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Physics 201- Fall 2007
Electrostatics Solutions
September 10
You are spending the summer working for a chemical company. Your boss has asked you to determine
where a chlorine ion of effective charge -e would situate itself near a carbon dioxide ion. The carbon
dioxide ion is composed of 2 oxygen ions each with an effective charge -2e and a carbon ion with an
effective charge +3e. These ions are arranged in a line with the carbon ion sandwiched midway between
the two oxygen ions. The distance between each oxygen ion and the carbon ion is 3.0 x 10-11 m. Assuming
that the chlorine ion is on a line that is perpendicular to the axis of the carbon dioxide ion and that the line
goes through the carbon ion, what is the equilibrium distance for the chlorine ion relative to the carbon ion
on this line? For simplicity, you assume that the carbon dioxide ion does not deform in the presence of the
chlorine ion.
Conceptual
I suppose the first thing to think about in this problem is the meaning of
“equilibrium”. Just as with mechanics, this word signals to us that we want the object to
experience zero net force. Here our object is the chlorine atom, no the oxygen or carbon
atoms. We don’t care about the forces on these other atoms as we’re assuming that the
carbon dioxide molecule does not deform.
Another fact to remember from mechanics is that our forces can be broken into
components. We need each component to separately sum to zero. From the picture
below we can see that the chlorine atom must lie on the axis of symmetry (directly below
the carbon atom). If this were not he case the forces due to the oxygen atoms would not
balance each other.
For electrostatic systems, the force can be calculated via the Coulomb force
equation. We only need to know the charge and separation distances to find the forces.
(Actually, here we want the position, so we’ll use the given charges and desired net force,
zero, to find the distance.)
I’m going to guess that the chlorine is a distance from the carbon that is similar to
the distance between the carbon and oxygens. In other words, I’m guess that the chlorine
is roughly 3 x 10-11m from the carbon. Why? Well, I suspect that all of these forces are
of similar strength, so the distance will have to be similar (that and charge magnitudes are
the only things that affect the force).
Pictorial
Let x= 3 x 10-11 m
x
x
O
-2e
C
+3
e
O
We’ll assume that the atoms can be
-2e
treated as point charges (despite what my
θ
sketch shows). Are we justified in doing
this? Well, the typical diameters of the
y
-10
atoms are around 10 m, so this
r
r
assumption isn’t looking too good as the
y
atoms are actually larger than the spacing
between their centers- meaning the
electron clouds overlap. But, we’re saved
x
Cl
by something we’ll see a short while-e
when we’re outside of spherically
symmetric charge distributions, the forces they generate are the same as those generated
by point charges which lie at the center. This is just like what we saw with gravitational
forces in mechanics. (BTW: This assumption of point charges is necessary if we wish to
use the Coulomb force equation.)
Here is a free body diagram for the chlorine atom:
FC
Cl
-e
FO
FO
Mathematical
From our symmetry, we see that as long as the chlorine is directly under the carbon, the x
component of the net force will be zero.
Looking at the y components.
For the carbon, the force is attractive, or in the positive y direction. For the oxygens, the
force will be repulsive, pushing the chlorine in the negative y direction.
Keep in mind that we need to break the forces on the chlorine due to the oxygens into
components. This means that the magnitude of the force is multiplied by the appropriate
trig term
Total magnitude of the oxygen force on the chlorine (no direction, just magnitude):
k (2e)e
FO =
r2
The y component of this is simply
!
FOy =
!
k (2e)e " y %
$ '
r2 # r &
(The y/r factor is the sine of θ, where we know that this angle is the
same as the one in the first sketch.)
θ
FO
Now we sum the three y components on the chlorine:
"
k (2e)e # y & k (2e)e # y & k ( 3e)e
=0
% ("
% (+
r2 $ r '
r2 $ r '
y2
We can simplify this a bit
!
!
!
!
4ke 2 " y % 3ke 2
$ '= 2
r2 # r &
y
and canceling the common factors on each side:
4y 3
=
r3 y 2
Then moving the factor y2 to the left and 4 to the right.
y3 3
=
r3 4
We want to determine y, so we need to eliminate r, which is also unknown but related to
y via the Pythagorean theorem.
r2 = y 2 + x 2
We can now substitute this into our force equation and solve for y.
!
Taking the 2/3 root of our force equation:
2
" y 3 % 3 y 2 " 3 %2 3
$ 3 ' = 2 =$ '
#4&
r
#r &
substitute in the expression for r2
!
FOy
" 3%
y2
=$ '
2
2
y + x #4&
2
3
Now solve for y2
!
2
" 3% 3
y = $ ' (y 2 + x 2)
#4&
" " %2 3 % " %2 3
3
3
2
y $$1( $ ' '' = $ ' x 2
# #4& & # 4&
2
2
" 3% 3 2
$ ' x
#4&
2
y =
2
" 3% 3
1( $ '
#4&
!
!
Taking the square root of both sides:
1
" 3% 3
$ ' x
# 4&
y=
2
" 3% 3
1( $ '
#4&
Finally, we substitute in the value of x and perform the final calculations:
y= 6.52 x 10-11m.
Evaluation
Does this answer make sense? We may not be able to say if it should be 6.52 x 10-11m or
5.52 x 10-11m, but the fact that it is the same order of magnitude as the carbon- oxygen
separation does give the solution some credibility.
