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HW 7 solutions
1. (a) When the negative rod is brought between
the two spheres, the free electrons in each sphere
are repelled by the rod, so the electrons move to
the edge farthest from the rod. While the positive
charges don’t actually move, this creates
polarization in each sphere, with a net positive
charge on the side closest to the rod and a net
negative charge on the side farthest from the rod.
(But remember, each sphere is still neutral overall,
because no charges can flow into or out of the
spheres.) This happens similarly in both the
hollow and solid spheres, because in a solid metal
sphere, the excess charge is only near the surface,
and the interior of the solid sphere has zero net charge.
(b) The solid sphere is much more massive than the
hollow sphere, since they are made of the same
material. So, the solid sphere feels a much greater
force due to gravity. Each sphere feels roughly the
same electrostatic force (initially), so they both start
accelerating towards the rod. Each sphere will reach
equilibrium when the force of tension from the string
exactly balances the gravitational and electrostatic
forces. That happens at a much greater angle for the
hollow sphere than for the solid sphere. (see free body
diagrams) Additionally, as each sphere gets closer to
the rod, the electrostatic force will increase slightly,
which will have more of an effect on the alreadycloser hollow sphere, which will also contribute to it
swinging farther.
(c) Look at the first diagram, which shows what happens when both spheres are present. The
hollow sphere feels a strong attractive force from the rod and also a much weaker repelling force
from the solid sphere. Why? Well, even though the solid sphere is electrically neutral, it has an
excess of positive charge on the left side and an excess of negative charge on the right side.
Assuming that the distance between the sphere is larger than the radius of the spheres, the solid
sphere repels the hollow one, because the positive sides are closer; the repulsion of like-charge
parts overcomes the attraction from the unlike-charged parts. So, when the solid sphere is
removed, that small repelling force is taken away, so the hollow sphere will swing slightly closer
to the rod.
2. (a) The disk is still neutral. Even though charges are moving around inside the disk, there is no
way for charge to flow into or out of the disk, so the disk is still neutral.
(b) There is an attraction between the disk and the ball. Because the ball is positively charged, it
causes the free electrons in the disk to move to the interior of the disk. This results in an excess of
negative charge in the interior of the disk and an excess of positive charge on the outer edge of
the disk. The negative interior is closer to the ball than the positive outer edge, so the aluminum
disk is attracted to the ball.
(c)
(d) Ions can flow through the thin watery layer on your skin, your finger is slightly conductive
and has free flowing negative ions. When you touch the outer edge of the disk, the positive outer
edge strongly attracts negative charges from your finger, so negative charges flow onto disk.
Now, the disk has a negative net charge, which causes it to be even more strongly attracted to the
positively charged ball, causing it to “jump” toward the ball.
3.
(a)
The bead accelerates toward the large ball, since it has opposite charge. To calculate the
bead’s initial acceleration, we’ll use Coulomb’s law to find Felec, and then we’ll use Fnet =
ma. Since the bead begins r= 0.50 m from the center of the ball, Coulomb’s law gives us
(
)
9
2
2
−6
−6
k | qbead || qball | 9.0 ×10 N ⋅ m /C 0.75×10 C −6.0 ×10 C
Felec =
=
= 0.144 N .
2
r2
( 0.50 m )
To find the bead’s acceleration, we divide by the bead’s mass to get
a= Fnet/m = (0.144 N)/(0.010 kg) = 14 m/s2.
(b)
The derivation of the expression ∆x = v0t + (1/2)at2 assumes constant acceleration. That’s
not the case here: intuitively, the force on the bead becomes stronger and stronger as the
bead gets closer to the ball. Coulomb’s law agrees: as r decreases, Felec increases. Therefore,
the bead’s acceleration is not constant, which means we can’t use that expression for Δx.
(c)
The actual time is less than the 0.21 s that the student calculates. His procedure gives us the
bead’s travel time assuming a constant acceleration of a = 14 m/s2. In reality, the
electrostatic force—and hence, the acceleration—increases as the bead approaches the ball.
So, for most of its journey the bead’s acceleration is greater than 14 m/s2. Since the bead
gains speed more quickly than the student assumed, it reaches the large ball more quickly
than he calculated.
