4. OBJECTIVES
At the end of this lesson, you will able to:
1. Calculate electric flux
2. Use Gauss’s law to infer electric field
due to uniformly distributed charges on
long wires, spheres, and large plates
4
5. ELECTRIC FLUX
Electric flux is the measure of the strength
of the electric field penetrating a surface.
5
wherein:
Φ = Electric flux in Nm2/C (Uppercase Greek
Letter Phi)
E = Electric field in N/C
A⊥ = Area perpendicular to the electric field
in m2
Φ =EA⊥
6. Treat area as a vector
6
Normal vector a
vector which is
perpendicular to
the surface at a
given point.
8. Electric flux base on how the area is positioned
8
The area vector is
positioned perpendi
cular to the electric
field In this
position, electric
field does not pass
through the hoop,
thus, there is no
flow or flux.
The area vector is
positioned at an
angle between 0 and
90 from the electric
field. In this
position, the
amount of electric
field that will pass
through will be less
than the 1st scenario.
The area vector is
positioned parallel
to the electric
field. In this
position, a
maximum amount
of electric field
will pass through.
9. ELECTRIC FLUX
9
Φ =EAcos(θ)
wherein:
Φ = Electric flux in Nm2/C (Uppercase Greek Letter Phi)
E = Electric field in N/C
A = Area of the surface in m2
θ = angle between the electric field and the area vector in degrees (0°≤θ≤90°)
Note: Electric flux is a scalar quantity because it is the dot product of two
vector quantities: electric field and the area vector (sometimes called as
surface vector).
10. Example 1
10
A uniform electric field E = 600 N/C is passing through a flat square
area A = 10 m2 as shown. Determine the electric flux.
Given:
E = 600 N/C
A = 10 m2
Solution:
Φ=EA⊥
Φ=(600
𝑁
𝐶
)(10𝑚2)
Φ=6000
𝑁𝑚2
𝐶
,Answer
11. Example 2
11
Calculate the electric flux through a rectangle of sides 7
cm and 11 cm kept in the region of a uniform electric field
of magnitude 150 N/C.
a. The angle theta θ is 0°
b. The angle theta θ is 56°
c. The angle theta θ is 90°
12. Example 2.a
12
Calculate the electric flux through a rectangle of sides 7
cm and 11 cm kept in the region of a uniform electric field
of magnitude 150 N/C.
a. The angle theta θ is 0°
Given:
l = 11 cm or 0.11m
w = 7 cm or 0.07m
Area of rectangle = lw
Area of rectangle = (0.11m)(0.07m)
= 0.0077 𝑚2
Given:
A = 0.077 𝑚2
E = 150N/C
Θ = 0
Solution:
Φ=EAcos(θ)
Φ=(150
𝑁
𝐶
)(0.0077𝑚2
) cos(0)
Φ=1.155
𝑁𝑚2
𝐶
,Answer
13. Example 2.b
13
Calculate the electric flux through a rectangle of sides 7
cm and 11 cm kept in the region of a uniform electric field
of magnitude 150 N/C.
c. The angle theta θ is 56°
Given:
A = 0.077 𝑚2
E = 150N/C
Θ = 56
Solution:
Φ=EAcos(56)
Φ=(150
𝑁
𝐶
)(0.077𝑚2
) cos(56)
Φ=0.65
𝑁𝑚2
𝐶
,Answer
14. Example 2.c
14
Calculate the electric flux through a rectangle of sides 7
cm and 11 cm kept in the region of a uniform electric field
of magnitude 150 N/C.
c. The angle theta θ is 90°
Given:
A = 0.077 𝑚2
E = 150N/C
Θ = 90
Solution:
Φ=EAcos(56)
Φ=(150
𝑁
𝐶
)(0.077𝑚2
) cos(90)
Φ=0
𝑁𝑚2
𝐶
,Answer
15. Example 3
15
The figure shows a circular surface of radius 2 m
immersed in a uniform downward electric field of
magnitude E = 9 N/C. What is the electric flux in the
circular surface?
