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PHY 204: Elementary Physics II -- Slides PHY 204: Elementary Physics II (2021)
2020
01. Introduction. Electric charge. Coulomb force. Electric 5eld 01. Introduction. Electric charge. Coulomb force. Electric 5eld
Gerhard Müller
University of Rhode Island
, [email protected]
Robert Coyne
University of Rhode Island
, [email protected]
Follow this and additional works at: https://digitalcommons.uri.edu/phy204-slides
Recommended Citation Recommended Citation
Müller, Gerhard and Coyne, Robert, "01. Introduction. Electric charge. Coulomb force. Electric 5eld" (2020).
PHY 204: Elementary Physics II -- Slides.
Paper 26.
https://digitalcommons.uri.edu/phy204-slides/26
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Mechanics
Kinematics:
• position: ~r =
Z
~v dt
• velocity: ~v =
d~r
dt
=
Z
~a dt
• acceleration: ~a =
d~v
dt
=
d
2
~r
dt
2
Dynamics:
~
F = m~a (cause and effect)
Modes of motion: translation, rotation, oscillation
Conservation laws: energy, momentum, angular momentum
Effective forces: elastic, contact, friction, weight, ...
Fundamental interaction: gravitational, electromagnetic, strong, weak
tsl1
Electricity and Magnetism
Particles generate fields:
• Massive particle generates gravitational field.
• Charged particle generates electric field.
• Moving charged particle generates electric field and magnetic field.
Fields exert force on particles:
• Gravitational field exerts force on massive particle.
• Electric field exerts force on charged particle.
• Magnetic field exerts force on moving charged particle.
Dynamics:
• Cause and effect between particles and fields and among fields.
Energy and momentum:
• Particles carry energy (kinetic, potential) and momentum.
• Fields carry energy (electric, magnetic) and momentum.
tsl2
Atomic Structure of Matter
Periodic table: ∼100 elements.
Building blocks of atoms: fundamental particles.
particle charge mass
electron q
e
= −e m
e
= 9.109 × 10
−31
kg
proton q
p
= +e m
p
= 1.673 × 10
−27
kg
neutron q
n
= 0 m
n
= 1.675 × 10
−27
kg
• SI unit of charge: 1C (Coulomb).
• Elementary charge: e = 1.602 × 10
−19
C.
• Atomic nuclei (protons, neutron) have a radius of ∼ 1fm = 10
−15
m.
• Atomic electron shells have a radius of ∼ 1
˚
A = 10
−10
m.
• Atoms are electrically neutral (equal numbers of electrons and protons).
• Ions: atoms with one or several electrons added or removed.
• Isotopes: atoms differing in the number of neutrons.
• Positively (negatively) charged objects have a deficiency (surplus) of electrons.
tsl3
Periodic Table of Elements
tsl368
Charging Up Insulators and Conductors
tsl369
Charging Up Insulators and Conductors
tsl369
Coulomb’s Law (1)
Electrostatic force between two charged particles:
F =
1
4πe
0
|q
1
q
2
|
r
2
= k
|q
1
q
2
|
r
2
Permittivity constant: e
0
= 8.854 × 10
−12
C
2
N
−1
m
−2
Coulomb constant: k = 8.99 × 10
9
Nm
2
C
−2
+
+
+
q q
1 2
F F
1221
r
+
Action-reaction pair of forces:
~
F
21
= −
~
F
12
.
Newton’s law of gravitation (for comparison)
Gravitational force between two massive particles:
F = G
m
1
m
2
r
2
Gravitational constant: G = 6.673 × 10
−11
Nm
2
kg
−2
F
21
F
12
r
m
1
m
2
tsl4
Coulomb’s Law (1)
Electrostatic force between two charged particles:
F =
1
4πe
0
|q
1
q
2
|
r
2
= k
|q
1
q
2
|
r
2
Permittivity constant: e
0
= 8.854 × 10
−12
C
2
N
−1
m
−2
Coulomb constant: k = 8.99 × 10
9
Nm
2
C
−2
+
+
+
q q
1 2
F F
1221
r
+
Action-reaction pair of forces:
~
F
21
= −
~
F
12
.
Newton’s law of gravitation (for comparison)
Gravitational force between two massive particles:
F = G
m
1
m
2
r
2
Gravitational constant: G = 6.673 × 10
−11
Nm
2
kg
−2
F
21
F
12
r
m
1
m
2
tsl4
Coulomb’s Law (2)
Coulomb’s law for electrostatic force in vector form:
~
F
12
= k
q
1
q
2
r
2
12
ˆ
r
12
,
~r
12
.
