Transcript If I bring a charged rod to a leaf electrometer: A] nothing will happen

Real Insulators (Dielectrics)
If I bring a charged rod to a leaf electrometer:
A] nothing will happen
B] nothing will happen until I touch the electrometer with the rod then the leaves will spread apart
C] the leaves will spread apart when I get the rod close to the
electrometer
What is the direction of the electric force on a negative
charge at 1? B
A] to the right
B] to the left
C] up
D] down
What is the direction of the electric force on a positive
charge at 2? B
If I put a + charge at 3., what is the direction of the electric
force?
A] straight down,
directly toward the - charge
B] straight up,
directly away from the - charge
C] a little to the right from straight down
D] a little to the left from straight down
What could be the charges here?
Assume that all field lines are shown.
A] Q1 = +3C, Q2 = -2 C
B] Q1 = +18C, Q2= -10 C
C] Q1= -18C, Q2=+10 C
D] Q1 = +9 C, Q2 = +5 C
E] none of these
Electric field is defined by the force on a “test charge”.
Since forces add by superposition, fields do also.
Field lines:
• Field lines never cross each other
• Field lines are perpendicular to conductors at their
surfaces
• Field lines share the symmetry of the sources & sinks
• Density of field lines (in 3D) is proportional to field
strength
Which picture shows the field lines from an infinite sheet
containing a uniformly dense negative charge? D
What is wrong with picture B?
A] field lines are
symmetric
top/bottom
B] field lines cross
C] field lines are
symmetric right/left
D] field lines should
start at sheet
What figure shows the E field lines for an electric dipole? B
What’s wrong with Figure C?
A] it has the wrong
symmetry
B] it has the wrong
direction of field
lines
C] field lines have
sharp turns
Two source charges have equal magnitude. One is + and the
other -, as shown.
What vector could represent the electric field at the point
shown caused BY THE POSITIVE CHARGE ALONE? B
Given that B is the field from the + charge alone, what
vector could be the total field from both charges? C
If charge +3q is changed to -3q, what is the direction
of the total E field at p?
A] upward and to the left
B] straight to the left
C] downward and to the left
D] straight downward
E] the E fields from +q and -3q perfectly cancel
Calculus: how do we find E fields from extended
charge distributions?
Use symmetry
Apply Coulomb’s Law to infinitesimal parts
Take components
Use calculus to sum up all contributions
for either the x or the y component


Two plates extend the same distance into the page.
Their heights, projected onto the vertical axis, are the same.
What is the area of plate 2, compared with the area of plate 1?

A] same
B] A1 cos
C] A1 /cos
D] A1 /sin 
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If the dot product of E and the area of plate 1 is EA1, what is the
dot product of E and the area of plate 2?

A] same
B] EA1 cos
C] EA1 /cos
D] EA1 /sin 

What would happen to the flux if we were to double the field?

A] same
B] double
C] half
D] quadruple

The flux is proportional to the field, and to the perpendicular area.
We know the density of field lines (#lines per perpendicular area) is
proportional to the field.


So the flux is proportional to the number of field lines
penetrating a surface area.
Note: the sign depends on how we “orient” the
surface!
Does the total flux through a spherical surface at radius
R change if we move the charge off-center?
• A] yes
• B] no
Does the total flux through a closed surface surrounding a
point charge change is we distort the surface (while keeping
it closed)?
• A] yes
• B] no
Gauss’ Law is useful for finding E fields when there is
symmetry.
Sphere of charge
Line of charge
Sheet of charge
Gauss’ law also shows that excess charge on a conductor is
all at the surface!
A small sphere and a short rod have the same charge Q.
Gauss’ Law allows us to conclude that the electric field a
distance r away from each (if looked at separately) would be
the same.
• A] True
• B] False
FALSE! There is not enough symmetry to use
Gauss with the finite rod!!!!!!
What is the charge on the inner surface of the conducting shell?
A] 0
B] +Q
C] -Q
D] +2Q
E] cannot determine
If I put no excess charge on the conductor, the net charge on
its inner surface will be:
a] 0
b] +Q
c] -Q
d] -2Q
e] cannot determine
Because there is zero E field in a conductor, the inner
surface must hold a charge -Q.
What is the charge on the outer surface?
a] 0
b] +Q
c] -Q
d] -2Q
e] cannot determine
With no net charge on the outer conductor, the inner surface
gets -Q and so the outer surface gets +Q.
Suppose, instead, I put +8Q on the outer conductor. What
then is the charge on its inner surface?
a] 0
b] +Q
c] -Q
d] +3Q
e] +4Q
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