Chapter 11
Electric Charge
and
Electric Field
1
Electric Charge:
A property involving the interactions between particles, e.g. rub amber
with wool to get the amber charged.
Electrostatics: the interactions between electric charges that are at rest
+
Positive charge
_
Negative charge
2 positive charges :
repel each other
2 negative charges :
repel each other
1 positive charge, 1 negative charge :
+
+
_
attract each other
_
_
+
Like charges repel and unlike charges attract
2
+
_
Electric field (E)
points away from
a positive charge
and points toward
a negative
charge.
+
+
3
Structure of atoms:
electron (-e)
electron
proton
neutron
: -e negative charge
: +e positive charge
: uncharged
e
proton (+e)
Nucleus: A very dense core making up by protons and neutrons in an atom
and with a size of about 10-15 m
Neutral atom: The number of electrons = the number of protons in the
nucleus. There is no net electric charge.
Atomic number:
Positive ion:
The number of electrons or protons in a neutral atom of
an element.
an atom that has lost one ore more electrons thru ionization
Negative ion: an atom that has gained one ore more electrons
4
Conservation of charge: The sum of all the electric charges in any closed
system remains constant.
•
Charge is not created or destroyed
•
Charge is conserved
•
Charge transfers from one object to another
Conductor: Materials that allow electric charge to move freely from one
point to another, e.g. metals
Insulator:
Material that cannot allow electric charge and free electrons to
transfer, e.g. non-metals, plastics, glass
Semi conductor: The electrical properties of the materials that are
between conductors and insulators, e.g silicon
Induction:
The process to give an object a charge
5
Coulomb’s Law – Charles Augustin de Coulomb (1736 – 1806)
The electric force between two stationary point charges is
•
•
proportional to the product of the charges
inverse proportional to the square of the distance between two point
charges
→
F =k
Where
r
Q1Q2
r2
F2 on 1
+
F = magnitude of electric force between two
point charges (a vector and is always
positive) [Newton]
k = proportional constant (~ 8.988x109
Nm2/C2)
Q1 and Q2 = electric charges (either positive
or negative); [Coulomb or “C”]
r = separation distance between two point
charges [m]
+
Q1
F1 on 2
Q2
r
Q1
F
F2 on 1 1 on 2
Point charges have opposite sign ⇒ attractive force
Point charges have same sign ⇒ repulsive force
+
Q2
6
Alternative Form of Coulomb’s Law
→
F=
where
1
Q1Q2
4πε 0
r2
k=
1
4πε 0
= 8.988 × 109 Nm 2 / C 2
ε0 = 8.854 x 10-12 C2/(Nm2) : another constant
electron (-e)
e
proton (+e)
e ~ 1.60217733 x 10-19 C
1 C ~ 6 x 1018 electron/proton
1 Coulomb ~ a total charge (negative) of about 6 x 1018 electrons
7
Electric Force
→
F = k2
Q1Q2
Both attractive force and
repulsive force
r2
Analogy to
→
Gravitational Force
M 1M 2
Fg =G
r2
Attractive force only
r
M2
M1
8
Example 11.1: Source: University Physics
Two small spheres spaced 20 cm apart have equal charge. How many excess
electrons must be present on each sphere if the magnitude of the force of repulsion
between them is 4.57 x 10-21 N.
Solution:
First find the total charge on the spheres:
1 Q2
F=
4π ∈0 r 2
⇒ Q = 4π ∈0 Fr 2
= 4π ∈0 ( 4.57 ×10−21 )(0.2)2
= 1.43 ×10−26 C
and therefore, the total number of electrons requires is:
n = Q / e = 1.43 x 10-16 C/1.60 x 10-19 C = 890.
9
Example 11.2: Source: University Physics
A negative charge –0.55 µC exerts an upward 0.2 N force on an unknown charge
0.3 m directly below it. a) What is the unknown charge (magnitude and sign)? b)
What are the magnitude and direction of the force that the unknown charge exerts
on the –0.55 µC charge?
1 µ = 10-6
Solution:
(a)
1 q1q2
1 (0.550 ×10 −6 C )q2
F=
⇒ 0.2 N =
2
4π ∈0
(0.30 m) 2
4π ∈0 r
⇒ q2 = +3.64 ×10 −6 C.
(b) F = 0.200 N, and is attractive (because the unknown charge is placed
below the negative charge and the force exerted on the unknown charge is an
upward force).
10
Principle of Superposition
In the presence of three or more point charges, the total electric
force on any one of the charges is equal to the vector sum of the
forces exerted by each of the individual charges.
F1 = F2 on 1 + F3 on 1 + ... + FN on 1
or
F1 = F21 + F31 + ... + FN1
11
Example 11.3: Determine the magnitude and direction of the resultant electric force
on Q3. Given that Q1 = 4 µC, Q2 = 4 µC, Q3 = 5 µC, and the distances between
charges are shown in the figure.