(d)
Except for Coulomb’s law, nothing was new: we used the same procedures we used when
examining other force-and-motion situations. Namely, we drew a force diagram to find the
net force, then used Fnet to find the acceleration, and finally used kinematics relations.
4. A small plastic ball of mass m = 10 g, charged with q = 5.0 μC, hangs from a
very light vertical spring of equilibrium length l = 0.40 m. The top of the
spring is attached to a bar located L = 1.00 m above the desktop. Attached to
the desktop directly below the ball and spring is a Q = –6.0 μC charge. The
ball hangs motionless, 0.35 m above the desktop.
(a) Find the spring constant of the spring.
Here’s the idea: the net force on the ball hanging
from the spring must be zero since the ball isn’t
accelerating. The magnitude of the forces acting
up on the ball (the spring pulling) must equal the
magnitude of the forces acting down on the ball
(gravity and the charge on the table attracting
down). I’m not going to worry about +/- signs,
because I’m comparing magnitudes of forces.
(b) The –6.0 μC charge is removed. Briefly describe
what happens to ball over the next few seconds.
Suddenly, the force pulling down is reduced so there’s a net force pulling up. The ball will go up,
speeding up, until the net force is zero. However, the ball won’t stop here since it has velocity, so actually
the ball will go above the new equilibrium and oscillate on the spring. If this were an ideal situation
without damping, this oscillation would happen forever, but I’m going to assume that this is a real-life
system which will eventually come to rest. Depending on which of these you assumed, your answers in
c) and d) will be different
(c) Eventually the plastic ball comes to rest at its new equilibrium position. Without doing any
calculations, quickly determine whether it settles a distance greater than, less than, or equal to 0.35 m
above the desktop.
I think it will settle higher than .35 m above the table. Since the force down is less, that means the spring
can exert less force and still hold the ball in equilibrium, which means the spring will be stretched less.
(d) Now calculate the plastic ball’s distance from the desktop.
Now there’s no electric force, so gravity is balanced by the spring force. We can figure out how much the
spring is stretched out after the charge is removed.
mg = KΔx so Δx = mg/K = 1.1 cm
If the spring is stretched 1.1 cm and it’s equilibrium length is 40 cm, then
the ball is 59 cm above the table.
5. Q1 = 10 μC and Q2 = 20 μC are nailed fixed in place. A frictionless
particle of charge Q3 = 10 μC is released from rest from the origin at t =0.
(See picture; the entire set-up is horizontal)!
(a) Before doing any calculations, draw an arrow showing the
approximate direction of the net force on Q3 at t = 0.
The charges are all repelling so the force on Q3 is down and to the right.
Since the charge at Q2 is more and they are the same distance, I would say that the force is angled
towards the x axis. (see picture)
(b) Find the magnitude of the force on Q3 at t = 0.
I’m going to calculate the force from each charge, Q1 and Q2,
first and then add them up.
F13 = kQ1Q3/(1)2 = 0.9 N (down)
F23 = kQ1Q3/(1)2 = 1.8 N (right)
To find the magnitude of the force, you need the pythagorean
theorem: Fon Q3 = [(0.9)2 + (1.8)2]1/2 = 2.0 N
(c) Find the direction of that force, expressed as an angle with the x-axis.
If θ is the angle between the force and the x-axis, then tanθ = 0.9/1.8, so θ = 27 degrees.
(d) After Q3 has slid a few tenths of a meter, is the angle of the force greater than, less than, or equal to the
value you calculated in part (c)? Explain your reasoning. (Note: Although you don’t have sufficient
information to calculate the new angle, you can still make a good argument about whether that angle gets
bigger or smaller.)
This turned out to be more difficult that expected, but, as always, I’m
focused on your ability to make quality arguments. Here are two that
came up when I (Eric) discussed with Ayush:
1) Imagine that Q3 will move along the arrow of its net force. This isn’t
really true, because as the charge moves, the force on it will change
direction, but this is an approximation that might get us a qualitative
answer. At this new location, draw what the force due to each
charge should be. Because of the angle that the charge moved
at, it should now be farther away from Q2 than Q1. The force
from Q2 will then reduce more than the force in Q1. Now, I’ll
make an attempt to draw both net forces together so we can
compare them (the dotted lines are before it moved, the dotted
lines with circles are after it moved). From this picture, it looks
like the angle of the net force from the x-axis gets greater.
