Physics 2112
Unit 2: Electric Fields
Today’s Concepts:
A) The Electric Field
B) Continuous Charge Distributions
Unit 2, Slide 1
Fields
What if I remove q2? Is
there anything at that point
in space?
2
q2
q1
q1
F12
MATH:
 kq q
F12  r
1 2
2
12
rˆ12
 F
E2 q
12
2

kq 1
2
12
r
rˆ 12
Unit 2, Slide 2
Vector Field
If there are more than two charges present, the total force on any
given charge is just the vector sum of the forces due to each of
the other charges:
q2
q2
F4,1
F4,1
q1
F1
F2,1
F1
F3,1
F3,1
q4
q3
F2,1
q1
q4
F1
q3
F1
F2,1
F2,1
F3,1
 kq q
F1  r
MATH:
1 2
2
12
F4,1
rˆ12 +
kq 1 q 3
2
13
r
+q1 -> -q1
rˆ 13 +
kq 1 q 4
2
14
r
rˆ14
F3,1
direction reversed
 F
E q
1
1

F4,1
kq 2
2
12
r
rˆ 12 +
kq 3
2
13
r
rˆ13 +
kq 4
2
14
r
rˆ14
Unit 2, Slide 3
Electric Field
“What exactly does the electric field that we calculate mean/represent? “
“What is the essence of an electric field? “


F
E 
q
The electric field E at a point in space is simply
the force per unit charge at that point.
Electric field due to a point charged particle
Superposition

E 
Qi
k r
i
2
i
Field points toward negative and
Away from positive charges.
rˆi
q2
E4
E2
E
E3
Field point in the direction of the
force on a positive charge.

Q
E  k 2 rˆ
r
q4
q3
Unit 2, Slide 4
CheckPoint: Electric Fields1
A
x
+Q
-Q
B
Two equal, but opposite charges are placed on the x axis. The
positive charge is placed to the left of the origin and the negative
charge is placed to the right, as shown in the figure above.
What is the direction of the electric field at point A?
A.
B.
C.
D.
E.
Up
Down
Left
Right
Zero
Unit 2, Slide 5
CheckPoint: Electric Fields2
A
x
+Q
-Q
B
Two equal, but opposite charges are placed on the x axis. The
positive charge is placed to the left of the origin and the negative
charge is placed to the right, as shown in the figure above.
What is the direction of the electric field at point B?
A.
B.
C.
D.
E.
Up
Down
Left
Right
Zero
Electricity & Magnetism Lecture 2, Slide 6
Example 2.1 (Field from three charges)
+q
1
Calculate E at
point P.
P
d
-q
2
+q
d
3
Electricity & Magnetism Lecture 2, Slide 7
CheckPoint: Magnitude of Field (2 Charges)
In which of the two cases shown below is the magnitude of the electric
field at the point labeled A the largest?
A
+Q
+Q
A
-Q
Case 1
A.
B.
C.
+Q
Case 2
Case 1
Case 2
Equal
Electricity & Magnetism Lecture 2, Slide 8
+
+
+
_
_
_
Electricity & Magnetism Lecture 2, Slide 9
CheckPoint Results: Motion of Test Charge
A positive test charge q is released from rest at distance r
away from a charge of +Q and a distance 2r away from a
charge of +2Q. How will the test charge move immediately
after being released?
A.
B.
C.
D.
To the left
To the right
Stay still
Other
Electricity & Magnetism Lecture 2, Slide 10
Example 2.2 (Zero Electric Field)
q1
(0,0)
Q2
x
(0.4m,0)
A charge of q1 = +4uC is placed at the
origin and another charge Q2 = +10uC is
placed 0.4m away. At what point on the
line connected the two charges is the
electric field zero?
Unit 2, Slide 11
Continuous Charge Distributions
“I don't understand the whole dq thing and lambda.”
Summation becomes an integral (be careful with vector
nature)