=~r
2
−~r
1
,
ˆ
r
12
.
=
~r
12
r
12
.
Electric force in hydrogen atom:
Average distance: r = 5.3 × 10
−11
m.
Elementary charge: e = 1.60 × 10
−19
C.
F = k
|q
1
q
2
|
r
2
=
(8.99 × 10
9
Nm
2
/C
2
)(1.60 × 10
−19
C)
2
(5.3 × 10
−11
m)
2
= 8.2 × 10
−8
N.
tsl370
Coulomb’s Law (2)
Coulomb’s law for electrostatic force in vector form:
~
F
12
= k
q
1
q
2
r
2
12
ˆ
r
12
,
~r
12
.
=~r
2
−~r
1
,
ˆ
r
12
.
=
~r
12
r
12
.
Electric force in hydrogen atom:
Average distance: r = 5.3 × 10
−11
m.
Elementary charge: e = 1.60 × 10
−19
C.
F = k
|q
1
q
2
|
r
2
=
(8.99 × 10
9
Nm
2
/C
2
)(1.60 × 10
−19
C)
2
(5.3 × 10
−11
m)
2
= 8.2 × 10
−8
N.
tsl370
Coulomb Force in One Dimension (1)
Find net force on charge q
0
due to charges q
1
and q
2
.
Consider x-component of force.
F
0
= +k
|q
1
q
0
|
(3.5m)
2
− k
|q
2
q
0
|
(1.5m)
2
= +3.67 × 10
−7
N − 7.99 × 10
−7
N = −4.32 × 10
−7
N.
Find net force on charge q
2
due to charges q
1
and q
0
.
F
2
= −k
|q
1
q
2
|
(2.0m)
2
+ k
|q
2
q
0
|
(1.5m)
2
= −5.62 × 10
−7
N + 7.99 × 10
−7
N = +2.37 × 10
−7
N.
tsl5
Coulomb Force in One Dimension (2)
Three particles with charges of magnitude 1C are positioned on a straight line with two equal spacings.
q
q q
1 2 3
(a) Find the direction (left/right) of the net forces
~
F
1
,
~
F
2
,
~
F
3
on each particle.
(b) Which force is the strongest and which force is the weakest?
tsl6
Coulomb Force in One Dimension (3)
Four point charges equal magnitude are lined up in three different configurations.
The Coulomb force between nearest neighbors is 4N.
(a)
(b)
(c)
Find direction and magnitude of the net force experienced by the green particle in each configuration.
tsl7
Coulomb Force in One Dimension (4)
Three charged particles are positioned along a straight line with two equal spacings .
The net Coulomb force on charge q
3
happens to vanish.
q = ?
q =1
2
1
q = 2
3
F = 0
3
d d
Cµ Cµ
What is the value of q
1
?
tsl8
Coulomb Force in One Dimension (5)
How are the forces
~
F
1
and
~
F
2
in (a) affected by the changes made in (b) and (c)?
What changes if the charge q
2
is made negative?
tsl347
Coulomb Force in Two Dimensions (1a)
Find the magnitude and direction
of the resultant force on charge q
0
.
• Magnitude of individual forces:
F
1,0
= k
|q
1
q
0
|
r
2
1,0
= 5.62 × 10
−7
N, F
2,0
= k
|q
2
q
0
|
r
2
2,0
= 6.74 × 10
−7
N.
• Components of individual forces:
F
x
1,0
= F
1,0
cos 45
◦
, F
y
1,0
= F
1,0
sin 45
◦
, F
x
2,0
= 0, F
y
2,0
= −F
2,0
.
tsl9
Coulomb Force in Two Dimensions (1b)
• Components of resultant force:
F
x
= F
x
1,0
+ F
x
2,0
= 3.97 × 10
−7
N, F
y
= F
y
1,0
+ F
y
2,0
= −2.77 × 10
−7
N.
• Magnitude of resultant force: F =
q
F
2
x
+ F
2
y
= 4.84 × 10
−7
N.
• Direction of resultant force: θ = arctan
F
y
/F
x
= −34.9
◦
.
tsl325
Coulomb Force in Two Dimensions (2)
The unknwon point charges q
1
, q
2
exert a force F
0
= 2N on the known point charge q
0
= 1nC. This force is
directed in the positive y-direction as shown.
Determine first whether q
1
, q
2
are positive or negative. Then determine the values of the two point charges.
q
0
q
1
q
2
y
x
3m
3m
F
0
q
0
q
1
q
2
y
x
3m
3m
F
0
F
F
2
1
0
0
tsl10
Coulomb Force in Two Dimensions (2)
The unknwon point charges q
1
, q
2
exert a force F
0
= 2N on the known point charge q
0
= 1nC. This force is
directed in the positive y-direction as shown.
