Download Exam I, vers. 0001 - University of Colorado Boulder

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Exam I, vers. 0001 - Physics 1120 - Fall, 2006
SOLUTIONS (SJP)
1. Two charged particles are fixed on a line. The charge on the left has positive charge +2q, and
the charge on the right has positive charge +q, as shown below. An electron is placed somewhere
on the line in between the two charges as indicated on the diagram.
+2q
+q
The electron is placed on the
line above in this region
If the electron is released from rest, it will initially
A) accelerate to the left, no matter where it is placed.
B) accelerate to the right, no matter where it is placed
C) accelerate right, left, or remain stationary, depending on where it is placed.
D) remain stationary, no matter where it is placed.
If you put it RIGHT NEXT to the +q, the 1/r^2 of the very nearby charge would win out,
because it goes to *infinity* (no matter WHAT the particular charge is), and that would
make a net attraction (to the right). But if you put it right next to the +2q, now that 1/r^2
"blowup" would make the force be to the left. So as you move it from near the +2q to
near the +q, you steadily change the net force from "strong left" to "strong right" with
everything inbeetween! You can get forces to the left, to the right, or even, at one sweet
spot, balance!
2. A block of mass m and positive charge +q is placed on a
frictionless ramp that makes an angle θ with the horizontal as
shown. The magnitude E of a horizontal electric field is adjusted
until the mass does not slide.
A student draws the following correct force diagram for this
r
situation, where N represents the normal force of the ramp
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Frictionless ramp
m,+q
θ
r
E
2
N
mg
qE
What is the magnitude of the field E?
A)
mg
q sin !
B)
mg tan !
q
C)
q sin !
mg
D)
q
mg tan !
E)
mg
q
This is mostly math and geometry! What you have to convince yourself is that in the triangle
shown, the UPPER angle is the same as "theta" in the figure. (Thing about the limit theta -> 0,
the triangle would get tall and skinny, it's the UPPER angle that would vanish). So, from that
triangle then
qE / mg = tan(theta), or E = mg tan(theta)/q.
Does it make sense? Units are correct, E should be (force / charge), which (mg / q) is.
If theta -> 0, then the ramp is flat, and you don't need ANY E AT ALL to hold it still. Our answer
should go to 0 (and that's what tan(0) is).
If theta -> 90, the ramp is vertical. This is a problem, without friction, there is NO WAY you can
make it "stick". There's nothing to cancel gravity? So E would blow up, it would have to go to
infinity. That's just what tan(90) does.
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The next two questions refer to the following situation. Two negative charges are fixed on the yaxis and a positive charge is fixed on the x-axis as shown in the
y –Q
diagram. The three charges form an equilateral triangle with
sides of length r. Point A is the origin. An electron is fired from
r
o
60
an electron gun to point A along the dotted path shown.
r
Consider only electrostatic forces which act in the x-y plane.
30o
A
+Q x
3. When the electron is at point A, in what direction is its
Electron
instantaneous acceleration?
r
path
A)
B)
C)
D)
E)
-Q
the acceleration is zero, so there is no direction
upwards, but slanted towards the right
upwards, but slanted towards the left
directly left (in the negative x direction)
directly right (in the positive x direction)
Think first about the E field at that point. It arises from three charges. But the two -Q's are
symmetric, they yield zero net field at point A. Only +Q contributes, making an E field at that
point which points straight LEFT. But electrons are negatively charged, so the force (and thus
acceleration) would be directly right.
In the end, the eletron is repelled equally from the two -Q's, but attracted straight towards the
+Q. (But only at point A!)
4. The magnitude of the force on the electron at point A would be
A)
3kQe
r2
B)
kQe
r2
C)
kQe
r sin 2 (60 ° )
2
D)
kQe 2
sin (30 ° )
r2
E)
3kQe 2
sin (30 ° )
r2
F = qE, and E = the vector sum of the three E's arising from the three big charges.
The upper and lower E's cancel out, leaving just the E from the single +Q:
E = kQ/(distance)^2, so we just need to look at the picture and find the distance from A to +Q.