You've been hired to design the hardware for an ink-jet printer. You know that these printers use an
electrode to deflect the droplets, causing them to form letters on a page. The basic mechanism is that
uniform ink drops of about 2x10-6m radius are charged to varying amounts after being sprayed out towards
the page at a speed of about 15 m/s. Along the way to the page, they pass into a region between two
deflecting plates that are 1.5 cm long. The deflecting plates are 1.0 mm apart and produce a uniform
electric field between them of 1.5 x 106 N/C. You measure the distance from the edge of the plates to the
paper and find that it is one-half inch. Assuming an uncharged droplet forms the bottom of the letter, how
much charge is needed on the droplet to form the top of a letter (standard 12 point font).
Pictorial
Here is my sketch of the system (not really to scale):
1mm
1.5cm
Path of uncharged
drop
1/2”
Paper
Electric field
Why should we believe this sketch? The uncharged droplet is undeflected, and more
importantly the charged droplet is pushed upward due to the electric force. We know that
for a positively charge drop this force will be up because the electric field is shown in the
direction. The important aspect of this figure is that the electric field is perpendicular to
the droplet’s initial velocity. The other important aspect to notice is the shape of the path
(charged drop). Between the plates, it is a parabola indicating that the drop is
accelerating there and outside the field, the trajectory is straight.
Now I’d like to redraw the figure labeling all of the distances.
y1
y2
yttot
d1
•
•
•
•
•
•
•
•
d2
d1= 1.5cm
d2= 0.5”
y1 is the change in vertical distance
while in the plate
y2 is the change in
vertical distance
while outside field
ytot= 4mm
Conceptual
Again, what’s going on physically? Within the plates, there is a force on the drop, which
will cause a deflection. Also, this force changes the drop’s velocity. Once it exits the
field then the drop continues in the same direction as it had upon exiting the field. (there
are no forces to deflect it.) We will write down equations for y1 and y2 and then sum
these to get the desired shift. Within these equations there will be terms involving the
acceleration. This acceleration is proportional to the charge of the drop.
Some of our assumptions & conditions:
• We’ll treat the drop as a point charge. It will also be assumed to be a perfect sphere.
• We will ignore gravity in this problem. Why can we do this? Not because the force
will be small- the mass will also be small, so the acceleration will still be 9.8 m/s2. At
this stage of the game this is more of a simplification/ intuitive guess. If we can later
show that the acceleration due to the electric field is much bigger than the
acceleration due to gravity, then we will have made our case.
• We’ll assume that the electric field exists only between the plates. This is a big
simplification. In reality the field will extend beyond the plates, and it will be nonuniform.
•
•
•
Other than the electric force, we’ll assume that there are no other forces on the drop.
There probably are air drag forces in real life, but let’s ignore those for now just so
we can get a feel for what is going on.
We’ll assume that 12 point lettering is 1/6” (there are 72 points in an inch). You can
also simply measure a print out of text to find that the letters are roughly 4mm tall (if
you go from the top of an “A” to the bottom of a “y”.
We’ll assume that the ink’s mass density is the same as water’s.
What sort of charge do you expect on the drop? It should be small- a small fraction of a
coulomb. Perhaps the total charge is around 1 nC. (Later in the semester we’ll see how
to calculate the total possible charge that a drop of this size could have.)
Mathematical
The total vertical distance is:
4mm= y1 + y2
What are useful expressions for y1 and y2? We must look back to our constant
acceleration equations from mechanics.
1
y1 = at12
2
y 2 = v y t2
!
t1 is the time it takes for the drop to travel the length of the plates, and t2 is the time it
takes to go from the plates to the paper. a is the acceleration when inside the field. vy is
the speed of the drop in the vertical direction when it is exiting the field (and the rest of
the way until it hits the paper).
The times only depend on the horizontal speed. (vx= 15 m/s)
d1
vx
d
t2 = 2
vx
t1 =
!
What is the speed of the drop as it exits the field? Again, we’ll use our constant
acceleration equations for this.
v y = at1
!
now let’s insert our expressions into the total distance equation:
1
4mm = at12 + v y t 2
2
2
1 " d1 %
= a$ ' + ( at `) t 2
2 # vx &
2
" d %" d %
1 " d1 %
= a$ ' + a$ 1 '$ 2 '
2 # vx &
# v x &# v x &
!
Now we can solve this to get the drop’s acceleration within the field. It turns out to be…
3000 m/s2. Wow. This is big, but that’s okay; if anything it’s good. We now see that the
acceleration due to the electric field is much more than the acceleration due to gravity.
We are justified in ignoring gravity in our calculations. Also, it is big simply because we
are giving this drop a large sideways push over a small time/ distance.
Now, how do we find the charge on the drop? This is where Newton’s Second Law and
the definition of an electric field will come into play.
What is the mass of the drop? This is where our assumption about the drop’s density will
come in handy.
qE = F = ma
qE
a=
m
!
!
The mass is given by:
" g %4
m = $1 3 ' ( r 3
# cm & 3
Here we use the drop’s given radius of 2 x 10-6m to find that the mass is approximately
3.35 x 10-14 kg.
Now we can solve for the charge of the drop, q. (We know that the magnitude of the
field is 1.5 x 106 N/C.)
q= 6.6 x 10-17C which means that roughly 415 electrons had to be removed from the drop.
Evaluation
We expected a small charge, but honestly this is even smaller than I guessed. Given the
large electric field (which is pushing the limits of possible electric fields in air), it
shouldn’t be surprising that we don’t need much of a charge to deflect the drop.
One aside… If you happen to work out y1- the distance that the drop is suppose to shift
while in between the plates, you’ll find that y1= 1.5mm. This is a larger distance than the
plate separation. Somebody needs to redesign the plate assembly.