pe 0 r
-p / 2
• Conclusion:
1
r
r'
++++++++++++++++ x
dx

The Electric Field produced by an infinite
line of charge is:
– everywhere perpendicular to the line
–
is proportional to the charge density
– decreases as 1/r.
Lecture 2, ACT 4
• Consider a circular ring with a uniform charge
distribution (l charge per unit length) as
shown. The total charge of this ring is +Q.
• The electric field at the origin is
(a)
(b)
zero

1 2pl
4 pe0 R
(c)

1
pRl
4 pe 0 R2
y
+ +++
+
+
+
+
+
+
+
++
+
R +
+
+
+
+
+
+ +
x
Summary
Electric Field Distibutions
Dipole
Point Charge
Infinite
Line of Charge
~ 1 / R3
~ 1 / R2
~1/R
Motion of Charged Particles in
Electric Fields
• Remember our definition of the Electric Field,
F = qE
• And remembering Physics 1501,
qE
F = ma  a =
m

Now consider particles moving in fields.
Note that for a charge moving in a constant field
this is
just like a particle moving near the
earth’s surface.
ax = 0
ay = constant
vx = vox
vy = voy + at
x = xo + voxt
y = yo + voyt + ½ at2
Motion of Charged Particles in
Electric Fields
• Consider the following set up,
++++++++++++++++++++++++++
e-------------------------For an electron beginning at rest at the bottom plate, what will be its
speed when it crashes into the top plate?
Spacing = 10 cm, E = 100 N/C, e = 1.6 x 10-19 C, m = 9.1 x 10-31 kg
Motion of Charged Particles in
Electric Fields
++++++++++++++++++++++++++
e--------------------------
vo = 0, yo = 0
vf2 – vo2 = 2aDx
Or,
qE 
2
v f = 2 Dx
 m 

(1.6x10 -19 C )(100N /C )
(0.1m)
v 2f = 2
-31
9.1x10 kg




v f = 1.9x106 m / s
Recap of today’s lecture
• Define Electric Field in terms of force on "test
charge"
• Electric Field Lines
• Example Calculations
• Continuous charge distributions => integrate
• Moving charges: Use Newton’s law
• Homework #1 on Mastering Physics
– From Chapter 20