From the upper triangle, that's r sin(60), so
F = kQe/(r sin(60))^2
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5. Two large, parallel planes of charge made from nonconducting material are shown edge-on. The left plane has
positive charge density ! , and the right plane has positive
charge density 2! . The planes are so large that they can be
considered infinite in extent. What is the magnitude of the
electric field in region III, which lies to the right of both
planes?
A)
3"
!0
B) 0
C)
"
2! 0
!
I
D)
"
!0
2!
II
E)
III
3"
2! 0
This is the superposition principle. A nonconducting sheet makes an E field which is sigma/(2
epsilon0) AWAY. If you are in region III, you must superpose the field from the "sigma" sheet
(2" )
"
and the field from the "2 sigma" sheets. The first is
, the second is
, and they both
2#0
2! 0
point in the SAME direction (to the right), so they add up to 3 sigma / (2 epsilon0)
6. A small negative test charge -q is released from rest near a
larger positive fixed charge +Q. Initially, as –q starts to move!
towards +Q, the magnitude of –q’s acceleration
A)
B)
C)
D)
increases with time
decreases with time
stays the same (but non-zero) as time passes
stays the same, exactly zero, as time passes
-
+++
+++
+++
-q
Q
The force is simply Coulomb's law. As time passes, -q will accelerate to the right, it will get
closer and closer to +Q! Since the force is kqQ/r^2, if r is getting smaller and smaller, the force
must be getting bigger and bigger. Bigger force means bigger acceleration. This is NOT a
constant acceleration (you couldn't use the old 1110 kinematics equations!)
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The next two questions refer to the diagram below. The diagram shows electric field lines
arising from two small charged particles P and Q. There are no other charges besides P and Q.
C
B
Q
P
A
7. Choose the best ranking for the electric fields at the three points A, B, and C, as indicated on
the diagram above:
A) E A > E B > EC
C) EC > E B > E A
B) E A = E B > EC
D) EC > E A = E B
E) E B > EC > E A
From the picture, where the lines are most dense, the field is the strongest. The lines are closely
spaced at B, a little farther apart at C, and MOST far apart at A. S
It should be
B>C>A
8. Which of the following statements is correct (only one is correct)
A)
B)
C)
D)
The electrostatic force on P is smaller than that on Q
The electrostatic force on P is greater than that on Q
The electrostatic force on P is equal to that on Q
There is not enough information given to determine whether the electrostatic force on
P is greater than or less than that on Q
This is Newton's third law, the Force of any object (A) on another one (B) will be equal and
opposite the force of B on A. That's a law of nature, holds for electricity like any other force!
(Note that the field line diagrams really don't give you much information about this question....
they tell you about the forces THESE two charges exert on OTHER charges elsewhere in the
universe! You have to think about connecting what we're learning now with older, fundamentally
Newtonian ideas, to make full sense of this whole business!)
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9. An insulating bar of length l and width w has a charge +Q
uniformly distributed on its upper half, and a charge –Q
uniformly distributed on its lower half. The point P lies
adjacent to the midpoint of the bar, as shown. What is the
direction of the electric field at the point P?
A) up
B) right
C) down D) left
E) Some other direction or zero.
A
w
P
D
B
C
l
2
+Q
l
2
-Q
The best way for me to think about this one is to imagine breaking that rod up into small chunks.
Think of a symmetric PAIR of dQ's, one above, one below. That's just a little electric dipole! The
upper + dQ will make an arrow at P which is "down and left", the lower - dQ will make an
arrow at P which is "down and right" (attractive). And those two arrows add up to a small
STRAIGHT down arrow. But now, pair by pair, each pair of dQ's adds more STRAIGHT DOWN
arrows at P. That's the result,( just like you would get from a single dipole! )
10. A plastic rod is charged up by rubbing it with a wool cloth, and brought near an initially
uncharged insulating sphere. The rod does not touch the sphere. The rod
A)
B)
C)
D)
is repelled by the sphere.
is attracted to the sphere.
cannot possibly feel any force due to the sphere.
could be attracted or repelled, depending on whether a plastic rod rubbed by a wool cloth
becomes positively charged or negatively charged.
