# Real Insulators (Dielectrics)

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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
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
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”
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
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
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] C] D]

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] C] D]

What would happen to the flux if we were to double the field?
A] same B] double C] half D] quadruple

Note: the sign depends on how we “orient” the surface!
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
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
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?
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
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
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

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