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Discussion Question 3A
P212, Week 3
Electric Flux
Electric Flux refers to the net amount of electric field flowing through a surface. A way to write this is:
Qenc
flux,
E dA . Why do we care about flux? Because Gauss’ law states that flux = E dA
,
0
which gives us an easy way to calculate electric fields for charge distributions. So, as first step, let’s
see how to find E dA , or the net flux, through some surfaces … [Remember that you care about net
flow through a surface, which is the difference between flow in and flow out]
(a) An infinitely thin imaginary plane (a “Gaussian surface”) is placed in
a uniform electric field as shown. Is there a net flux through the plane?
There is a net flux.
E
0
Would there be a net flux if the plane were rotated 90º?
If the plane were rotated 90º, there would be no net flux through the plane.
(b) An imaginary sphere is now placed in the uniform field. Is there a
net flux through the sphere? Is there a net field in the sphere?
There is no net flux through the sphere (the same number of lines enter
as leave). The net field in the sphere is E .
E0
Is Gauss’ Law relevant for finding the field in this case? Why?
Gauss’ Law is not relevant because net flux is zero.
(c) A Gaussian sphere encloses charges Q1 and Q2 at its center, where
Q1 + Q2 < 0. Draw the electric field lines due to these charges.
Field lines, shown in red, point toward negative charges.
Now use arrows to show the direction of dA .
dA , shown in green, always points out normal to surface.
Is the net flux through the sphere positive, negative, or zero?
Because dA and E point in different directions, the net flux is negative.
(d) What is the surface area, dA , of the sphere in part (c) if it has a radius R?
Is Gauss’ Law relevant here? Gauss’ Law is relevant since there is a net flux.
Find E at the surface of the sphere.
(Q1 Q2 )
(Q1 Q2 )
E dA E 4 R 2
E
2
0
04 R
1
Q 1 ,Q 2
dA 4 R 2
Discussion Question 3A
P212, Week 3
Electric Flux
(e) The sphere is now placed in a uniform electric field, E0 .
Draw an arrow showing the direction of the net field at point A.
How does the magnitude of the field at point A compare to the
field at point B?
Equal in magnitude (though opposite in direction)
How does the magnitude of the field at point C compare to the
field at point D?
Field at D > field at C
Write an expression for E at point C.
By superposition, E
E0
Q1 Q2
4 0 R2
2
A
E0
Q 1, Q 2
C
D
B
Discussion Question 3B
P212, Week 3
Application of Gauss’ Law
Follow the steps outlined on the previous page, find the electric field E at the indicated field
point x due to the three charged objects shown below. In all cases, the variable r denotes the
distance from the center of the object to the field point.
Basic Object #1: A solid sphere of radius R carrying a volume
charge density
find E at the indicated field point.
x
E=
r
R3
radially outward.
3 0r 2
R
Basic Object #2: An infinitely long line carrying a line
charge density
find E at the indicated field point.
x
E=
2
o
r
radially outward
r
Basic Object #3: A plane of infinite area carrying a surface
charge density
find E at the indicated field point.
r
E = /2
o
in positive x-direction.
x
Discussion Question 3C
P212, Week 3
Electric Field due to Charged Sheets and Slabs
In the tutorial, you calculated the electric field of a charged plane of infinite area. The field
turned out to be independent of distance. This property makes it easy to analyze systems of
many charged sheets and slabs using superposition.
Here’s a warmup exercise: consider the thick slab of infinite area shown at right. The slab is
made of a non-conducting material. It carries a net charge per unit area of , which is distributed
uniformly throughout the interior of the slab.
(a) Write down the electric field E at the point x.
E = /2
o
x
x
(b) Would the field change if the slab were moved further to the left?
no
(c) Would the field change if the thickness of the slab were increased?
no
(d) Would the field change if the slab were replaced by two slabs, each carrying charge density
/2?
no
x
See how easy superposition is with infinite planes?
Let’s also remind ourselves how to deal with conductors. Suppose that the slab described above
were made of metal instead of non-conducting material … what would change?