Objects will always polarize in the presence of electric charges nearby. It doesn't matter if they
are insulators or conductors, even insulators still have atoms in them which can individually
polarize! Induced (polarization) charges always attract (because the opposite charge is always
nearer). It doesn't matter the sign of the plastic rod!
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11. An electric dipole is placed in an electric field
as shown to the right. At the instant shown, what
direction is the net force on the dipole?
A) up
B) right
C) down
D) left
E) The force is exactly zero
A
D
B
Dipole
+
C
The field is STRONGER on the left, and WEAKER
on the right (because the lines spread out).
The + charge in that little dipole is pushed right,
STRONGLY, and the - charge is pulled left, but
more WEAKLY. "dipoles", by definition, have exactly equal + and - charges on the two ends, so
these two forces do NOT exactly balance. The repulsion of the + "wins", and the net force is to
the right.
For you to think about - after awhile, if this dipole were "jostled" a little bit, what would end up
happening to it?
12. Two concentric spherical surfaces are labeled S1 and S2. The
volume within the inner surface S1 is called region A. The
volume between S1 and the outer surface S2 is region B. The
total flux through S1 is zero and the total flux through S2 is
negative. Which of the following is correct for QA (the total net
charge in region A), and for QB (total net charge in the region B
between the two surfaces)?
[Hint: what does Gauss’ Law tell you about the connection
between the Flux and the enclosed charge?]
A
S1
B
A) QA < 0, QB < 0
S2
B) QA = 0, QB > 0
C) QA < 0, QB > 0
D) QA > 0, QB > 0
E) QA = 0, QB < 0
Gauss' law says "Flux = Q(enclosed)/epsilon0. So, Since S1 has ZERO flux, it must enclose
ZERO net charge. So QA=0.
Gauss' law says if S2 has negative flux, it must enclose total negative charge. But we just argued
QA=0, so QB MUST be negative then!
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13. A thin, uniformly-charged rod has total
charge Q and length L. A point P is a distance
d from the end of the rod along a line parallel
to the rod. What is the correct expression for
the magnitude dE of electric field at point P
due to a small segment dx of the rod a
distance x from the end, as shown in the
diagram?
A)
point P d
L
dx
x
Total charge Q
kQ dx
kQ dx
kQ dx
kQ dx
kQ
B)
C)
D)
E)
2
2
2
2
Lx
xL
d2
L( x ! d )
(L + d )
This is just Coulomb's law. dE = k dQ / (distance)^2.
What is distance? Distance from point P to the charge, which from the figure is just "x"
What is dQ? that would be (charge/length)* (little length),
Or dQ = (Q/L) * (dx).
Putting it together, dE = k Q dx / (L * x^2)
14. Four charged particles (two having a charge +Q and two having a charge –Q are distributed
as show below. Each charge is an equal distance d from the origin. What is the magnitude of the
net electric field at the origin? (Hint: cos 45o = sin 45o = 1/ 2 .)
y
4kQ
d2
B) 0
2kQ
C)
d2
A)
+Q
+
d
d
2 2kQ
-Q
D) This one:
d
d2
3kQ
E)
d2
At the origin, the upper +Q makes a straight down
field, of magnitude E0 = kQ/d^2. But, they ALL
make fields of this same magnitude! It's just directions that are different.
d
+
+Q x
-Q
The two + charges make arrows "down" and "left" respectively.
If you add a down arrow and a left arrow, (draw the picture!) you get a "down-left" arrow of
length Sqrt[E0^2 + E0^2] = E0 Sqrt[2]
Now, the two minus charges ALSO make E fields "left" and "down", so they do the exact same
thing. So the result is what I just got, doubled!
E = 2 Sqrt[2] E0 = 2Sqrt[2] k Q/d^2, this is answer D
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15. Four charges, –Q, -2Q, +Q, and +2Q, are arranged in a square, centered on the origin, as
shown. What is the direction of the electric field at the
y
origin?
