Download Gauss`s Law and Electric Flux

Gauss’s Law and Electric Flux
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Gauss’s Law relates number of E field lines entering and
leaving a surface to the net charge inside the
surface.
Consider imaginary spheres
centered on:
a.) +q (blue)
b.) -q (red)
c.) midpoint (yellow)
Number of lines exiting:
a.) (blue)
24
b.) -q (red)
-24
c.) midpoint (yellow)
0
More on Gauss’s Law later.
c
a
b
Electric Flux
• Flux:
Let’s quantify previous discussion about fieldline “counting”
Define: electric flux ΦΕ through the closed
surface S
r r
Φ E ≡ ∫ E • dA
S
E
dA
S
“S” is surface
of the box
Electric Flux
r r
ΦE ≡ ∫ E • dA
S
•What does this new quantity mean?
• The integral
r r is over a CLOSED SURFACE
• Since E • dA is a SCALAR product, the electric flux is a SCALAR
quantity
r
• The integration vector dA is normal to the surface and points OUT
r
r
of the surface. E • d A is interpreted as the component of E which is
NORMAL to the SURFACE
• Therefore, the electric flux through a closed surface is the sum of
the normal components of the electric field all over the surface.
• The sign matters!!
Pay attention to the direction of the normal component as it
penetrates the surface… is it “out of” or “into” the surface?
• “Out of” is “+” “into” is “-”
Electric Flux
Special case: uniform E perpendicular to plane
surface A.
ΦE
E
r
r
≡ ∫ E • d A = ∫ E dA cos θ = E ∫ dA = EA
S
S
constant
dA
s
S
=1
Another: uniform E at angle to plane surface A.
constants
ΦE
s
θ
dA
r
r
≡ ∫ E • d A = ∫ E dA cos θ = E cos θ ∫ dA = EA cos θ
S
S
S
E
CLICKER QUESTION
A flat disk with radius r = 0.1m, is oriented with its normal unit vector
relative the a constant E field as shown above. What is the electric flux
through the disk area?
r
2 r
o
A) π (0.1) E
2 r
o
B ) π (0.1) E cos 30
2 r
o
C ) π (0.1) E sin 30
( )
( )
( )
( )
D) 2π (0.1) E cos 30
r
E ) 2π (0.1) E sin 30o
How to think about flux
• We will be interested in net
flux in or out of a closed
surface like this box
w
z
• This is the sum of the flux
through each side of the box
“A” is surface
of the box
y
x
– consider each side separately
surface area vector:
A
r
A = area • yˆ
= w 2 yˆ
• Let E-field point in y-direction
r
– then E and A are parallel and
• Look at this from on top
– down the z-axis
r r r 2
E • A =| E | w
A
UI3PF3:
6) A cube is placed in a uniform
electric field. Find the flux through the
bottom surface of the cube.
a) Фbottom < 0
b) Фbottom = 0
c) Фbottom > 0
Exercise UI3A2
2A
•Imagine a cube of side a positioned in a
region of constant electric field as shown
•Which of the following statements
about the net electric flux ΦE through
the surface of this cube is true?
(a) ΦE = 0
2B
(b) ΦE ∝ 2a2
a
a
(c) ΦE ∝ 6a2
• Consider 2 spheres (of radius R and 2R)
drawn around a single charge as shown.
– Which of the following statements
about the net electric flux through the
2 surfaces (Φ2R and ΦR) is true?
R
2R
(a) ΦR < Φ2R
(b) ΦR = Φ2R
(c) ΦR > Φ2R
Exercise UI3A2
2A
•Imagine a cube of side a positioned in a
region of constant electric field as shown
•Which of the following statements
about the net electric flux ΦE through
the surface of this cube is true?
(a) ΦE = 0
(b) ΦE ∝ 2a2
a
a
(c) ΦE ∝ 6a2
r r
• The electric flux through the surface is defined by: Φ ≡ ∫ E • dA
r r
• ∫ E • dA is ZERO on the four sides that are parallel to the electric field.
r r
• E • dA on the bottom face is negative. (dA is out; E is in)
∫r r
• ∫ E • dA on the top face is positive. (dA is out; E is out)
• Therefore, the total flux through the cube is:
r r
Φ ≡ ˜ E • dS
+Φ
+ Φtop = 0 − Ea 2 + Ea 2 = 0
A =Φ
sides
bottom
Exercise UI3A2
2B
• Consider 2 spheres (of radius R and 2R)
drawn around a single charge as shown.
– Which of the following statements
about the net electric flux through the
2 surfaces (Φ2R and ΦR) is true?
(a) ΦR < Φ2R
(b) ΦR = Φ2R
(c) ΦR > Φ2R
R
2R
• Look at the lines going out through each circle -- each circle has the
same number of lines.
• The electric field is different at the two surfaces, because E is
proportional to 1 / r 2, but the surface areas are also different. The
surface area of a sphere is proportional to r 2.
• Since flux =
∫
r
r
E • d A , the r 2 and 1/r 2 terms will cancel, and the two
S
circles have the same flux!
• There is an easier way. Gauss’ Law states the net flux is proportional
to the NET enclosed charge. The NET charge is the SAME in both
cases.
• But, what is Gauss’ Law ???
--You’ll find out next lecture!
c
a
b
r r
qenclosed
∫ E • dA = Φ E =
ε0
Fundamental Law
of Electrostatics
• Coulomb’s Law
Force between two point charges
OR
• Gauss’ Law
Relationship between Electric Flux
and charges
Gauss’ Law
• Gauss’ Law (a FUNDAMENTAL LAW):
The net electric flux through any closed surface is
proportional to the charge enclosed by that surface.
r r
qenclosed
∫ E • dA = Φ E =
ε0
• How do we use this equation??
• The above equation is ALWAYS TRUE but it may not
be easy to use.
• It is very useful in finding E when the physical
situation exhibits massive SYMMETRY.
UI4PF4:
dA
1
2) A positive charge is contained inside a spherical shell.
How does the electric flux dФE through the surface
element dA change when the charge is moved from
position 1 to position 2?
a) dФE increases
b) dФE decreases
c) dФE doesn’t change
dA
2
dA
1
3) A positive charge is contained inside a spherical shell.
How does the flux ФE through the entire surface change
when the charge is moved from position 1 to position 2?
a) ФE increases
ФE decreases
c) ФE doesn’t change
b)
dA
2
dA
1
3) A positive charge is contained inside a spherical shell.
How does the flux ФE through the entire surface change
when the charge is moved from position 1 to position 2?
a) ФE increases
ФE stays the same
c) ФE = 0
b)
dA
2