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
12212
21 r̂r
qkq
MATH:
12F
= 12212
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
14214
41132
13
31122
12
21 ˆˆˆ rr
qkqr
r
qkqr
r
qkq
MATH:
1F
= 14214
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 andAway 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
ii
i rr
QkE ˆ
2
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. UpB. DownC. Left D. RightE. Zero
CheckPoint: Electric Fields1
Unit 2, Slide 5
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. UpB. DownC. Left D. RightE. Zero
CheckPoint: Electric Fields2
Electricity & Magnetism Lecture 2, Slide 6
A
Bx+Q -Q
Example 2.1 (Field from three charges)
Electricity & Magnetism Lecture 2, Slide 7
Calculate E at point P. d
P
d
-q +q
+q1
2
3
CheckPoint: Magnitude of Field (2 Charges)
Electricity & Magnetism Lecture 2, Slide 8
In which of the two cases shown below is the magnitude of the electric field at the point labeled A the largest?
A. Case 1B. Case 2C. Equal
+Q
+Q
+Q
-Q
A
A
Case 1 Case 2
CheckPoint Results: Motion of Test Charge
Electricity & Magnetism Lecture 2, Slide 10
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 leftB. To the rightC. Stay stillD. 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 11
(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 12
Linear Example:
charges
pt for E
dq = l dx
i
ii
i rr
QkE ˆ
2
r
r
dqkE ˆ
2
Clicker Question: Charge Density
Electricity & Magnetism Lecture 2, Slide 13
Linear (l = Q/L) Coulombs/meter
Surface (s = Q/A) Coulombs/meter2
Volume (r = Q/V) Coulombs/meter3
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.
“I would like to know more about the charge density.”
Some Geometry
24 RAsphere
334 RVsphere LRVcylinder
2
RLAcylinder 2
334 RVQA
24 RAQB R
R
R
Q
Q
B
A
3
1
4 2
334
“Please go over infinite line charge.”
Example 2.3 (line of charge)
Electricity & Magnetism Lecture 2, Slide 14
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.
y
a
h
P
x
r
dq = l dx
Clicker Question: Calculation
Electricity & Magnetism Lecture 2, Slide 15
2x
dxA) B) C) D) E)
What is ? 2r
dq
22 ha
dx
22 ha
dx
22 hx
dx
2x
dx
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
Clicker Question: Calculation
Electricity & Magnetism Lecture 2, Slide 16
We know:
222 hx
dx
r
dq
xx dEE
What is ?
A) B) C) D)
xdE
1cosdE 2cosdE1sindE 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
q1
q2
dq = l dx
xdE
dEy
We know:
sinq2 DEPENDS ON x!
Clicker Question: Calculation
Electricity & Magnetism Lecture 2, Slide 17
2sindEdEE xx222 hx
dx
r
dq
What is ?
A) B)
C) neither of the above
xE
222sinhx
dxk
a
hx
dxak
0222sin
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
q1
q2
dq = l dx
xdE
dEy
We know:
Clicker Question: Calculation
Electricity & Magnetism Lecture 2, Slide 18
2sindEdEE xx222 hx
dx
r
dq
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)?
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
q1
q2
dq = l dx
xdE
dEy
We know:
xdE
Calculation
Electricity & Magnetism Lecture 2, Slide 19
222 hx
dx
r
dq
2cosdEdEE xx
What is ?
22
2
0 )tan*(
sin*sec*)(
hh
hdkPEf
x
)(PEx
1)(
22 ah
hkPEx
rr
dqkE ˆ
2
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)?
x = h*tanQ
xa
P
x
r
q1
q2
dq = l dx
xdE
dEy
dx = h*sec2 Q dq
Exercise for student:
Change variables: write Q in terms of x
Result: do integral with trig sub
Observation
Electricity & Magnetism Lecture 2, Slide 20
xa
P
x
r
q1
q2
dq = l dx
xdE
dEy
1
22 ah
h
1)(
22 ah
hkPEx
since
Ex < 0
make since?
We had:
xdE
Back to the pre-lecture
Electricity & Magnetism Lecture 2, Slide 21
22
2
0 )tan*(
sin*sec*)(
hh
hdkPEf
x
“Please go over infinte line charge. How does R get outside the intergral?”
xa
P
x
r
q1
q2
dq = l dx
xdE
dEy
How would this integral change if the line of charge were infinite in both directions?
hdkf
sin
0
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
For an infinite line of charge, we had:
xdE
Back to the pre-lecture
Electricity & Magnetism Lecture 2, Slide 22
)(PEx
xa
P
x
r
q1
q2
q2
dq = l dx
xdE
dEy
How would this integral change we wanted the y component instead of the x component?
hdk
sin2/
2/
A) The limits would be Q1 to –Q1
B) The limits would be +/- infinity
C) The limits would be - /2p to /2p
D) sinQ would turn to tanQ
E) sinQ would turn to cosQ
CheckPoint: Two Lines of Charge
Electricity & Magnetism Lecture 2, Slide 24
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 25
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
Example 2.5 (E-field above a disk)
Unit 2, Slide 26
a
h
P
What is the electric field a distance h above the center of disk of uniform charge Q and radius a?
y
x
dipoles
Unit 2, Slide 27
q q
d
- +
Dipole moment = p = qd
Points from negative to positive.(opposite the electric field.)
q qCl Na
Torque on dipole
Unit 2, Slide 28
q
q
-+ t = 2*(qE X d/2)
= p X E
dU= -dW = tdQ = pE*sin *Q dQDU= -pEcosQ
Define U = when Q = p/2
U= -p E
Q