–Q
-2Q
A) directly up (in the positive y direction)
B) directly left (negative x direction)
C) upwards, slanted towards the left
D) upwards, slanted towards the right
E) the net field is zero, so there is no direction
+Q
+2Q
At the origin, the +Q makes a SMALL arrow up and left.
The +2Q makes a doubly large arrow up and right. Since
the distnace of the corners from the origin is the same,
you're adding two arrows, and the doubly big components will "win". So those two add up to a
force which is "up and right".
The -Q makes a small arrow up and left. The -2Q makes a bigger arrow up and right. This is
exactly the same story, it just doubles up what I just got!
Conclusion: the result is "up and right"
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10
16. Two charges are released from rest at the positions shown on the x-axis. The one on the left
(labeled 1) has mass m and charge +Q, and the one on the right (labeled 2) has mass 3m and
charge +4Q. Compare the magnitudes of the accelerations of the two charges immediately after
releasing them from rest:
m, +Q
A) a1 = 4a 2
D) a1 = 3a 2
B) a1 = a 2 / 4
E) a1 = 4a 2 / 3
3m, +4Q
+
+
1
2
C) a1 = a 2
The forces are equal and opposite (by Newton's III law), and remember, F= ma.
So F1 = m a1, and F2 = (3m) a2.
But the magnitudes of F1 = F2, which means m a1 = (3m) a2 (in magnitudes)
So a1 = (3m) a2 / m = 3 a2.
The Light one feels the SAME force, but accelerates three times as much!
17. The electric field throughout a region of space is given by the formula
ur
E = A y xˆ + B x yˆ ,
r
where (x,y) are the coordinates of a point in space, and A, B are constants. What is E ! xˆ ?
A) Ay
B) Bx
C) Ax
D) By
E) none of these
When you dot, x(hat) dotted with x(hat) gives 1, but y(hat) dotted with x(hat) gives 0.
So the answer is just Ay, nothing more to it...
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18. In Figure I, two positive charges +Q and + Q0 are a distance b apart. In Figure II, the
situation is the same but an additional neutral metal cube is moved next to the charges in the
position shown. Compare the magnitude of the net electric force on the charge + Q0 in Figure I
with the magnitude of the net electric force on the charge + Q0 in Figure II.
+Q
I
+Q
+ Q0
b
II
+ Q0
neutral
metal cube
b
The magnitude of the net electric force F (on + Q0 ) in Figure I is
A) greater than
B) less than
C) equal to
the magnitude of the net electric force F (on + Q0 ) in Figure II.
In Figure I, there is a rightward arrow on Q0, F0 (on Q0, by Q)
In Figure II, there is still the EXACT same right arrow on Q0, namely again F0 (on Q0, by Q),
but IN ADDITION, there is another force, the attraction to the neutral metal cube! (Why is there
an attraction? Because the cube must polarize. It becomes --- on the left, and +++ on the right,
and the --- is closer. So it attracts Q0 a little. )
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Question 1 (10 points total). Consider the following two arrangements of point charges fixed in space.
+Q
+q
A
θ
θ
y
+Q
x
+q
+2Q
x
B
Is the magnitude of the electric force on +q due to the two +Q charges in case A greater than, less than, or
equal to, the magnitude of the electric force on +q due to the single +2Q charge in case B?
(a) (1 point) Write greater than, less than, or equal to: ________LESS THAN__________________
(b) (4 points) Carefully explain your reasoning. Please be sure to include ALL reasons you can think of:
There are TWO reasons, and both are important (and we needed to see them both for full credit!)
First of all, note that the hypotenuse of any triangle is the biggest side, so the distance from q to Q is
BIGGER in figure A. That means the Coulomb force will be SMALLER due to each of those two +Q's. So
this makes me argue the force on +q is SMALLER in A.
But there's another effect. The force from each of the two charges is a vector, which points down and left at
+q (from the upper one) and up and left (from the lower one). Only the x components of these arrows will
add up, the y-components CANCEL. So that cancellation FURTHER reduces the effective force.