E 
Qi
k r
i
i
2
rˆi

E 
k
dq
r
2
rˆ
WHAT DOES THIS
MEAN ?
Integrate over all charges (dq)
r is vector from dq to the point at which E is defined
Linear Example:
l  Q/L
dE
pt for E
r
charges
dq  l dx
Electricity & Magnetism Lecture 2, Slide 12
Clicker Question: Charge Density
“I would like to know more about the charge density.”
Linear (l  Q/L) Coulombs/meter
Surface (s  Q/A) Coulombs/meter2
Volume (r  Q/V) Coulombs/meter3
Some
Geometry
A sph ere  4  R
V sp h ere 
4
3
Acylinder  2 RL
2
R
V cylinder   R L
3
2
What has more net charge?.
A) A sphere w/ radius 2 meters and volume charge density r = 2 C/m3
B) A sphere w/ radius 2 meters and surface charge density s = 2 C/m2
C) Both A) and B) have the same net charge.
Q A  rV  r
4
3
R
Q B  s A  s 4 R
3
2
QA
QB

r 43  R
3
s 4 R
2

1 r
3s
R
Electricity & Magnetism Lecture 2, Slide 13
Example 2.3 (line of charge)
“Please go over infinite line
charge.”
P
y
Let’s do one slightly different.
r
h
dq  l dx
x
Charge is uniformly distributed along
the x-axis from the origin to x  a.
The charge density is l C/m. What is
the x-component of the electric field
at point P: (x,y)  (0,h)?
a
Electricity & Magnetism Lecture 2, Slide 14
Clicker Question: Calculation
y
P
Charge is uniformly distributed along
the x-axis from the origin to x  a.
The charge density is l C/m. What is
the x-component of the electric field
at point P: (x,y)  (0,h)?
r
h
dq  l dx
x
x
a
We know:

E 
What is
A) dx
x
2
dq
r
2
k
dq
r
2
rˆ
?
dx
B)
a +h
2
2
C)
l dx
a +h
2
2
D)
l dx
x +h
2
2
E)
l dx
x
2
Electricity & Magnetism Lecture 2, Slide 15
Clicker Question: Calculation
Charge is uniformly distributed along
the x-axis from the origin to x  a.
The charge density is l C/m. What is
the x-component of the electric field
at point P: (x,y)  (0,h)?
dE
y
dE x P
2
r
1
x

E 
We know:
k
dq
r
2
rˆ
x
a
dq  l dx
dq
r
2

l dx
x +h
2
2
Ex 
 dE
x
What is dE x ?
A) dE cos  1
B) dE cos  2
C)dE sin  1
D) dE sin  2
Electricity & Magnetism Lecture 2, Slide 16
Clicker Question: Calculation
y dE
Charge is uniformly distributed along
the x-axis from the origin to x  a.
P
The charge density is l C/m. What is
the x-component of the electric field dE x
2
at point P: (x,y)  (0,h)?

E 
k
dq
r
2
1
rˆ
r
2

x
a
dq  l dx
l dx
x +h
2
2
Ex 
 dE
x

 dE sin 
2
E x?

A)
x
We know:
dq
What is
r
k l sin  2
x
-
dx
2
+h
2
B)
a
k l a  sin  2 
0
dx
x +h
2
2
C) neither of the above sin DEPENDS ON x!
2
Electricity & Magnetism Lecture 2, Slide 17
Clicker Question: Calculation
y
Charge is uniformly distributed along
the x-axis from the origin to x  a.
The charge density is l C/m. What is
the x-component of the electric field
at point P: (x,y)  (a,h)?
dE
P
dE x
2
r
1
x

E 
k
dq
r
2
rˆ
x
a
dq  l dx
We know:
dq
r
2

l dx
x +h
2
2
Ex 
 dE
x

 dE sin 
2
What is x in terms of Q ?
A) x = h*tanQ
2
B) x = h*cosQ2
a
Q
C) x = h*sin
a2 + h2 2
a
D) x = h / 2cosQ
22
(a - x) + h
Electricity & Magnetism Lecture 2, Slide 18
Calculation
Charge is uniformly distributed along
the x-axis from the origin to x  a.
The charge density is l C/m. What is
the x-component of the electric field
at point P: (x,y)  (a,h)?
dE
y
dE x
P
dE x
2
r
1