Determine first whether q
1
, q
2
are positive or negative. Then determine the values of the two point charges.
q
0
q
1
q
2
y
x
3m
3m
F
0
q
0
q
1
q
2
y
x
3m
3m
F
0
F
F
2
1
0
0
tsl10
Coulomb Force in Two Dimensions (3)
Point charges of equal magnitude are positioned at the corners of an equilateral triangle.
q = 1nC
1
q = −1nC
2
q = −1nC
3
• Copy this configuration and indicate by arrows the direction of the resultant force on each point charge.
• Which point charge experiences the strongest force?
tsl11
Coulomb Force in Two Dimensions (4)
Point charges of equal magnitude are positioned at the corners of a square.
• Copy this configuration and indicate by arrows the direction of the resultant force on each point charge.
• If the force between nearest-neighbor charges is 1N, what is the strength of the resultant force on each
charge?
tsl12
Coulomb Force in Two Dimensions (5)
Two identical small charged spheres, each having a mass m = 30g, hang in equilibrium at an anlge of θ = 5
◦
from the vertical. The length of the strings is L = 15cm.
L L
θ
θ
q
q
mg
F
e
T
• Identify all forces acting on each sphere.
• Find the magnitude of the charge q on each sphere.
tsl13
Coulomb Force in Two Dimensions (5)
Two identical small charged spheres, each having a mass m = 30g, hang in equilibrium at an anlge of θ = 5
◦
from the vertical. The length of the strings is L = 15cm.
L L
θ
θ
q
q
mg
F
e
T
• Identify all forces acting on each sphere.
• Find the magnitude of the charge q on each sphere.
tsl13
Electric Field of a Point Charge
electric
charge
electric
field
electric
charge
generates
locally
exerts force
exerts force over distance
(1) Electric field
~
E generated by point charge q:
~
E = k
q
r
2
ˆ
r
(2) Force
~
F
1
exerted by field
~
E on point charge q
1
:
~
F
1
= q
1
~
E
(1+2) Force
~
F
1
exerted by charge q on charge q
1
:
~
F
1
= k
qq
1
r
2
ˆ
r (static conditions)
• e
0
= 8.854 × 10
−12
C
2
N
−1
m
−2
• k =
1
4πe
0
= 8.99 × 10
9
Nm
2
C
−2
+
q
r
r
source point
r
r
pointfield
point
field
^
^
E
E
tsl14
Superposition Principle for Electric Field
Electric field on line connecting two point charges:
+
q
E
q
E E
E
EE
1
2 2
2
1
1
Electric field of point charges in plane:
+
E
E
E
E
E
E
q
q
2
1
1
2
1
2
tsl15
Vector Field and Electric Field Lines
• The electric field is a vector field:
~
E(~r) =
~
E(x, y, z) = E
x
(x, y, z)
ˆ
i + E
y
(x, y, z)
ˆ
j + E
z
(x, y, z)
ˆ
k
• Electric field lines: graphical representation of vector field.
• Properties of electric field lines (electrostatics):
• Electric field lines begin at
positive charges or at infinity.
• Electric field lines end at
negative charges or at infinity.
• The direction of
~
E is tangential to
the field line going through the field point.
• Electric field lines bunched together
indicate a strong field.
• Electric field lines far apart
indicate a weak field.
• Field lines do not intersect.
tsl22
Electric Field on Line Connecting Point Charges (1)
Consider the x-component of the electric field.
• Electric field at point P
1
:
E = E
1
+ E
2
=
kq
1
(7m)
2
+
kq
2
(3m)
2
= 1.47N/C + 12.0N/C = 13.5N/C.
• Electric field at point P
2
:
E = E
1
+ E
2
=
kq
1
(3m)
2
−
kq
2
(1m)
2
= 7.99N/C − 108N/C = −100N/C.
20 1
3
2
4 5 6
7
x [m]
2
P
1
P
q = +8nC
1
q =+12nC
tsl19
Electric Field on Line Connecting Point Charges (2)
Four particles with charges of equal magnitude are positioned on a horizontal line in six different
configurations.
(a)
(b)
(c)
(d)
(e)
(f)
Determine for each configuration the direction of the resultant electric field (left/right/zero) at the location
indicated by ×.
tsl16
Electric Field on Line Connecting Point Charges (3)
• Is the unknown charge positive or negative?
• What is the value of the unknown charge?
(a)
(b)
?
9nC
2m
3m
3m
2m
9nC
E=0
E=0
?
