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Chapter 22
Gauss’ Law
Mathematics Reminder
u = aiˆ + bĵ
v = ciˆ + dĵ
u × v º ac + bd = u v cos f
where f is the angle between the vectors
Math: Area Vector
Gauss’ Law
F=
qenclosed
e0
F : Electric flux through a closed surface
qenclosed : Total charge enclosed by the closed surface
But how do you find F?
Through which closed surface?
Why it makes sense
All capture the same number of field lines.
Flux by Integration
dA º n̂dA
where n̂ is the unit normal vector.
Gauss' Law:
qenclosed
F=
e0
Relates the outside to the inside
F=
Flux through the area
of the Gaussian surface
q
e0
Charge inside the
Gaussian surface
A spherical Gaussian surface (#1)
encloses and is centered on a point
charge +q. A second spherical
Gaussian surface (#2) of the same
size also encloses the charge but is
not centered on it.
Compared to the electric flux
through surface #1, the flux
through surface #2 is
+q
Gaussian
surface #1
A. greater.
B. the same.
C. less, but not zero.
D. zero.
Gaussian
surface #2
E. not enough information given to decide
Two point charges,
+q (in red) and –q
(in blue), are
arranged as shown.
Through which
closed surface(s) is
the net electric flux
equal to zero?
A. surface A
B. surface B
C. surface C
D. surface D
E. both surface C and surface D
Area vector of a cube
Convention: Area vector of a close surface points outward.
Sign of Flux
• Positive when E field is going out
• Negative when E field is going in
F>0
F<0
Sign of Flux (Example)
F=0
Example (constant E)
Six surfaces:
Ftotal = F1 + F2 + F 3 + F 4 + F 5 + F 6
= -EA + EA + 0 + 0 + 0 + 0
Þ Ftotal = 0
qenclosed
but Ftotal =
e0
Þ qenclosed = 0
 is difficult to find in general
… so we need to be smart.
+
E here is different
than E there.
You are allowed to change the shape of the
Gaussian surface to make the surface integral
easy.
Spherical Gaussian Surface
Simplify the integral
E parallels to dA everywhere
Þ E × dA = E cos(0 )dA = EdA.
By spherical symmetry:
E is constant everywhere on the sphere
Therefore the flux becomes:
F=
ò E × dA = ò EdA = E ò dA = EA
Outside Uniform Sphere of Charge
F=
(Electric field) ´ (Area)
Area of sphere = 4p r
q
e0
q
2
Þ F = (E)(4p r )
2
(E)(4p r ) =
2
ÞE=
q
e0
q
4pe 0 r 2
Gaussian
surface
r
x
What to write in the exam
In the exam, just a few words of
explanation is enough in simplifying the
flux integral:
Þ (E)(4p r ) =
2
ÞE=
q
4pe 0 r 2
q
e0
Must include diagrams
When answering questions on Gauss’ law, you
MUST include a diagram to indicate the
Gaussian surface you picked.
Without defining the Gaussian surface with the
diagram, the flux cannot be defined, and your
integral will have no meaning.
You will lose half your points without a diagram.
Linear fly density
L=4m
There are N = 8 flies on the rod. The number of flies per unit length is:
N 8 flies
l= =
= 2 flies / m.
L
4m
How many flies n are on l = 2m of the rod?
n
l=
l
Þ n = ll = (2 flies / m)(2m) = 4 flies.
Linear charge density, 
+++++++++++++++++++++++++++++++++++++
L=4m
There are Q = 8C on the rod. The charge per unit length is:
Q 8C
l= =
= 2C / m.
L 4m
How much charge q is on l = 2m of the rod?
q
l=
l
q = ll = (2C / m)(2m) = 4C.
Surface charge density 
Q = 10C on a disc of radius R = 4m.
What is the charge per unit area?
++++++
++++++++++++++++++
++++++
Q
Q
10C
2
s= =
=
=
0.2C
/
m
A p R 2 p (4m)2
How much charge q is in a circle of r = 2m? (red circle)
q
s= 2
pr
Þ q = sp r 2 = (0.2C / m 2 )(p )(2m)2 = 2.5C
Volume charge density 
Q q
r= =
V v
R
Þ q = rv,
where q is the charge on a volume v,
Q is the total charge on the whole volume V.
Unit: C / m 3
Charge Density Summary
1D (l is linear charge density, C / m) :
q = l ´ length
2D (s is surface charge density, C / m 2 ) :
q = s ´ area
3D ( r is volume charge density, C / m 3 ) :
q = r ´ volume
[But watch out for regions where the charge density drops to zero!]
A Line of Charge
l : Linear charge density (charge per unit length)
Unit: C/m
Simplify the integral
Ftotal = F1 + F2 + F 3
=
ò
S1
E × dA + ò E × dA + ò E × dA
S2
S1
S3
By cylindrical symmetry:
E points radially outward
Therefore the flux on S1 and S3 are zero:
F1 = ò E × dA =
ò
F = ò E × dA = ò
S1
3
S3
F1 = F 3 = 0
S3
S1
E cos 90 dA = 0
S3
E cos 90 dA = 0
S2
Simplify the integral
E parallels to dA everywhere on S2
Þ E × dA = E cos(0 )dA = EdA.
S3
S1
S2
By cylindrical symmetry:
E is constant everywhere on S2
Therefore the flux on the curved surface becomes:
F2 = ò E × dA =
S2
ò
S2
EdA = E ò dA = EA2
S2
Ftotal = F1 + F2 + F 3 = 0 + EA2 + 0 = EA2
F=
(Electric field) ´ (Area)
Area of cylinder A2 = 2p rl
Þ F = (E)(2p rl)
q
e0
q = ll
Charge enclosed
E(2p rl) = ll / e 0 ; l : Linear charge density
l
ÞE=
2 e 0p r
An Plane Sheet of Charge
A Plane Sheet of Charge
F=
(Electric field) ´ (Area)
q
q = sA
e 0 Charge enclosed
σ is the charge per area
F = EA + EA = 2EA
sA
s
2EA =
ÞE=
e0
2e 0
area A
r
x
Gaussian surface
(cylinder, pillbox)
Two Parallel Plates
E = σ/ ε0
Derivation with Gauss Law
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
F = EA, q = s A
E
where s is charge per unit area
q
F=
e0
sA
Þ EA =
e0
s
ÞE=
e0
No E inside a conductor
A Solid Conducting Sphere
Outside:
qenclosed = Q
E(4p r ) =
2
Þ Eout =
Q
Q
Q
e0
Q
4pe 0 r
Inside:
qenclosed = 0
Þ Ein = 0
2
Useful fact
E field is always zero inside a conductor at
equilibrium (no movement of charges).
Since F=qE, if E is non-zero, the electric force
will push the charges around. Therefore the only
way equilibrium can be established is if F=0,
which implies E=0.
An Uniformly Charged Sphere
Total charge: Q
Radius: R
We can compute the volume charge density:
Q
r=
4 3
pR
3
R
Outside (r > R)
Total charge inside the Guassian surface of radius r:
qenclosed = Q
F = E(4p r 2 )
Þ E(4p r ) =
2
Þ Eout =
Q
e0
Q
4pe 0 r 2
R
Inside (r < R)
r=
Total charge inside the Guassian surface of radius r:
qenclosed
4 3
= r pr
3
4 3
r pr
Þ E(4p r 2 ) = 3
e0
rr
Qr
ÞE=
=
3e 0 3e ( 4 p R 3 )
0
3
Q
r
Þ Ein =
4pe 0 R 3
R
Q
4 3
pR
3
R