Physics 2112 Unit 2: Electric Fields Today’s Concepts: A) The Electric Field B) Continuous Charge Distributions Unit 2, Slide 1
Physics 2112
Unit 2: Electric Fields
Today’s Concepts:
A) The Electric Field
B) Continuous Charge Distributions
Unit 2, Slide 1
Fields
Unit 2, Slide 2
122
12
21 r̂r
qkq
MATH:
12F
122
12
1
2
12 r̂r
kq
q
F
2E
q1
2
What if I remove q2? Is
there anything at that point
in space?
q1q2
F12
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:
+q1 -> -q1 direction reversed
Vector Field
Unit 2, Slide 3
F2,1
F3,1
F4,1
F1
q1
q2
q3
q4
F2,1
F3,1
F4,1
F1
F2,1
F3,1
F4,1
F1
F2,1
F3,1
F4,1
F1
q1
q2
q3
q4
142
14
41132
13
31122
12
21 ˆˆˆ rr
qkqr
r
qkqr
r
qkq++
MATH:
1F
142
14
4132
13
3122
12
2
1
1 ˆˆˆ rr
kqr
r
kqr
r
kq
q
F++
E
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
E2
E3
E4
EField points toward negative and
Away from positive charges.
Field point in the direction of the
force on a positive charge.
“What exactly does the electric field that we calculate mean/represent? “
“What is the essence of an electric field? “
Electric Field
Unit 2, Slide 4
q4
q2
q3
q
FE
rr
QkE ˆ
2
i
i
i
i rr
QkE ˆ
2
Electricity & Magnetism Lecture 2, Slide 5
We never talk about the “gravitational field” in 2111.
We just talked about a “gravitational force”.
Let’s say we had. What is the magnitude of the
gravitational field close to the surface of the earth?
A) g
B) mg
C) m/g
D) g/m
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. Up
B. Down
C. Left
D. Right
E. Zero
CheckPoint: Electric Fields1
Unit 2, Slide 6
A
Bx+Q -Q
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. Up
B. Down
C. Left
D. Right
E. Zero
CheckPoint: Electric Fields2
Electricity & Magnetism Lecture 2, Slide 7
A
Bx+Q -Q
Question
Electricity & Magnetism Lecture 2, Slide 8
P
d
What is the direction of the
electric field at point P, the
unoccupied corner of the
square? -q +q
+q
d
A)
B)
C)
D)Need to know d
Need to know
sign of charge
placed at point P
E)
0E
Example 2.1 (Field from three charges)
Electricity & Magnetism Lecture 2, Slide 9
Calculate E at
point P. d
P
d
-q +q
+q1
2
3
CheckPoint: Magnitude of Field (2 Charges)
Electricity & Magnetism Lecture 2, Slide 10
In which of the two cases shown below is the magnitude of the electric
field at the point labeled A the largest?
A. Case 1
B. Case 2
C. Equal
+Q
+Q
+Q
-Q
A
A
Case 1 Case 2
Two charges q1 and q2 are fixed at points (-a,0) and (a,0) as
shown. Together they produce an electric field at point (0,d) which
is directed along the negative y-axis.
x
y
q1 q2
(-a,0) (a,0)
(0,d)
Which of the following statements is true:
A) Both charges are negative
B) Both charges are positive
C) The charges are opposite
D) There is not enough information to tell how the charges are
related
Question: Two Charges
Unit 2, Slide 11
E
_ _+ +
_+
Electricity & Magnetism Lecture 2, Slide 12
A 6uC charge is placed in
an electric field and feels a
0.18N force. If the charge
is replaced by a 9uC
charge, what force will the
new charge feel?
Question
Unit 2, Slide 13
E q1
A) 0N
B) 0.09N
C) 0.18N
D) 0.27N
E) 0.36N
CheckPoint Results: Motion of Test Charge
Electricity & Magnetism Lecture 2, Slide 14
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. To the left
B. To the right
C. Stay still
D. Other
xq1 Q2
(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?
Example 2.2 (Zero Electric Field)
Unit 2, Slide 15
(0,0)
l Q/L
Summation becomes an integral (be careful with vector
nature)
“I don't understand the whole dq thing and lambda.”
WHAT DOES THIS
MEAN ?
Integrate over all charges (dq)
r is vector from dq to the point at which E is defined
r
dE
Continuous Charge Distributions
Electricity & Magnetism Lecture 2, Slide 16
Linear Example:
charges
pt for E
dq l dx
i
i
i
i rr
QkE ˆ
2
r
r
dqkE ˆ
2
“Please go over infinite line
charge.”
Example 2.3 (line of charge)
Electricity & Magnetism Lecture 2, Slide 17
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)?
Let’s do one slightly different.
x
y
a
h
P
x
r
dq l dx
Question: Calculation
Electricity & Magnetism Lecture 2, Slide 18
2x
dxA) B) C) D) E)
What is ? 2r
dq
22 ha
dx
+22 ha
dx
+
l22 hx
dx
+
l2x
dxl
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)?
x
y
a
h
P
x
r
dq l dx
We know:
rr
dqkE ˆ
2
Question: Calculation
Electricity & Magnetism Lecture 2, Slide 19
We know:
222 hx
dx
r
dq
+
l xx dEE
What is ?
