The document discusses the concept of electric fields created by charged particles, detailing their properties, calculations, and the effects on test charges placed within the field. It explains Gauss's Law, the behavior of electric field lines, and the characteristics of conductors in electrostatic equilibrium. Examples are provided to illustrate calculations involving electric fields, flux, and surface charge density in various scenarios.

1. ELECTRIC FIELD

2. Particle 2 sets up an electric field at all points in
the surrounding space, even if the space is a vacuum. If
we place particle 1 at any point in that space, particle 1
knows of the presence of particle 2 because it is
affected by the electric field particle 2 has already set
up at that point.

3. Electric Field
➢At any point in space, it is defined as the force exerted on
a tiny test charge placed at that point divided by the
magnitude of the test charge.
𝐸 =
Ԧ
𝐹
𝑞𝑜
➢It is the “alteration in space” caused by a charge at rest.
➢The SI unit of electric field is newton per coulomb (N/C).
Note: Electric force is a push or pull. Electric field is an
abstract property set up by a charged object.

4. A test charge is used to investigate the
electric field surrounding a charge particle or
group of charges by measuring the force
acting on it. It is so small/imaginary that the
force it exerts does not significantly alter the
distribution of those other charges that
create the field.
Thus, the electric field describes only the
effect of the charge(s) creating the [net]
electric field at that point P.
According to Michael Faraday, an electric field extends
outward from every charge and permeates all of space.

5. To strengthen the idea that E does not depend on
a test charge q, we’ll use Coulomb’s law to
determine the magnitude of the electric field at a
distance r from a single charge Q.
𝐸 =
Ԧ
𝐹
𝑞𝑜
=
𝑘
𝑞𝑜𝑄
𝑟2
𝑞𝑜
= 𝑘
𝑄
𝑟2
For its magnitude,
𝐸 = 𝑘
𝑄
𝑟2

6. ELECTRIC FIELD LINES
Electric field lines
extend away from
positive charge
(where they originate)
and toward negative
charge (where they
terminate).

7. (a) If q is a positive test charge under an electric field of
a single positive charge, F and E point in the same
direction.
(b) If q is negative, F and E point in the opposite
directions.

8. FOUR IMPORTANT PROPERTIES OF ELECTRIC FIELD LINES:
1. The electric
field lines must be
tangent to the
direction of the
field at any point.

9. 2. The greater the
line density (the
closer to the
charge), the greater
the magnitude of
the electric field.

10. 3. The lines always start
from positively charged
objects and end on
negatively charged objects.

11. 4. The electric
field lines must
never cross.

12. This tells us that electric fields also obey the principle of
superposition. If you want the net electric field at a given
point due to several particles, find the electric field due to
each particle and then sum the fields as vectors.

13. Example #1: The Presence of E of Q
Calculate the magnitude and direction of
the electric field at point P which is 30 cm from
charge 𝑄 = −3.0𝜇𝐶.

14. Example #2: What’s in the Halfway?
Two charges, 𝑄1 = +2.0 × 10−8𝐶 (left)
and 𝑄2 = +3.0 × 10−8𝐶 (right), lying on a
horizontal line are 50mm apart. What is the
electric field halfway, align with the two
charges, between them?

15. Example #3: E in the 2-D
Calculate the total
electric field (a) at
point A and (b) at point
B due to both charges,
Q1 and Q2.

16. Properties of Conductors
When there is no motion of electrons in
these materials, the conductor is said to be in
an electrostatic equilibrium. An isolated
conductor, say floating in a vacuum, has the
following properties:
1. The electric field inside a conductor is zero in
the static situation, from its core to its surface.

17. The electric field lines extending
to the outer surface creates an
electric field as if the metal were
not there.
2. Any extra [or net charge] charge on an isolated
conductor resides entirely on its surface.

18. 3. The electric field of a conductor is always
perpendicular to the outside surface of that material.

19. 4.The charge accumulates at sharp points on an irregularly
shaped conductor, where the radius of curvature of the
surface is the smallest.

20. Example #4 (Concept):
A neutral hollow metal
box is placed between two
parallel charged plates as
shown in the figure. What is
the field like inside the box?

21. Answer:
The external field does not changed
since the electrons in the metal can move
just as freely as before to the surface.
Hence, the field inside the box is zero.
If the box is solid and hollow, free
electrons in the box would be redistributed
along the surface until the individual fields
will cancel each other inside the box. Then,
the net field inside is zero.

