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Chapter 12
Gauss’s Law
1
Electric Flux
(Flux of an electric field)
Electric flux is to indicate the flow of electric field through a surface.
Φ E = ∫ E cos φ dA
= ∫ E⊥ dA
r r
= ∫ E • dA
Φ: Greek letter “phi”
The electric flux through an area = (the magnitude of
electric field) x (area)
r r
Surface integral of E⊥ over A or of E • dA
Where,
ΦE
: electric flux (Nm2/C)
E
: electric field (N/C)
E⊥
: perpendicular component of the electric field E
φ
: the angle between the area vector and E
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Electric field E
r
E
A (Area)
r
E
φ
φ
A
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General form of electric flux:
Electric flux can be non-uniform
r r
r r
Φ E = ∑ Ei ⋅ dAi = ∫ E • dA
i
Uniform electric flux over a flat surface:
E
E
A
r r
ΦE = E • A
Φ E = E⊥ A
= EA cos φ
A
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Gauss’s Law
Karl Friedrich Gauss (1777 – 1855)
Gauss’s Law is to show the relationship between electric field and electric
charge. It is equivalent to Coulomb’s Law.
The net (total) electric flux through any closed surface is proportional to
the total electric charge inside the surface.
r r Qenc
Φ E = ∫ E • dA =
ε0
Surface integral
Electric flux
Total charge enclosed by
the surface
r r Qencl
Φ E = ∫ E • dA =
ε0
Component of
E normal to
surface A
Constant
(electrical
permittivity)
The closed surface is also called “Gaussian surface “
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A surface (S) with a number of point charges below it:
E2
3
E = ∑ Ei
E3
E1
i =1
dA
3
3
∫ E • dA = ∫ ∑ E • dA = ∑ ∫ E • dA
S
S
3
=∑
i =1
=
i =1
S
i
i =1
S
i
Q1
Q3
Q2
Qi
ε0
Qencl
ε0
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Principle of superposition
Total flux = sum of the flux from each charge
ΦE =
Q1
ε0
M
ΦE = ∑
i =1
+
Q2
ε0
+
Q2
ε0
+ ...
Qi
ε0
Q4
QM
Q3
S
Q5
Q1
Q2
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Gauss’s Law vs Coulomb’s Law
The flux is independent of R
E
r r Qenc
∫ E • dA =
Gauss’s Law
ε0
ε 0 ∫ EdA = Qenc
ε 0 E (4πR ) = Q
2
dA
R
surface area of
a sphere
Point charge Q (+)
1
Q
E=
2
4πε 0 R
A sphere
Coulomb’s Law
Magnitude of electric
field of a point charge
F0 =
1
QQ0
4πε 0 r 2
F0
E=
Q0
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When there is excess charge, it resides on the surface of conductor (at
rest). The net (total) electric field in the material of the conductor is
zero.
For a single point charge (Q), the magnitude of electric field (E) is:
1
Q
E=
4πε 0 r 2
r is the distance from the point charge
For the charge Q on the surface of conductor with a radius R,
1
Q
E=
4πε 0 r 2
Outside the sphere where r > R
E=0
Inside the sphere where r < R
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Example 12.1: The cube has sides of length L = 10 cm. The electric field is
uniform, has a magnitude E = 4 x 103 N/C, and is parallel to the xy-plane at an
angle of 36.9o measured from the +x-axis toward the +y-axis. a) What is the
electric flux through each of the six cube faces S1, S2, S3, S4, S5 and S6. b) What is
z
the total electric flux through all faces of the cube?
r r
Solution: ( a ) Φ = E ⋅ A = EA cos θ
= −
n
j (left) Φ S1 =
S1
r
where A = Anˆ
S6 (back)
-(4 x 10 3 N/C)(0.1 m) 2 cos(90 - 36.9 o) = -24
N.m2/C
S3 (right
side)
S1 (left side)
= − k(top) Φ S 2 =
n
S2
-(4 x 10 3 N/C)(0.1 m) 2 cos 90 o = 0
= +
n
j (right) Φ S3 =
S3
+(4 x 10 3 N/C)(0.1 m) 2 cos (90 o - 36.9 o) = +24
N.m2/C
L
L
L
x
= + k(bottom) Φ S 4 =
n
S4
(4 x 10 3 N/C)(0.1 m) 2 cos 90 o = 0
= + i(front) Φ S 5 =
n
S5
+(4 x 10 3 N/C)(0.1 m) 2 cos 36.9 o = 32 N.m2/C
= − i(back) Φ S6 =
n
S6
-(4 x 10 3 N/C)(0.1 m) 2 cos 36.9 o = -32 N .m2 /C
b)
S2 (top)
S5
(front)
y
S4
(bottom)
90 - 36.9o
y
o
x 36.9
The total flux through the cube must be zero; any flux entering the cube must also
leave it.
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Example 12.2: A metal sphere of radius 0.45 m carries a net charge of 0.25 nC
uniformly distributed on its surface. Find the magnitude of the electric field a) at a
point 0.1 m outside the surface of the sphere; b) at a point inside the sphere, 0.1 m
below the surface.
Solution:
(a)
1 Q
1 (2.50×10−10 C)
E (r = 0.450 m + 0.1 m) =
=
= 7.44 N / C.
2
2
4πε0 r 4πε0 (0.550m)
r
(b) E = 0
(inside the sphere)
inside of a conductor or else free charges would move under the
influence of forces, violating our electrostatic assumptions (i.e.,
that charges are not moving).
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Example 12.3: A point charge q1 = 4 nC is located on the x-axis at x = 2m, and a
second point charge q2 = -6 nC is on the y-axis at y = 1 m. What is the total electric
flux due to these two point charges through a spherical surface centered at the
origin and with radius a) 0.5m? b) 1.5m? c) 2.5m?
Solution:
y
(a) No charge enclosed so Φ = 0. (r =
0.5m < 1m or 2m where the charges
are located.)
1
x
0
− 6.00 ×10 −9 C
2
(b) Φ =
=
=
−
678
Nm
/ C.
−12 2
2
ε 0 8.85 ×10 C / Nm
q2
(c ) Φ =
q1 − q2
ε0
(4.00 − 6.00) ×10 −9 C
2
=
=
−
226
Nm
/ C.
−12 2
2
8.85 × 10 C / Nm
1
2
(Only q2 is enclosed)
(Both q1 and q2 are
enclosed)
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