+
+
tsl17
Electric Field of Point Charges in Plane (1)
Determine magnitude of
~
E
1
and
~
E
2
and identify directions in plane:
E
1
=
k|q
1
|
(3m)
2
=7.99N/C, E
2
=
k|q
2
|
(5m)
2
=4.32N/C.
Determine x- and y-components of
~
E
1
and
~
E
2
and of the resultant field
~
E:
E
x
1
= 0, E
y
1
= 7.99N/C;
E
x
2
= −3.46N/C, E
y
2
= 2.59N/C;
E
x
= −3.46N/C, E
y
= 10.6N/C.
Determine magnitude and direction of
~
E:
E =
q
E
2
x
+ E
2
y
= 11.2N/C, θ = arctan
E
y
E
x
= 108
◦
.
20 1
3 4
[m]
x
1
2
3
2
q =+12nC
1
q = +8nC
E
2
E
1
E
θ
E
tsl18
Electric Field of Point Charges in Plane (2)
(a) Find the electric charge q
2
.
45
o
E
q = ?
2
q = 3nC
1m
1m
1
+
(b) Find the angle θ.
1m
E
1m
q = 4nC
2
q = 2nC
1
θ
= ?
tsl20
Electric Field of Point Charges in Plane (3)
Two point charges, one known and the other unknown, produce a horizontal electric field as shown.
What is the value of the unknown charge?
1nCq = ?
5m
5m
E
tsl21
Electric Field of Point Charges in Plane (4)
Consider four triangles with point charges of equal magnitude at two of the three corners.
a
a
a a a
a
a a
a
a
(1)
(2)
(3) (4)
(a) Determine the direction of the electric field
~
E
i
at the third corner of triangle (i).
(b) Rank the fields E
i
according to strength.
tsl326
Electric Field of Point Charges in Plane (5)
Find magnitude and direction of the resultant electric field at point P.
• E
1
=
k|q
1
|
8m
2
= 3.38 N/C.
• E
2
=
k|q
2
|
4m
2
= 6.75 N/C.
• E
3
=
k|q
3
|
8m
2
= 3.38 N/C.
• E
x
= E
1
cos 45
◦
+ E
3
cos 45
◦
= 4.78 N/C.
• E
y
= E
2
= 6.75 N/C.
• E =
q
E
2
x
+ E
2
y
= 8.27 N/C.
• tan θ =
E
y
E
x
= 1.41.
• θ = arctan 1.41 = 54.7
◦
.
x
q = +3nC q = −3nC
32
q = +3nC
1
2m2m
2m
E
E
3
1
E
E
E
y
x
E
2
θ
P
tsl371
Intermediate Exam I: Problem #1 (Spring ’05)
The electric field
~
E generated by the two point charges, 3nC and q
1
(unknown), has the direction shown.
(a) Find the magnitude of
~
E.
(b) Find the value of q
1
.
2m
3nC
4m
E
45
o
y
q
1
x
Solution:
(a) E
y
= k
3nC
(2m)
2
= 6.75N/C,
E
x
= E
y
,
E =
q
E
2
x
+ E
2
y
= 9.55N/C.
(b) E
x
= k
(−q
1
)
(4m)
2
,
q
1
= −
(6.75N/C)(16m
2
)
k
= −12nC.
tsl331
Intermediate Exam I: Problem #1 (Spring ’05)
The electric field
~
E generated by the two point charges, 3nC and q
1
(unknown), has the direction shown.
(a) Find the magnitude of
~
E.
(b) Find the value of q
1
.
2m
3nC
4m
E
45
o
y
q
1
x
Solution:
(a) E
y
= k
3nC
(2m)
2
= 6.75N/C,
E
x
= E
y
,
E =
q
E
2
x
+ E
2
y
= 9.55N/C.
(b) E
x
= k
(−q
1
)
(4m)
2
,
q
1
= −
(6.75N/C)(16m
2
)
k
= −12nC.
tsl331
Intermediate Exam I: Problem #1 (Spring ’05)
The electric field
~
E generated by the two point charges, 3nC and q
1
(unknown), has the direction shown.
(a) Find the magnitude of
~
E.
(b) Find the value of q
1
.
2m
3nC
4m
E
45
o
y
q
1
x
Solution:
(a) E
y
= k
3nC
(2m)
2
= 6.75N/C,
E
x
= E
y
,
E =
q
E
2
x
+ E
2
y
= 9.55N/C.
(b) E
x
= k
(−q
1
)
(4m)
2
,
q
1
= −
(6.75N/C)(16m
2
)
k
= −12nC.
tsl331