Given:
r = 2m
Area of circle = π𝑟2
= 4 π 𝑚2
Given:
A = 4 π 𝑚2
E = 9N/C
Θ = 30
Solution:
Φ=EAcos(θ)
Φ=(9
𝑁
𝐶
)(4 π 𝑚2
) cos(30)
Φ=97.95
𝑁𝑚2
𝐶
,Answer
Area of circle = π(2𝑚)2
17. 17
Gauss's law
the net flux of an electric field in a closed surface is directly
proportional to the enclosed electric charge.
18. 18
Gauss's law
the net electric flux going through a closed surface is the
sum of all charges Q inside that closed surface divided by the
permittivity of free space ε0.
Φ𝑛𝑒𝑡=
Q𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
ℇ0
wherein:
Φ = net flux of the closed surface in Nm2/C
Qenclosed = net charge inside the closed
surface (sum of all charges inside) in C
ε0 = permittivity of free space (read as
Epsilon naught and has a constant value of
8.85x10^-12 C2/Nm2)
Note: permittivity the ability of a medium to store electric energy.
19. Example 1
19
What is the net flux inside a closed spherical surface if a
7 nC charge is inside it?
Given:
Q = 7nC / 7 × 10−9𝐶
ℇ0 = 8.85 × 10−12
C2/Nm2
Solution:
Φ𝑛𝑒𝑡=
Q𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
ℇ0
Φ𝑛𝑒𝑡=
7 × 10−9𝐶
8.85 × 10−12𝐶2/𝑁𝑚2
Φ𝑛𝑒𝑡= 790.96
𝑁𝑚2
𝐶
20. Example 2
20
Two objects O1 and O2 have charges of +1.0 µC and -2.0 µC respectively,
and a third object, O3, is electrically neutral.
a. What is the electric flux through the gaussian surface A1 that
encloses all three objects?
b. What is the electric flux through the gaussian surface A2 that
encloses the third object only?
21. Example 2
21
Two objects O1 and O2 have charges of +1.0 µC and -2.0 µC respectively,
and a third object, O3, is electrically neutral.
a. What is the electric flux through the gaussian surface A1 that
encloses all three objects?
Given:
q1 = 1uC / 1 × 10−6
𝐶
q2 = -2uC / -2 × 10−6𝐶
q3 = 0c
ℇ0 = 8.85 × 10−12
C2/Nm2
Solution:
Φ𝑛𝑒𝑡=
Q𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
ℇ0
Φ𝑛𝑒𝑡=
1 × 10−6𝐶+(−2 × 10−6𝐶)+0c
8.85 × 10−12𝐶2/𝑁𝑚2
Φ𝑛𝑒𝑡= -1.1 × 105 𝑁𝑚2
𝐶
Note: negative flux
means number of
field lines going
inside the surface is
greater that the
number of fields
lines moving
outward.
22. Example 2
22
Two objects O1 and O2 have charges of +1.0 µC and -2.0 µC respectively,
and a third object, O3, is electrically neutral.
b. What is the electric flux through the gaussian surface A2 that
encloses the third object only?
Given:
q3 = 0c
ℇ0 = 8.85 × 10−12 C2/Nm2
Solution:
Φ𝑛𝑒𝑡=
Q𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
ℇ0
Φ𝑛𝑒𝑡=
0𝑐
8.85 × 10−12𝐶2/𝑁𝑚2
Φ𝑛𝑒𝑡= 0
𝑁𝑚2
𝐶
23. Example 3
23
There are three charges q1, q2, and q3 having charges 1 C, 2 C,
and 3 C inside an enclosed space. Two charges q4 and q5 having
charges 4 C and 5 C are outside the enclosed space. What is the
net flux on the surface?
Given:
q1 = 1 C
q2 = 2 C
q3 = 3 C
q4 = 4 C
q5 = 5 C
ℇ0 = 8.85 × 10−12
C2/Nm2
Solution:
Φ𝑛𝑒𝑡=
Q𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑
ℇ0
Φ𝑛𝑒𝑡=
1𝑐+2𝑐+3𝑐
8.85 × 10−12𝐶2/𝑁𝑚2
Φ𝑛𝑒𝑡=6.78× 1011 𝑁𝑚2
𝐶