1µ = 10−6
QQ
F = k 12 2
r
Distance between Q1 and Q 3 = (0.4) 2 + (0.5) 2 = 0.64m
4 × 10 −6 C ⋅ 5 ×10 −6 C
F13 = 9 × 10 Nm / C
= 0.44 N
2
0.64
9
2
2
By symmetric, the magnitude of electric forces F23 =
F13 = 0.44N
( F13 ) x = F13 cosθ = 0.44 cos 38.7 = 0.343N
y
Q1
( F23 ) x = F23 cosθ = 0.44 cos 38.7 = 0.343N
+
0.4
0.5
θ = 38.7 o
tan θ =
( F3 ) x = ( F13 ) x + ( F23 ) x = 0.343 + 0.343 = 0.686 N 0.4 m
0
( F13 ) y = F13 sin θ = 0.44 sin 38.7 = 0.275 N
( F23 ) y = − F23 sin θ = 0.44 sin 38.7 = −0.275 N
( F3 ) y = ( F13 ) y + ( F23 ) y = 0.275 − 0.275 = 0 N
θ
0.5 m
Q3
F23
x
+
F13
0.4 m
+
Q2
12
Electric Field (E)
Electric field is created in the space when a charged object is
present and can be used to describe an electric force that acts at a
certain distance.
Electric field that is created by any charged objects would exert
electric force on another charged object.
→
→
E = electric force per unit charge =
Where
F0
Q0
E = electric field (N/C), electric field is a vector
F0 = electric force acting on a positive test charge (N)
Q0 = a positive test charge (C)
13
Electric force exerted on Q 0 by an electric field : F0 = Q0 E
Q=+
Q=-
: F0 has the same direction as E
: F0 has opposite direction as E
E
E
F0
Q0
Q0 +
-
F0
Magnitude of electric field of a point charge:
r
F0 =
1
QQ0
E=
4πε 0 r 2
E=
1
F0
Q0
-
4πε 0 r
+
E
Q
Q0
r
Q
2
^r
+
Q
^r
E
+
Q0
14
Using a unit vector ( r^ ) to indicate the direction, we have
→
E=
Q
1
4πε 0 r
2
^
r
Electric Field of a Point Charge
Note: The electric field (E) points away from a positive charge (same direction
as r) and points toward a negative charge.
+
_
+
+
15
Total electric field due to the effect of all the charges:
→
→
→
Qi ^
F0 → → →
E=
= E 1 + E 2 + E 3 + ... + E N = ∑ ke 2 ri
Q0
ri
i
Under the condition of electrostatics of which the charges
have no net motion, the electric field at every point within
the material of a conductor = 0
+
+
R
+
+
+
+
+
+
16
Example 11.4: Source: University Physics
A particle has charge –3.0 nC. a) Find the magnitude and direction of the electric
field due to this particle at a point 0.25 m directly above it. b) At what distance from
this particle does its electric field have a magnitude of 12.0 N/C.
1 nC = 10-9 C
Solution:
1 Q
1 (3.00 ×10 −9 C )
E=
=
= 432 N / C down toward the particle.
2
2
4π ∈0 r
4π ∈0 (0.250 m)
E = 12 N / C =
1 Q
4π ∈0 r 2
1 (3.00 ×10 −9 C )
⇒r=
= 1.50 m.
4π ∈0 (12.0 N / C )
+
E
F
-3nC
17
Electric Field Line
(Michael Faraday first introduced the concept of field lines)
An imaginary line or curve in space and its tangent at any point gives
the direction of the electric field vector at that point.
Field lines can be used to graphically illustrate the geometry of electric
fields.
Notes:
•
•
•
the electric field has a
unique direction at a point
one field line passing
through a point
field lines therefore will not
intersect
→
E1
2
→
1
E2
Electric field line
18
away from a +ve
charge
Field Maps
E
+
toward a -ve
charge
_
E
+
+
19
Uniform field
negative charge
E
positive charge
Field lines of a uniform field are:
• straight
• parallel
• evenly spaced
20
Electric Dipole
A pair of separated electric charges of equal magnitude but opposite sign.
Net force on an electric dipole in a
uniform external field = 0
+Q
r
F = F−Q + F+Q = −QE + QE = 0
+
p
F+Q
θ
F-Q
-
E
-Q
Electric dipole moment (p):
p = Qr
[unit: C•m]
Water molecule
-
Note: Defining the direction of p being from
negative charge toward positive charge, electric
dipole moment is a vector.
O
+
H
H
+
21
Torque of an electric dipole:
→
→
+Q
r
→
τ = p× E
p
F+Q
+
θ
τ = pE sin θ
F-Q
-
E
-Q
where
τ is the torque of an electric dipole [Nm];
p is the electric dipole moment [Cm]
E is the electric field [N/C]; and
θ is the angle between the direction of p and E.
τ = pE sin θ = QrE sin θ = (QE )(r sin θ )
+Q
+
r
θ
r sinθ
-Q
Potential energy (P) of an electric dipole in an electric field:
→
→
P = − p • E = − pE cos θ
In any displacement, the work done by a
conservative force = negative change in potential
energy. The conservative force always pushes the
system toward lower potential energy.
22