22. Faraday’s Ice-Pail Experiment
Result: The electrometer reads none when the charged
metal ball is still outside.

23. Result: The needle deflected due to the outer plate
becoming negatively charged (charge separation)
Faraday’s Ice-Pail Experiment

24. Result: The needle did not deflect either when the ball
touched the inner plate.
Faraday’s Ice-Pail Experiment

25. Result: When the ball was removed, the conductor
remains negatively charged.
Faraday’s Ice-Pail Experiment

26. Gauss’s Law
It is a more general version
of Coulomb’s law, developed
by Karl Friedrich Gauss.
Gauss’s law involves the
concept of electric flux, which
refers to the electric field
passing through a given area.

27. For a uniform electric field passing through an
area, the electric flux is defined as
Where the θ is the angle
between the electric field
direction and a line
drawn perpendicular to
the area.
Φ𝐸 = 𝐸𝐴 cos 𝜃

28. The flux can be written
equivalently as
Φ𝐸 = 𝐸⊥𝐴
or
Φ𝐸 = 𝐸𝐴⊥

29. Example #5: Flux in a
Cube
Calculate the net
electric flux through a
closed surface with
the given figure.

30. The number of field lines passing through unit
area perpendicular to the field is proportional to
the magnitude of the field:
Hence,
𝐸 ∝
𝑁
𝐴⊥
𝑁 ∝ 𝐸𝐴⊥ = Φ𝐸

31. Gauss’s law involves the total
flux through a closed surface –
a surface of any shape that
encloses a volume of space.
Then, for each tiny areas ∆Ai,
we have
Φ𝐸 = 𝐸1∆𝐴1 cos 𝜃1 + 𝐸2∆𝐴2 cos 𝜃2 + ⋯
Φ𝐸 = ෍ 𝐸Δ𝐴 cos 𝜃 = ෍ 𝐸⊥Δ𝐴

32. The number of field lines is proportional to the
magnitude of the charge. Hence, the net number
of lines pointing out any closed surface must be
proportional to the net charge enclosed by the
surface.
Φ𝐸 = ෍
𝑐𝑙𝑜𝑠𝑒𝑑
𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝐸⊥Δ𝐴 ∝ 𝑄𝑒𝑛𝑐𝑙

33. Finally, the Gauss’s law equation is given
with the constant of proportionality. Then,
we have
෍
𝑐𝑙𝑜𝑠𝑒𝑑
𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝐸⊥Δ𝐴 =
𝑄𝑒𝑛𝑐𝑙
𝜖𝑜

34. A “gaussian surface” to
enclosed the charge given in
order to measure the electric
field at all directions. Even if
the surface that encloses the
charge is irregular, the net
electric flux is still equal to
𝑄𝑒𝑛𝑐𝑙
𝜖𝑜
.

35. Example #6:
A thin spherical shell
of radius ro possesses a
total net charge Q that
is uniformly distributed
on it. Determine the
electric field outside
and inside of the shell.

36. Solution:
Outside Gaussian Surface - the
electric field will have the same
magnitude at all points on an
imaginary Gaussian surface. Because
the electric field is perpendicular to
this surface, then the cos θ cancels
out.
Inside Gaussian Surface - Due to no
charge enclosed, then Qencl = 0. This
satisfies the idea that there is no
electric field inside a conductor
under electrostatic equilibrium.

37. Example #7: E in a Ball
The electric field just outside a
3.50-cm-radius metal ball is 2.75 x
102 N/C and points toward the ball.
What charge resides on the surface
of the ball and its magnitude?

38. Electric Field of a Line Charge
In the given figure, the
electric charge is distributed
uniformly along an infinitely
long, thin wire. The charge
per unit length is 𝜆 =
𝑄
𝐿
(lowercase Lambda). The
selected field point must have
lesser distance to the wire
than the length of the wire.

39. Then, the electric field is
𝛷𝐸 = 𝐸𝐴 = 𝐸 2𝜋𝑟𝑙 =
𝜆𝑙
𝜖𝑜
𝐸 =
1
2𝜋𝜖𝑜
𝜆
𝑟
The electric field has the
same value everywhere on the
walls of the Gaussian
cylindrical surface.

40. Example #8: E in a Line
A very long uniform line of charge has a charge per unit length
4.80 μC/m and lies along the x-axis. A second-long uniform line of
charge has a charge per unit length –2.40 μC/m and is parallel to
the x-axis at y = 0.400 m. What is the net electric field (magnitude
and direction) at the following points on the y-axis: (a) y = 0.200 m
and (b) y = 0.600 m?

41. Example #8: E in a Line
A very long uniform line of charge has a charge per unit length
4.80 μC/m and lies along the x-axis. A second-long uniform line of
charge has a charge per unit length –2.40 μC/m and is parallel to
the x-axis at y = 0.400 m. What is the net electric field (magnitude
and direction) at the following points on the y-axis: (a) y = 0.200 m
and (b) y = 0.600 m?