We know: E 
dq
r
2

k
dq
r
l dx
x +h
2
2
rˆ
Ex 
2
x
x
a
dq  l dx
 dE
x

 dE cos 
2
x = h*tanQ
dx = h*sec2Q d
What is
Ex (P)

E x ( P )  kl
f
?
h * sec  * sin 
2
 d  (( h * tan  )
0
2
+h
2
)

E x ( P )  kl 


h
h +a
2
2

- 1


Electricity & Magnetism Lecture 2, Slide 19
Observation
y

E x ( P )  kl 


h
h +a
2
2

- 1


dE
P
dE x
2
since




h
h +a
2
2

 1


Ex < 0
r
1
x
a
x
dq  l dx
make since?
Exercise for student:
Change variables: write Q in terms of x
Result: do integral with trig sub
Electricity & Magnetism Lecture 2, Slide 20
Back to the pre-lecture
y
“Please go over infinte line
charge. How does R get
outside the intergral?”
dE
dE x
P
dE x
2
We had:

E x ( P )  kl
f
h * sec  * sin 
2
 d  (( h * tan  )
2
0

 kl
f
 d
0
r
+h
2
1
)
x
sin 
a
x
dq  l dx
h
to -
8
B) The limits would be +
8
How would this integral change if the line of charge were
infinite in both directions?
D) sinQ would turn to tanQ
A) The limits would be Q1 to –Q1
E) sinQ would turn to cosQ
C) The limits would be -/2 to /2
Electricity & Magnetism Lecture 2, Slide 21
Back to the pre-lecture
y
2
For an infinite line of
charge, we had:
 /2
E x ( P )  kl
 d
- / 2
dE
dE x
P
sin 
dE x
r
h
2
x
1
a
x
dq  l dx
How would this integral change we wanted the y component
instead of the x component?
A) The limits would be Q1 to –Q1
D) sinQ would turn to tanQ
B) The limits would be +/- infinity
E) sinQ would turn to cosQ
C) The limits would be -/2 to /2
Electricity & Magnetism Lecture 2, Slide 22
k in terms of fundamental constants
Note
k 
E line 
1
4 
l
r 2 
o

o
2lk
r
Unit 2, Slide 23
CheckPoint: Two Lines of Charge
Two infinite lines of charge are shown below.
Both lines have identical charge densities +λ C/m. Point A is
equidistant from both lines and Point B is located a above the top line
as shown.
How does EA, the magnitude of the electric field at point A,
compare to EB, the magnitude of the electric field at point B?
A. EA < EB
B. EA = EB
C. EA > EB
Electricity & Magnetism Lecture 2, Slide 24
Example 2.4 (E-field above a ring of charge)
P
y
x
What is the electric field a
distance h above the
center of ring of uniform
charge Q and radius a?
h
a
Unit 2, Slide 25
Example 2.5 (E-field above a disk)
P
y
x
What is the electric field a
distance h above the
center of disk of uniform
charge Q and radius a?
h
a
Unit 2, Slide 26
dipoles
-q
+q
Cl
q
Na
q
d
Dipole moment = p = qd
Points from negative to positive.
(opposite the electric field.)
Unit 2, Slide 27
Torque on dipole
+q
-q
 = 2*(qE X d/2)
Q
=pXE
dU= -dW = dQ = pE*sinQ*dQ
DU -pEcosQ
Define U = when Q = /2
U -p E
Unit 2, Slide 28
Example 2.6 (Salt Dipole)
+
-
20o
The two atoms in a salt (NaCl) molecule are
separated by about 500pm. The molecule is
placed in an electric field of strength 10N/C at
an angle of 20o.
What is the torque on the molecule?
What is the potential energy of the molecule?
Unit 2, Slide 29