(f) Would the field outside the slab (e.g. at the point x) change?
x
no,
is still the same outside
(g) Would the field inside the slab change?
yes, field inside is zero
(h) Would the distribution of charge within the slab change? How so?
yes, charge is all on surface of the plates
Now that we have all the ingredients, let’s analyze a more complicated system.
Two non-conducting slabs of infinite area are given a charge-per-unit area of a = -4 C/m2 and
2
b =+1.5 C/m respectively. A third slab, made of metal, is placed between the first two plates.
The charge density m on the metal slab is 0 (i.e. the slab is uncharged).
y
a
m
b
= -4 C/m2
2
b = +1.5 C/m
m=0
a
x
A
B
non-conducting
metal
C
non-conducting
(a) Find the magnitude and direction of the electric field at the points A, B, and C.
EA = EB = (-5.5 C/m2)/2
EC = (-2.5 C/ m2)/2
o
o
All in negative x-direction.
(b) Now let’s think physically about the metal slab. The defining property of a conductor is that
the charges within are free to move, and they will do so in response to an electric field. As a
result, surface charges L and R will accumulate on the left- and right-hand surfaces of the slab.
What are the signs of these surface charges?
2
L
– positive
R
- negative
(c) Go ahead and use Gauss’ Law plus your answers to question (a) to calculate the surface
charges L and R induced on the metal slab.
L
= 2.75 C/m2
R=
-2.75 C/m2
(d) Finally, let’s add a net charge per unit area of
the induced surface charges L and R?
L
= 4.25 C/m2
R=
m
= +3 C/m2 to the metal slab. Now what are
-1.25 C/m2
3
Discussion Question 3D
P212, Week 3
Electric Field Due to Charged Cylinders and Lines
Consider a non-conducting cylinder of infinite length with a hollow core. The inner radius is a,
the outer radius is b, and the solid region in between carries a uniformly-distributed volume
charge density .
(a) Using Gauss’ Law and the steps outlined earlier, calculate the electric field at a radius of r
from the axis of the cylinder, where a < r < b. (This one is a bit challenging.)
E2 r
E
r2
2
Q
a2
r
o
r2
b
a2 )
a
rˆ
(b) Now suppose a hollow metal cylinder of inner radius c and outer radius d is placed around
the non-conducting tube. The metal tube carries no net charge. What is the surface charge
density c induced on the inner surface of the metal cylinder?
uncharged
metal tube
Qin
Qin
Qtube
1
a2 )
b2
2 c
b2
b2
a2 )
2c
a2 )
a
b
c
d
Now for a challenge! Suppose we add an infinite line charge inside the hollow cylinder, but at
an off-center location: the line runs along the line x = d where d < a. The line carries a charge
per unit length . The values of all parameters are given in the figure.
y
a = 3 cm
b = 5 cm
= +4 C/m3
d = 2 cm
= -10 C/m
b
a
x
d
(c) What is the electric field
at the position (x,y) = (–b,0) (i.e. the point marked with
an x on the figure)? Be sure to give both magnitude and direction.
This one takes a little thought: the problem looks like it does not have enough symmetry for
you to apply Gauss’ Law, but you do have the tools to solve it … think superposition …
E
=
(b d )
2
1
8.85 x10
b2
xˆ ; Ecyl
1x10 7
12
) 0.07
a2 )
xˆ ; E
b
b2
1
b d
.052 .032 ) 4 x10
0.05
6
a2 )
x̂
b
18459 N/C xˆ
(d) And finally, calculate the components of the electric field at the point (x,y) = (0,b).
d
E
E=
b2
d
b2
d2
d2
xˆ
b2
1
d2
b2
E 12402 xˆ (7232 31006)yˆ
2
b
xˆ
b2
d2
a2 )
b
b2
d2
ŷ
b
b2
d2
(12402 xˆ 23774 yˆ ) N / C
; Ecyl
b2
a2 )
b
yˆ