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Transcript
Relative to the effort it would take to thoroughly read
the chapter once, how much effort did you put into
doing the reading assignment?
a) 0-20%
b) 20-50%
c) 50-90%
d) 90-110%
e) 110-150%
f) 150-200%
g) >200%
Consider a thin spherical shell of radius 9.0 cm made
of a perfectly conducting material centered on the
origin of a Cartesian coordinate system. Two charged
particles lie outside the ball on the x-axis of the same
coordinate system: a particle with -5.0 microcoulombs
of charge at x = -11 cm and a particle with charge
+5.5 microcoulombs of charge at x = +11 cm. What is
the direction of the electric field at the origin.
a) The electric field is in the +x direction at the
origin.
b) The electric field is in the -x direction at the
origin.
c) The electric field is 0 at the origin. It has no
direction.
1.00 Coulomb of charge is placed on a
solid sphere made of a perfectly
conducting material. Find the magnitude
of the electric field at a distance of
1.00 meters from the center of the
sphere. Assume the radius of the sphere
to be less than 1.00 meters
a) 8.99 x 109 N/C
b) 1.00 N/C
c) 0.00 N/C
d) None of the above.
Why is the interior of a car considered to
be a good place to be during a thunder
storm?
a) A typical car is a good
approximation to a conducting
shell. Inside a conducting shell,
the electric field is zero.
b) A typical car is a good
approximation to a conducting
shell. Inside a conducting shell,
the electric field is uniform (but not
zero).
Consider an empty spherical gold shell
having an inner radius 25.0 cm and
thickness 1.5 mm (and hence an outer
radius of 26.5 cm) and a charge of
+2.40 mC. What is the direction of the
electric field at a point on the interior
surface of the sphere?
a) Radially inward (toward the center
of the sphere).
b) Radially outward.
c) There is no direction because the
electric field is zero.
Consider an empty spherical gold shell
having an inner radius 25.0 cm and
thickness 1.5 mm (and hence an outer
radius of 26.5 cm) and a charge of
+2.40 mC. What is the direction of the
electric field at a point 1 nm inside the
outer surface of the sphere?
a) Radially inward (toward the center
of the sphere).
b) Radially outward.
c) There is no direction because the
electric field is zero.
Consider an empty spherical gold shell
having an inner radius 25.0 cm and
thickness 1.5 mm (and hence an outer
radius of 26.5 cm) and a charge of
+2.40 mC. What is the direction of the
electric field at a point 1 nm outside the
outer surface of the sphere?
a) Radially inward (toward the center
of the sphere).
b) Radially outward.
c) There is no direction because the
electric field is zero.
Suppose you calculate a force using
F=qE and then use that force in Newton’s
2nd Law to calculate an acceleration. For
the latter step, what kind of diagram is
required?
a) A before and after picture.
b) A free body diagram.
c) No diagram at all.
What is wrong with the following
expression? 
N
E  15
C
a) The quantity on the left is a vector
whereas the quantity on the right is
a scalar. A non-zero vector cannot
be equal to a scalar.
b) Nothing.
What is wrong with the following
expression?
N
E  15 Westward
C
a) The quantity on the left is a vector
whereas the quantity on the right is
a scalar.
b) The quantity on the left is a scalar
whereas the quantity on the right is
a vector.
c) Nothing.
What is wrong with the following
expression?

N
E  15
Southward
C
a) The quantity on the left is a vector
whereas the quantity on the right is
a scalar. A non-zero vector cannot
be equal to a scalar.
b) Nothing.
What is wrong with the following
expression?
N
E  15
C
a) The quantity on the left is a vector
whereas the quantity on the right is
a scalar.
b) The quantity on the left is a scalar
whereas the quantity on the right is
a vector.
c) Nothing.
Suppose, during a test, you need the mass
or charge of an electron or proton. Where
do you find it?
a) One doesn’t “find” any of these
values. They have to be memorized.
b) One finds the masses and the value
of e, in coulombs, on the formula
sheet. One has to know that the
charge of a proton is +1e and the
charge of an electron is -1e.