July 8, 2008 – Determining Charge
17
Name ________________________ Date ____________ Partners________________________________
DETERMINING CHARGE ON A BALL
Materials:
Hanging metal coated pith ball
stationary metal coated pith ball
Measuring scale
Teflon rod
silk
Source:
Fishbane, Gasiorowicz, and Thornton, Physics for Scientists and Engineers 3rd ed.
Prentice Hall, 2005.
Background:
In 1785, Coulomb developed his method for measuring the
electric force between two charged objects. Coulomb created a
torsion balance, which is an apparatus similar to the pith ball set
you will be using. Coulomb confirmed that the electric force is
proportional to the inverse square of the torsion in the fiber (Fig.
1). The electrostatic force of repulsion between the two charged
balls causes a rotation in the apparatus. If we know the angle
produced by the electrostatic force of repulsion between the two
balls, it is possible to measure the force of repulsion between them.
The two balls are charged by rubbing a material containing electric
charge on one of the balls. the electrical force between the spheres
is very large compared with the gravitational attraction; hence the
gravitational force can be neglected.
Figure 1
It has been verified that the electric force is inversely proportional to the square of
the separation between two particles or point charges and that the force acts along the
direct line of separation between the two charges. This force is proportional to the charge
of the two particles. Furthermore, this charge is an attractive force if the charges are
unlike in sign (positive to negative) and is a force of repulsion if the two charges are the
same in sign (positive to positive or negative to negative). Coulomb’s law can therefore
be expressed in the following relationship:
q1q2
(1)
r2
where ke is the Coulomb constant. Using the SI units the Coulomb constant can be written
as 8.9875 X 109 N m2 / C2 .
F = ke
University of Virginia Physics Department
PHYS 636, Summer 2008
18
July 8, 2008 – Determining Charge
Purpose:
In this experiment you will measure the force of repulsion between two equally
charged pith balls (each of mass ~0.28 ± 0.7 g) and use the force to determine the total
charge Q on one of the two equally charged pith balls.
Pith ball apparatus
Stationary apparatus
Figure 2
Procedure:
1. We shall call the apparatus with the pith ball hanging by two threads with the
scale on the base (Fig. 2) the pith ball apparatus. We shall call the other
apparatus in Fig. 2 with the long horizontal wooded and plastic rod the stationary
apparatus. Set the two apparatuses together with the pith balls touching as
shown in Fig. 3. The pith balls must be of the same size and they should be at
the same height. You may need to put something beneath the apparatus to have
them at the same height. The pith ball apparatus should be set with the threads
holding the pith ball hanging straight down (note that in Fig. 3 they are not quite
straight down!). The pith ball on the stationary apparatus should be placed so that
it touches the hanging pith ball. The stationary pith ball no longer moves. Note its
position as seen in Fig. 3.
Figure 3: The balls are at
rest and are touching. They
are ready to be charged.
University of Virginia Physics Department
PHYS 636, Summer 2008
July 8, 2008 – Determining Charge
19
2. Now you are ready to place charge on the two pith balls. To do this, you need to
wipe the Teflon rod vigorously with the silk or strike the silk with the Teflon rod
and bring the rod in contact with both pith balls simultaneously as shown in Fig.
4. Try not to touch the pith balls with your fingers, as you may ground them and
cause them to discharge. The pith balls need to either be in contact with each
other when you initially charge them or you need to let them momentarily touch
each other after you have charged them.
Figure 4: The two balls
are touching and are being
charged by the charged
Teflon rod.
Question 1: The pith balls are conductors because of the paint on them. When
two conductors are in contact, what do we know about their electric potential and
charge?
Question 2: When the two pith balls are touching, what has to be true for the two
pith balls to have the same charge?
University of Virginia Physics Department
PHYS 636, Summer 2008
20
July 8, 2008 – Determining Charge
Question 3: What do you expect the pith balls to do after they have been
charged? Why?
3. Now that the pith balls have like charges, they should repel. Let the hanging pith
ball stop swinging and measure the distance x between the two pith ball centers by
observing the scale that is underneath them. Look from the front with the two pith
balls between your eyes and the scale on the base. This is the view seen in Fig. 5.
The two pith balls should still be at about the same height.
Figure 5: Pith balls are both
charged and are repelling. The
pith ball on the right is
stationary, but the pith ball on
the left swings up and to the
left. We measure the distance
between the two pith balls
centers.
Distance between pith ball centers: x = _____________________
University of Virginia Physics Department
PHYS 636, Summer 2008
July 8, 2008 – Determining Charge
21
Question 4: Do not do this, but what happens to the pith balls if you touch them
with your hand?
Question 5: What would happen if you brought up a measuring scale next to the
pith balls in order to measure the distance x between the centers? Try it if you are
not sure. Explain.
Question 6: What would be the effect on the pith balls if the humidity in the room
were raised?
Data Analysis:
Now measure the charge on the pith balls.
1. First let’s look at the forces acting on the hanging pith ball (see Figure 6). There
are three forces acting directly on this body: the gravitational force (weight) FW,
the electrostatic force of repulsion between the two like charges FC, and the
tension in the string FT. The direction of the gravitational force acts downward
and the force of repulsion due to the stationary pith ball acts to the left. The
tension in the string provides a force acting in a straight line between the
suspension and the pith ball. While the actual pith ball apparatus has two strings
supporting each pith ball, it is sufficient for our 2-dimensional examination to
assume that there is just one string acting as a force of tension directly between
the bar and the pith ball. The effective (not actual length) length of the string is L.
University of Virginia Physics Department
PHYS 636, Summer 2008
22
July 8, 2008 – Determining Charge
FT
FC
Fw
Figure 6
2. Without the Coulomb force involved, we know that the orientation of the pith ball
will be completely vertical. With the Coulomb force, the pith ball rises in a
circular path with respect to the suspension because of the force of tension in the
string.
θ
L
L-h
FT
FC
x
h
θ
Fw
FC
Figure 7
3. As shown in Figure 7, as a result of the Coulomb force, the pith ball will rise to a
height h above its original position (exaggerated here). We can now calculate the
charge on the pith ball based on the forces involved.
University of Virginia Physics Department
PHYS 636, Summer 2008
July 8, 2008 – Determining Charge
23
a. From similar triangles we notice that
b. We solve for FC and find FC =
FC
x
=
.
FW L − h
(2)
x
FW
L−h
(3)
c. We substitute for both FC and FW and find
FC =
kQ 2
x
x
=
FW = mg
2
x
L−h
L
(4)
where we let L − h ≈ L in the denominator.
d. We solve Eq. (4) for Q2 and Q and find
Q 2 = mg
3
2
x3 mg x3 (0.0028 kg)(9.8 m/s 2 ) x 3
−12 x C
=
=
=
3.05
×
10
kL
k L ( 9.0 × 109 N ⋅ m 2 / C 2 ) L
L m2
(5)
Q = 1.75 × 10−6
x3 C
L m
Question 7: Explain precisely what distance you should measure for the length L.
Be careful here. Is it actually the length of the string?
5. Make sure you have measured x, the distance between the centers of the charged
balls, before measuring L. Then measure L.
L: __________________
6. Use the measurements for L and x and Eq. (5) to determine the charge on each
pith ball. It should be a few nanocoulombs.
University of Virginia Physics Department
PHYS 636, Summer 2008
24
July 8, 2008 – Determining Charge
Q = ___________________ nC
7. Determine the number of free electrons that were placed on the pith ball using the
elementary charge e = 1.602 x 10-19 C.
# free electrons = ______________________
University of Virginia Physics Department
PHYS 636, Summer 2008