A) B) C) D)
xdE
2cosdE2cosdE-
1sindE 2sindE
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)?
rr
dqkE ˆ
2
xa
P
x
r
1
2
dq l dx
xdE
dEy
2sindE-E)
h
We know:
sin2 DEPENDS ON x!
Question: Calculation
Electricity & Magnetism Lecture 2, Slide 20
- 2sindEdEE xx222 hx
dx
r
dq
+
l
What is ?
A) B)
C) neither of the above
xE
-+
-222sin
hx
dxak l +
-
a
hx
dxk
0
222sinl
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)?
rr
dqkE ˆ
2
xa
P
x
r
2
dq l dx
xdE
dEy
We know:
Clicker Question: Calculation
Electricity & Magnetism Lecture 2, Slide 21
2sindEdEE xx222 hx
dx
r
dq
+
l
rr
dqkE ˆ
2
What is x in terms of Q ?
A) B) C) 22 ha
a
+ 22)( hxa
a
+-D)x = h*tanQ2 x = h*cosQ2 x = h*sinQ2 x = h / cosQ2
xa
P
x
r
2
dq l dx
xdE
dEy
PROBLEM: We have two variables Q and x
Observation
Electricity & Magnetism Lecture 2, Slide 22
xa
P
x
r2
dq l dx
xdE
dEy
+1
22 ah
h
-
+ 1)(
22 ah
h
h
kPEx
l
Note:
Ex < 0
make sense?
For an infinite line of
charge, we had:
xdE
Back to the pre-lecture
Electricity & Magnetism Lecture 2, Slide 23
)(PEx
xa
P
x
r
dq l dx
xdE
dEy
How would this integral change we wanted the y component
instead of the x component?
-
f
dh
k
l
0
*sin
A) The limits would be Q1 to –Q1
B) The limits would be +/- infinity
C) The limits would be -p/2 to p/2
D) sinQ would turn to tanQ
E) sinQ would turn to -cosQ
We had:
xdE
Back to the pre-lecture
Electricity & Magnetism Lecture 2, Slide 24
( )22
2
0)tan*(
)cos(*sec*)(
hh
hdkPE
f
y+
--
l
“Please go over infinte line charge. How does R get outside the intergral?”
xa
P
x
r
dq l dx
xdE
dEy
How would this integral change if the line of charge were
infinite in both directions?
f
dh
k
l
0
*cos
A) The limits would be Q1 to –Q1
B) The limits would be + to -
C) The limits would be -p/2 to p/2
D) sinQ would turn to tanQ
E) sinQ would turn to -cosQ8 8
k in terms of fundamental constants
Unit 2, Slide 25
Note
o
kp4
1
o
liner
Ep
l
2
r
kEline
l2
CheckPoint: Two Lines of Charge
Electricity & Magnetism Lecture 2, Slide 26
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
Example 2.4 (E-field above a ring of charge)
Unit 2, Slide 27
a
h
P
What is the electric field a
distance h above the
center of ring of uniform
charge Q and radius a?
y
x
Question
Unit 2, Slide 28
In the previous question,
what should the electric
field be if h>>>>a?
a) k Q / h2
b) k Q / a2
c) k Q / (a*h)
d) k Q a*h
Example 2.4 (E-field above a ring of charge)
Unit 2, Slide 29
a
h
P
What is the electric field a
distance h above the
center of ring of uniform
charge Q and radius a?
y
x
A total charge Q is uniformly
distributed over a half ring
with radius R. The total
charge inside a small
element dθ is given by:
p dR
Q
2
A.
p dR
Q
B.
R
Q
p
C.
p dQ
D.
p dQ
2
E.
θ
dθ
Q
R
Question
A total charge Q is uniformly
distributed over a half ring
with radius R. The Y
component of electric field
created by a short element
dθ is given by:
p
sin2R
kQd
A.
p
cos2R
kQd
B.
p
sin3R
kQdC.
p
cos3R
kQdD.
y
x
Question
A total charge Q is uniformly
distributed over a half ring
with radius R. The total
electric field at point C is:
2R
kQC. D.
y
x
Question
2R
kQ
p
A.
2
2
R
Qk
p
B.
0
dipoles
Unit 2, Slide 33
q q
d
- +
Dipole moment = p = qd
Points from negative to positive.(opposite the electric field.)
q qCl Na
Water molecule has dipole moment of 620X10-30 Cm
= 2 * 1.6X10-19C * 190pm
Torque on dipole
Unit 2, Slide 34
q
q
-
+ = 2*(qE X d/2)
= p X E
dU= -dW = -dQ = pE*sinQ*dQ
DU - pEcosQ
Define U = when Q = p/2
U -p E
Q
Example 2.6 (Salt Dipole)
Unit 2, Slide 35
-
+
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?
20o+x
A. (a)
B. (b)
C. (c)
D. (d)
E. (c) and (d)
An electrically neutral dipole
is placed in an external field.
In which situation(s) does
the potential energy of the
dipole have the smallest
magnitude using the
standard definition of U?
(a) (b)
(c) (d)
+-
+ -
+- +
-
A. (a)
B. (b)
C. (c)
D. (d)
E. (c) and (d)
An electrically neutral dipole
is placed in an external field.
In which situation(s) does
the potential energy of the
dipole have the lowest value
using the standard definition
of U?
(a) (b)
(c) (d)
+-
+ -
+- +
-