2)Here’s another idea: move the charge in
the same direction, but this time, move it just a little bit so that the distances from
the other two charges are pretty much the same. My goal in doing this is to see
how the angles of the force changing affects the direction without changing the
magnitude of the force. The angle of the “vertical force” pointing down should
change more because of the direction that the charge is going to move in (see
picture). Now, I ‘ll try to draw the same picture above combining the directions
on the same graph, but this time I’ll keep the lengths the same and change the
angles accordingly. This picture says that the angle of the net force from the xaxis gets smaller.
The thing that came out of the discussion is that the direction of the force
depends on two things: how the components of the force change in magnitude
and how they change in direction. It seems like you’d have to know how both of
these things change exactly to know what the final change in angles. Still, I think
I can respond against the argument in 1), because I think that the change in
magnitude of the “horizontal force” will be much less than depicted, making the angle of the net force
decrease to less than before. Do you understand this response? Do you agree with it? Can you think of a
counterargument to your argument and respond to it? These are the skills that this problem is really
trying to get at.
6. A uniformly charged rod of length L = 1.0 meter and charge Q = 2.0
μC sits on the floor. At t=0, a small bead of mass m=1.0gram and
charge q = 100 nC is released from rest from a point d = 0.50 meters
vertically above the left end of the rod as shown in the picture.
(a) Find the bead’s vertical acceleration immediately after it is released. (Do not neglect gravity.) Try and
do this in symbols before you put in numbers.
The idea is to split the bar up into little tiny charges (dQ), find the vertical component of the force
between piece of the bar and the bead (dF cosθ), and then add up all the forces from each little piece of
the bar (integrate).
We know that the amount of charge (dQ) we have is how long our section of the bar is (dl) times the
charge density of the bar (Q/L). So we can rewrite (dQ) = Q/L (dl) and now we can integrate along the
length of the bar from 0 to L. Careful, the notation is really unfortunate, because d is the distance of the
charge from the end of the bar, but it is also the differential. When it’s a differential I’ll put it in
parenthesis, like (dQ).
Physics-wise, we are done. You’ve written the integral from the physical situation. Great! Now, it’s just
some math trickery to solve this integral: substitute in l = d tanθ and (dl) = d (dtanθ) = d sec2θ (dθ). Now
that we’re integrating along the angle it also changes the limits: now we’re integrating from an angle of 0
to Tan-1(L/d).
(b) Does the bead land on the left end of the rod? Explain.
The bead will not land on the rod, since there is a force in the x-direction pushing the charge left. If it
starts above the left end of the rod, then force in the x-direction will accelerate it to the left so whenever it
reaches the same height as the bar, it will be to the left of it.
(c) Immediately before it hits the rod, is the magnitude of the bead’s acceleration greater than, less than,
or equal to your part (a) answer? Explain.
Here again, it’s your ability to clearly argue one point that matters. Here’s my argument:
At t=0, the net force in the y-direction is mg - .0032 = .0068 N down and the acceleration is a = Fy/m = 6.8
m/s2. If this acceleration stays constant (which is won’t, because the electric force will change as the bead
moves, but it’s an approximation), then the time it takes the bead to fall to the same height as the bar is
found by d = 1/2 a t2 and t = .38 s.
My guess is that the magnitude of the electric force in the x-direction is less than the electric force in the ydirection. The part of the bar that exerts a force that is the most directed in the y-direction is the left end,
because the force due to it points straight up. The part of the bar that exerts a force that is the most
directed in the x-direction is the right end of the bar. As you go from the left end of the bar to the right,
the force gets more directed in the x direction, but it gets weaker. Therefore, I argue that the force in the
x-direction is less than the force in the y-direction.
Since the force in the x-direction is less than the force in the y-direction, I’m arguing that the bead will be
less than distance d from the left end of the bar when they’re at the same height. If this is true, then the
electric force in the x-direction will be bigger since the bar is approximately the same distance away, but
now all of its force on the bead is in the x-direction. There is no electric force in the y-direction.
So in the x-direction I’m saying the force increases. In the y-direction the force increases, because the
electric force opposing gravity gets to 0. So if both components of the force increase, then the net force
must increase, and so the acceleration must be greater.
Does this argument match or go against yours? If you agree, how could you build on it or connect it to
your own? If you disagree, how would you respond to this argument?