Both of these effects are in the same direction, they both make F(in A) smaller than F(in B)
Now consider a third arrangement consisting of the +q charge with a
charged insulating rod having length y and charge +2Q distributed
uniformly along it.
Rank the magnitude of the electric force on the +q in case C with that in
case A and case B. [ For example, if you thought the force was strongest
in case A, with B next strong, and with C the weakest, write A>B>C. ]
+q
C
θ
θ
+2Q y
x
(c) (1 point) Your rank: __________B >C>A_________________
(d) (4 points) Carefully explain your reasoning. Please be sure to include ALL reasons you can think of:
We already figured out B>A above. So we need to see where C "fits".
Let's compare C with B. It's the same reasoning as above, except now we have to split that rod up into a
bunch of little dq's. I would consider them in PAIRS. Each PAIR looks a lot like the arguments above: the
points are FARTHER away than x, and there is some cancellation. But, note that ALL those pairs are "closer
in" than the pair in case A. So the extra distance isn't as much, and the cancellation effect isn't as much
(because their angles are all LESS than the full theta). So although we get these same types of cancellation
in C as we did in A, we do not get AS MUCH. So C will be intermediate in force between A and B.
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Question 2 (15 points total). Consider an imaginary cubical surface with sides of length L surrounding a
positive point charge +q as shown below. Initially the charge is inside the cube in the exact center. This is
shown in case A below. The charge is then moved some distance to the right, so that it remains inside the
cube, but is now closer to the right hand face. The new position is shown in case B below.
L
Case A
L
Case B
the charge +q
is near the right
hand face
charge +q is at
the center
right hand face
right hand face
Is the flux through the right hand face in Case A greater than, less than, or equal to, the flux through the
right hand face in Case B?
(a) (1 point) Write greater than, less than, or equal to: ___LESS THAN____
(b) (4 points) Carefully explain your reasoning:
I could argue that since +q is closer, the E field is stronger, and flux = E dot dA must get bigger. But that's
NOT a complete argument, though, because there's a "cancelling" effect - namely, the field lines near the top
of the grey face in B are STEEPER now, and so that cos(theta) means LESS flux. So how do we decide which
"wins"? I would say, draw the field lines! In case A, a certain number of lines leave the charge, and exactly
1/6 go through each of the six faces. But in case B, the SAME number of lines leave the charge in exactly the
same patter, and any line that goes "right" is very likely to go through that right face. So you capture much
closer to HALF of the field lines now! That's a lot more than 1/6, so more flux!
Partial credit (but not full) for the "E is bigger" argument - reasoning needs to be compelling and complete!
Is the total flux through all six sides of the cube in Case A greater than, less than, or equal to, the
total flux through all six sides of the cube in Case B?
(c) (1 point) Write greater than, less than, or equal to: ___EQUAL TO______
(d) (4 points) Carefully explain your reasoning:
This is just Gauss' law! Total flux = Q(enclosed)/epsilon0, but Q(enclosed) is still the same.
OR you could argue the same total # of lines leaving the charge must exit the cube . (But both of
these arguments are NOT needed for full credit, either one is perfectly adequate and complete!)
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(e) (5 points) What is the total flux through the right hand face in Case A? Express your answer
ONLY in terms of quantities which are explicitly GIVEN (which might include some of q, L,
and/or constants of nature, but nothing else.) You must explain your work clearly for full credit.
Well... the TOTAL flux through all 6 faces of the cube is q/epsilon0. But look, the flux is
completely symmetric, the SAME flux must be exiting each of the 6 faces. So any one face must
have 1/6 the total flux, or q/(6 epsilon0).
That's it! Note that if you try to use Coulomb's law, or actually compute the flux by doing some
sort of integral, you're likely to get rather stuck. Just go back to Gauss' law, that's why it's so
cool!
If your answer involves L, it's probably not right, because the answer would be the same no
matter what L is! And if your answer involves something like "E*L^2", that's not at all right for a
variety of reasons. First of all, E is not uniform across the face, and neither is the angle between
E and the normal... and in any case, E is not one of the given quantities in the problem!
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