Transcript Lect06

c
a
b
Last Week Rules for Field Lines
+
-
•Field lines are a way of representing an electric field
•Lines leave (+) charges and return to (-) charges
•Number of lines leaving/entering charge  charge
•Tangent of line = direction of E
•Local density of field lines  local magnitude of E
•Lines from an isolated charge go to infinity
•Field lines cannot cross
Electric Flux
• Flux:
Let’s quantify previous discussion about
numbers of field-lines
Define: electric flux Fs through the closed
surface S
“S” is surface
of the box
Note: in Fishbane they use dA to refer to an
element of surface area. In these slides I use dS
for the same thing
Flux
• How much of something is passing
through some surface
Ex: How many hairs passing through your
scalp.
• Two ways to define
1. Number per unit area (e.g., 10 hairs/mm2)
This is NOT what we use here.
2. Number passing through an area of interest
e.g., 48,788 hairs passing through my scalp.
This is what we are using here.
Electric Flux
•What does this new quantity mean?
• The integral is over a CLOSED
SURFACE
• Since
is a SCALAR product, the
electric flux is a SCALAR quantity
• The integration vector
is normal to
the surface and points OUT of the
surface.
•
is interpreted as the component
of E which is NORMAL to the SURFACE
dS
Electric Flux
• 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 “-”
How to think about flux
surface area vector:
• Consider the flux through
this surface and define the
surface area vector S
• Let E-field point in ydirection


– then E and S are parallel
   2
and
ES  Ew
• Look at this from on top

S  Area  yˆ
w
w
 w 2 yˆ
How to think about flux
• Consider flux through two
surfaces that “intercept
different numbers of field
lines”
case 1
– first surface is the surface from
the previous slide
– Second surface rotated by an
angle q
Flux:
E-field surface area
case 1
case 2

E  Eo yˆ
w2

E  Eo yˆ
2
w
case 2
E S
Eo w 2
Case 2 is
smaller!
Eo w2  cos q
q
The Sign Problem
• For an open surface we
can choose the direction of
S-vector two different ways
left
right
– to the left or to the right
– what we call flux would be
different these two ways
– different by a minus sign
• For a closed surface we
can choose the direction
of S-vector two different
ways
– pointing “in” or “out”
– Choose “out”
– Integral of EdS over a closed
surface gives net flux “out,”
but can be + or -
A differential surface
element, with its vector
Question 1
Wire loops (1) and (2) are
placed in a uniform electric
field as shown. Compare
the flux through the two
surfaces.
a) Ф1 > Ф2
b) Ф1 = Ф2
c) Ф1 < Ф2
1
E
2
Question 1
1. a
2. b
3. c
81%
b
a
c
10%
9%
Question 1
Wire loops (1) and (2) are
placed in a uniform electric
field as shown. Compare
the flux through the two
surfaces.
a) Ф1 > Ф2
b) Ф1 = Ф2
c) Ф1 < Ф2
1
E
2
Question 2
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
Question 2
86%
13%
c
b
1%
a
1. a
2. b
3. c
Question 2
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
Question 3
•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 FE through
the surface of this cube is true?
(a) FE = 0
(b) FE  2a2
(c) FE  6a2
a
a
Question 3
85%
12%
c
b
3%
a
1. a
2. b
3. c
Question 3
•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 FE through the surface of this
cube is true?
(a) Fs = 0
(b) Fs  2a2
a
a
(c) Fs 6a2
• The electric flux through the surface is defined by:
•
•
is ZERO on the four sides that are parallel to the electric field.
on the bottom face is negative. (dS is out; E is in)
•
on the top face is positive. (dS is out; E is out)
• Therefore, the total flux through the cube is:
The Fundamental Law of
Electrostatics?
• Coulomb’s Law
Force between two point charges
OR
• Gauss’ Law
Relationship between Electric Fields
and charges
dS
1
• A positive charge is contained inside a spherical shell.
• When the charge moves from position 1 to position 2;
•The electric flux dФs through the surface element dS
increases
•BUT the electric flux through the entire surface stays
constant
dS
2
Gauss’ Law
Gaussian surface 4
• Consider the dipole flux passing through the ‘Gaussian’
surfaces
–
–
–
–
Surface 1 has +ve flux proportional to charge +q
Surface 2 has –ve flux proportional to charge –q
Surface 3 has zero flux, as many flux lines enter as leave
Surface 4 also has zero flux, all leaving lines return
Question 4
• 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 (F2R and FR)
is true?
(a) FR < F2R
(b) FR = F2R
(c) FR > F2R
Question 4
1. a
2. b
3. c
87%
b
a
c
9%
5%
Question 4
• 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 (F2R and FR)
is true?
(a) FR < F2R
(b) FR = F2R
(c) FR > F2R
•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 =
, the r 2 and 1/r 2 terms will cancel, and the two
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.
Gauss’ Law
• Gauss’ Law (a FUNDAMENTAL LAW):
The net electric flux through any closed surface is
proportional to the charge enclosed by that surface.
• How do we use this equation??
•The above equation is ALWAYS TRUE but it isn’t
always easy to use.
•It is very useful in finding E when the physical
situation exhibits strong SYMMETRY.
Gauss’ Law…made easy
•To solve the above equation for E, you have to be able to CHOOSE A
CLOSED SURFACE such that the integral is TRIVIAL.
(1) Direction: surface must be chosen such that E is known to be
either parallel or perpendicular to each piece of the surface;
If
then
If
then
(2) Magnitude: surface must be chosen such that E has the same
value at all points on the surface when E is perpendicular to the
surface.
Gauss’ Law…made easy
•With these two conditions we can bring E outside of the
integral…and:
Note that
is just the area of the Gaussian surface over which
we are integrating. Gauss’ Law now takes the form:
This equation can now be solved for E (at the surface) if we know
qenclosed (or for qenclosed if we know E).
Summary
• Electric Flux
• Gauss’ Law
• Read Chapter 23