42. The electric field just outside
the surface of any good conductor
of arbitrary shape is given by
𝐸 =
𝜎
𝜖𝑜
Where 𝜎 =
𝑄
𝐴
is the surface
charge density on the conductor
at that point.
Electric Field at a Selected Surface of a Conductor

43. The dashed boxes are
gaussian boxes. If the
field is going away from
the surface, the 𝜎 is
positive (this is due to
the positive charges at
the surface). Opposite
for the negative 𝜎.

44. Example #9: Use Gauss in the Earth
The Earth has a net electric charge. The resulting
electric field near the surface can be measured with
sensitive electronic instruments; its average value is about
150 N/C, directed toward the center of the earth. What is
the corresponding surface charge density? What is the
total surface charge of the Earth? Use Gauss’s law to solve
the unknown quantities. The radius of the earth is 𝑅𝐸 =
6.38 × 106
𝑚.

45. Electric Field of an Infinite Plane Sheet of Charge
The electric field caused by a thin, flat, infinite
sheet on which there is a uniform positive charge
per unit area σ (surface charge density) is
𝛷𝐸 = 2𝐸𝐴 =
𝜎𝐴
𝜖𝑜
𝐸 =
𝜎
2𝜖𝑜

46. Example #10: Two E Planes
Two very large, nonconducting
plastic plane sheets, each 10.0 cm thick,
carry uniform charge densities on their
surfaces. Use Gauss’s law to find the
electric field at the following points A
(5.00 cm from the surface with σ1), B
(1.25 cm from the surface with σ3), and
C (in the middle of the right-hand
sheet).
σ1 = -6.00 μC/m2 σ3 = +2.00 μC/m2
σ2 = +5.00 μC/m2 σ4 = +4.00 μC/m2

47. Solution:
A.Directions of electric fields: E1 (right), and
E2, E3 and E4 (left)
𝐸𝐴 = 𝐸𝑛𝑒𝑡 = 𝐸1 + 𝐸2 + 𝐸3 + 𝐸4
𝐸𝐴 =
𝜎1
2𝜖𝑜
−
𝜎2
2𝜖𝑜
−
𝜎3
2𝜖𝑜
−
𝜎4
2𝜖𝑜
𝐸𝐴 =
1
2𝜖𝑜
𝜎1 − 𝜎2 − 𝜎3 − 𝜎4
=
1
2𝜖𝑜
ቆ
ቇ
6.00
𝜇𝐶
𝑚2
− 5.00
𝜇𝐶
𝑚2
− 2.00
𝜇𝐶
𝑚2
− 4.00
𝜇𝐶
𝑚2
= −2.82 × 105
𝑁
𝐶

48. B. Directions of electric fields: E1, E3 and E4
(left), and E2 (right)
𝐸𝐵 = 𝐸1 + 𝐸2 + 𝐸3 + 𝐸4
𝐸𝐵 = −
𝜎1
2𝜖𝑜
+
𝜎2
2𝜖𝑜
−
𝜎3
2𝜖𝑜
−
𝜎4
2𝜖𝑜
𝐸𝐵 =
1
2𝜖𝑜
− 𝜎1 + 𝜎2 − 𝜎3 − 𝜎4
=
1
2𝜖𝑜
ቆ
ቇ
−6.00
𝜇𝐶
𝑚2
+ 5.00
𝜇𝐶
𝑚2
− 2.00
𝜇𝐶
𝑚2
− 4.00
𝜇𝐶
𝑚2
𝐸𝐵 = −3.95 × 105
𝑁
𝐶

49. C. Directions of electric fields: E1, E3 and E4
(left), and E2 (right)
𝐸𝐶 = 𝐸1 + 𝐸2 + 𝐸3 + 𝐸4
𝐸𝐶 = −
𝜎1
2𝜖𝑜
+
𝜎2
2𝜖𝑜
+
𝜎3
2𝜖𝑜
−
𝜎4
2𝜖𝑜
𝐸𝐶 =
1
2𝜖𝑜
− 𝜎1 + 𝜎2 + 𝜎3 − 𝜎4
=
1
2𝜖𝑜
ቆ
ቇ
−6.00
𝜇𝐶
𝑚2
+ 5.00
𝜇𝐶
𝑚2
+ 2.00
𝜇𝐶
𝑚2
− 4.00
𝜇𝐶
𝑚2
𝐸𝐶 = −1.69 × 105
𝑁
𝐶