Download UV practice

More Practice!
You will be given the following:
• One object such as from those below
or
• A region with e-field lines indicated as in the example below
• A location A and a location B
• Either a positive or a negative small test charge that is moving from A to B.
1.
2.
3.
4.
Answer the questions for the situation given:
Draw in the field lines if not given.
Decide the charge on the object if not given.
Assume the charge has a velocity heading toward B. Will the charge slow down or
speed up? Draw a force vector and decide based on what you know from your study
of motion.
5. Use the energy bar charts to help decide if Uelec increases or decreases. Negative
values are less than positive.
6. When charged particle moves from point A to point B the work done by the E-field
would be positive, negative, or zero?
7. From WE, figure out whether the charge must slow down, or speed up or neither.
-
- - - - - - - - - -
+ + + + + + + + + +
+
-
+
+
+
+
+
+
+
Example solution for the situation given and a positive test charge.
A
1. What charge must be causing the field? (Decide whether the
shaded area is positively or negatively charged.)
B
Since the E field lines point left, then the force on a positive test
charge will be to the left and the shaded area charge must be attracting the positive
test charge. That makes the shaded area negative.
2. Assume the charge has a velocity heading toward B. Will the charge slow down or
speed up? Draw a force vector and decide based on what you know from your study
of motion.
Force
acceleration
Vel. initial
Since the force is to the left then the acceleration is to the left. If vi is to the right, then
it must slow down.
3. Use the energy bar charts to help decide if
Uelec increases or decreases.
@B
@A
K
Ue
W
K
Ue
Since K decreases, U must increase. And Ue
is negative (from kq/r or from knowing it is
a potential well, deep hole). Note that W is
work by forces other than elec and grav
which is zero in this case.
4. When charged particle moves from point A to point B the
work done by the E-field would be positive, negative, or zero?
FE
Δx
From the equation Wby E=Fll by E•Δ (see vectors), work by
electric field is negative.
5. From WE, figure out whether the charge
must slow down, or speed up or neither.
Since work is negative, the K must decrease.
U
@B
@A
K
WbyE
V
6. Place UA and UB on an energy number line.
UB
UA
VB
VA
7. Place VA and VB on an energy number line.
K
Example solution for the situation given and a NEGATIVE test charge “on”.
A
1. What charge must be causing the field? (Decide whether the shaded area
is positively or negatively charged.)
B
Since the E field lines point left, then the force on a positive test charge
will be to the left and the shaded area charge must be attracting the
positive test charge. That makes the shaded area negative. Note that even though we
will later ask questions about the Neg “on” charge, we have not at all changed the
“by” charge on the plate and that is what causes the E-field. (For E directions, think
forces on + “on”charges always!!)
2. Assume the charge has a velocity heading toward B. Will the charge slow down or
speed up? Draw a force vector and decide based on what you know from your study
of motion.
Force
acceleration
Vel. initial
Since the force is to the right then the acceleration is to the right. If vi is to the right,
then it must speed up. Notice that force is opposite the direction of the E-field since it
is “on” a neg charge.
3. Use the energy bar charts to help decide if
Uelec increases or decreases.
@B
@A
K
Ue
W
K
Ue
Since K increases, U must decrease. And Ue
is positive (from kq1q2/r).
4. When charged particle moves from point A
to point B the work done by the E-field would be positive,
negative, or zero?
FE
Δx
From the equation Wby E=Fll by E•Δ (see vectors), work by
electric field is positive.
5. From WE, figure out whether the charge
must slow down, or speed up or neither.
@B
@A
K
WbyE
Since work is positive, the K must increase.
U
V
UA
UB
VB
VA
6. Place UA and UB on
an energy number line.
7. Place VA and VB on an energy number line.
Remember that V=U/qon and qon is neg this time.
K