Transcript With insulators, you can often use Gauss` law in combination with

With insulators, you can often use Gauss’ law in
combination with the principle of superposition.
(PoS is that the field from two sources is the vector
sum of the fields from each source.)
Two examples:
1) Two infinite planes, one with charge density +, one
with charge density - .
2) Subtractive superposition: a sphere with a spherical
(off center!) void
What is the direction of the electric field at the point shown?
A] NE
B] SE
C] NW
D] SW
E]it varies, depending on distances a & b
What is the magnitude of the electric field at the point shown?
A]
B]
C]
D]

20

0
2
0
2
20
E] 0
Use extreme caution when applying Gauss’ law, when
conductors are present.
It is important not only that the “initial state” have
sufficient symmetry, but also that the final state maintain
the symmetry.
For example, consider an infinite plane with initial
charge density , and then add a point charge above it.
If the plane is insulating, you can solve this problem.
If the plane is conducting, you can’t solve this problem!
Electric Fields & Conductors
(in statics…)
No field in the conductor (or else charges would move)
So, by Gauss’ law, no excess charge inside conductor
(all excess charge is on surfaces)
Field lines must contact the conductor perpendicularly
The surface charge density is directly proportional to the
field just outside the conductor (by Gauss’ law)
E = 0
Let’s draw some field lines for
a conducting plane with -Q total charge, and an
external point charge of +Q
& conducting shells with charges inside.
There still must be no field in the conductor. So the inner
surface is still -Q. Thus, the outer surface will have +9Q, so
that the total is 8Q (=+9Q - 1Q).
Given the peanut shaped geometry and the fact that the
charge +Q is off-center, is the charge density on the inner
surface uniform (the same everywhere on that surface)?
a] yes
b] no
The negative charge on the inner surface will be
concentrated close to the positive charge. (The E field next to
the surface is stronger there !)
If there is +9Q of charge on the outer surface, will the charge
density on the outer surface be uniform?
a] yes
b] no
There is no electric field in the conductor. So the outside of
the conductor cannot “know” that there is an off-center
charge in the middle!
So the charge density on the outer surface is uniform.
a] yes
A metal box with no net charge is placed in an initially
uniform E field, as shown.
What is the total charge on the inner surface ? Assume this
surface has area A.
A] 0
B] 0 EA


C] 0 EA
D] 20 EA
E] cannot determine

No charge on the inner surface. What is the total charge on
the outer surface?
A] 0
B] 0 EA


C] 0 EA
D] 20 EA
E] cannot determine

No charge on the inner surface. Since the total charge is 0,
there can be no net charge on the outer surface.
What is the charge density on the left outer face of the box?
Assume the external field is uniform, as shown.
A] 0
B] 0 E


C] 0 E
D]
20 E
E] cannot determine

What is the electric flux of the field
through the rectangle in the xz plane between x=[0,L1], z=[0,L2]
A] 0
B] L12L2
C] larger than answer B
D] smaller than answer B
What is the electric flux of the field
through the half pipe shown:
A] 0
B] L12L2
C] larger than answer B
D] smaller than answer B



What is the flux of the electric field through the
surface shown?
A] 0
L
B]
0
L
C]
0
D] L
80
E] 2 L
0
A point charge +Q is a distance d above an insulating sheet
with charge density .
What is the field at point P?
Insulating sheet: superposition of fields gives answer B.
Suppose, instead, that a conducting sheet with charge
density =  is brought from far away (far down, in the
picture) to a distance d away from the charge +Q, then what
is the field at P?
With a conducting sheet, the charge +Q will cause the charges to
redistribute. Cannot determine! (Need Physics 400 level)
Suppose, instead, that a conducting sheet with charge density = 0 is
brought from far away (far down, in the picture) to a distance d away
from the charge +Q, then what is the field at P?
In which case does the electric potential energy increase?
A
Or C: both cases
B
D: neither case
In which case does the electric potential energy increase?
A
Or C: both cases
B
D: neither case
Work done by electric force (source: fixed charges)
on a test charge
depends only on endpoints, not on path.
(You can see this easily for a single fixed charge… it holds in general
because of superposition.)
Electric forces are “conservative” - We can define a potential energy.
When a + charge moves “down the field”, the electric force does work
on it, increasing its kinetic energy (or putting energy elsewhere).
When a + charge moves “up the field”, it either loses kinetic energy, or
some other force must push it up.
The electrical potential energy of a system of charges is the work
necessary to assemble the charges from “infinity”.
(For point charges, we take U=0 at infinity.)
This will include all pairs of interactions.
Two equal + charges are initially stationary and separated by r0.
If they are allowed to fly apart (to infinity), what will be the kinetic energy
of each?
q
A] 40 r0
q2
B] 40 r0 2

1 q2
C]
4 0 
r0
1 q2
D]
8 0 r0


1
1
Three equal + charges are initially stationary and at the vertices of an
equilateral triangle with side r0.
If they are allowed to fly apart (to infinity), what will be the kinetic energy
of each?
q
A] 40 r0
q2
B] 40 r0 2
1 q2
C]
4 0 
r0
1 q2
D]
8 0 r0
1

1

Just as we can define electric field as the force felt by a test charge
We define “potential” as potential energy of a test charge.
1 q pc
Vpc 
40 r
dV 
1 dq
40 r
Just as a conservative force is:
(minus) the derivative of the potential energy

 W  U   dU 
The electric field is
(minus) the derivative of the potential.

 Fdx
Eqiupotential surfaces
Equipotential surfaces are perpendicular to field lines
In 2D pics, equipotentials look like lines, but they are surfaces.
Note: just because V=0 doesn’t mean E=0!
A function can be zero but have a non-zero derivative.
Also: it’s time to think in 3D. The derivative can be taken w.r.t
x, y, or z.
dV  E  dl
dV  E x dx  E y dy  E z dz
V
Ex  
x
E  V
This means: hold y, z constant, so dy=dz=0
Note that E is a vector, but V is a scalar.
3 ways to calculate E fields
A) Direct sum of sources, using Coulomb + Calculus (+ Components!)
B) Gauss’ Law & Symmetry
C) The negative of the derivative of the electric potential (if given)
2 Ways to calculate the electric potential
A) The negative of the integral of the E field (if given)
B) Sum of the sources, using Calculus if necessary
Note: by sum of sources, I mean use the result from integrating the
Coulomb field for a point source, V = kq
r

Last time, we found the potential from a ring of charge.
Here’s another example of integrating over sources to find V:
A line of charge.
Let’s find V by integrating E for a line of charge.
A perfectly insulating plane has charge density
+2 C/m2.
What is the magnitude of the E field a distance x
above the plane (in terms of x and?
A] 
B] 1/
C] 2/
D] 2x
F] 2/(x
Where is the electrical potential higher?
A] farther from the plane of positive charge
B] closer to the plane of positive charge
C] it’s the same everywhere, since the plane is infinite
The field is 1.13 x 1011 N/C, regardless of x. (Gauss)
What is the magnitude of the difference in potential
between a point in the plane and a point 10 m above
the plane?
A] 1.13 x 1011 V
B] 1.13 x 1010 V
C] 1.13 x 109 V
D] 0 V
E] potential difference does not exist in this problem
since V ≠ 0 at infinity
E (none)
Which graph could be the potential from an infinite plane
of positive charge density, where x = distance from
plane?