Elihu S. Conant-Haque
662020532
Physics II Lab 1, Monday, January 10: Coulomb's Law
Thinking and calculating: 1. If two excess electrons are placed on a long straight copper wire, where are
they most likely to be found? (Anywhere? On one end? On opposite ends?) Explain your answer.
Since like charges repel, and the two electrons are like charges, each electron will feel a repulsive
force from the other. Since the copper wire is quite conductive, the electrons will be free to move along its
length. They will tend to move very far apart as the repulsive force pushes them. If the wire is not
extremely long the electrons will probably move all the way to its ends. Since air is much less conductive,
they will stop moving away from each other once there is no more wire along which to travel, and will
stay at the ends. (I imagine that if it were long enough the electrons would become so distant that the
repulsive force would no longer be enough to overcome the material’s resistance to current flow, and they
wouldn’t be able to get to the ends).
2. If 106 electrons are placed on a long straight copper wire, how will they be distributed? (Anywhere? On
one end? On opposite ends? Evenly throughout the volume? Over the outside surface?) Explain your
answer.
Since electrons are like charges and between any like charge and all other like charges there is a
repulsive force, all electrons will feel repulsive forces but will be free to move only within the copper
wire since the surrounding air is so much less conductive. Since the repulsive force is inversely
proportional to distance between charges, the electrons will tend to arrange such that the distances
between each electron and each other electron are all as small as possible. That arrangement will be an
even distribution throughout the volume.
3. The conducting rod to the right has no net charge. A negatively charged object is brought near the right
end of the rod. What happens to electrons in the conducting rod?
Electrons within the conducting rod will flow to its left end, as long as the material’s conductivity
allows, tending to achieve charge neutrality/electric field neutrality in the space between the rightmost
end of the conductive rod and the leftmost end of the negatively charged object.
4. Would the charged object exert a force on the neutral rod?
Yes.
If yes, in what direction is the force on the rod? (Attract or Repel?) Explain.
Ans: Attract. Explanation: Since, once the electrons have flowed away from the negatively
charged object, the rightmost end of the conductive rod is positively charged and very near the leftmost
end of the negatively charged object, those two ends will each feel an attractive force from the other.
Elihu S. Conant-Haque
662020532
Situation
Observations
i) Charges and net charge on knob
ii) Charges and net charge on leaves
iii) Net charge on whole electroscope
iv) Position of leaves.
Example: (not correct) Initial condition with rod
and hand far from knob.
i) 100 positive, 100 negative; net charge=0
ii) 200 positive, 200 negative; net charge=0
iii) net charge=0
iv) leaves together
a) Rod positive; Rod near; Hand far away
1)
2)
3)
4)
10-, 6+, net = 42-, 6+, net = 4+
12-, 12+, net = 0
Leaves apart
b) Rod positive; Rod far; Hand touching.
1)
2)
3)
4)
6-, 6+, net = 0
6-, 6+, net = 0
12-, 12+, net = 0
Leaves together
c) Rod (positive) near; Hand touching
1) 11-, 6+, net = 5
2) 6-, 6+, net = 0
3) 17-, 12+, net = 54) Leaves together
d) Rod (positive) near; Hand moved away
5) 11-, 6+, net = 5
6) 6-, 6+, net = 0
7) 17-, 12+, net = 58) Leaves together
e) Rod (positive) far; Hand moved away.
1)
2)
3)
4)
8-, 6+, net = 29-, 6+, net = 317-, 12+, net = 5Leaves apart
f) Rod (positive) far; Hand touching
1)
2)
3)
4)
6-, 6+, net = 0
6-, 6+, net = 0
12-, 12+, net = 0
Leaves together
g) Rod (positive) far; Hand moved away
5)
6)
7)
8)
6-, 6+, net = 0
6-, 6+, net = 0
12-, 12+, net = 0
Leaves together
h) Rod (positive) near; Hand far.
1) 10-, 6+, net = 42) 2-, 6+, net = 4+
3) 12-, 12+, net = 0
Elihu S. Conant-Haque
662020532
i) Explain your observations for situation “h” in the table above using the concepts of charge carriers in
conductors, the sequence of situations, and Coulomb’s Law as discussed in Chapter 21.02 of your
textbook.
With the positive rod touching, the electroscope, containing equal amounts of - and + charges,
undergoes a rearrangement of charges so that the - charges can get as far away as possible from the
positive rod that repels them. When the ground comes along, it gives or takes however many electrons are
needed to restore charge neutrality between the rod and the electroscope and stop the electrostatic force
from acting. In this case the ground has to “give” 5 electrons to the electroscope. With the rod still in
proximity to the electroscope, the electroscope will tend to hold on to those 5 extra electrons. So when the
ground leaves the two other objects, it has no reason to take back the 5 electrons that it gave. This leaves
the electroscope with an imbalance of charges, which only causes a “neutral-looking” leaf position
because that positive rod is near. When the rod is moved away, The imbalance of - and + charges causes
repulsive forces between the excess electrons, which move as far apart as possible. Since the greatest
distance between the tips of the deflected leaves is greater than the greatest distance the electrons could
achieve within the knob, they all move to the leaves, which are deflected by the electrons’ tendency to
repel from one another. The observations for situation ‘h’ are thus explained.
1. Complete the sketch below of the physical situation represented by equation 21a when both charges are
positive. Draw a Free Body Diagram for particle 0 (which we will designate as the left hand particle).
Add and label the vector 𝑟⃗10 to your sketch.
2. What is the relationship between k and the electric permittivity in the front of your textbook? What is
the numerical value of k and its units in the SI system?
K is
1
4𝝅ε
where epsilon is the electric permittivity of free space: 8.86*10^-12
k = 9.0*10^9 N*m^2/C^2
3. How does equation 21a change if you want to find the force exerted on particle 1 by particle 0?
(Rewrite it and explain the change.)
𝐹
=𝑘
1 𝑜𝑛 0
𝑞 0*𝑞
(𝑟
1
2
)
10
𝑟
|𝑟
10
|
Gives the magnitude of the force between these two charges, but specifies a
10
direction indicating the member of the Newton’s third law pair which acts on 0 and is exerted by 1. To
express this same force magnitude as it is exerted by 0 on 1, the unit vector must be changed to go from 0
to 1 instead of from 1 to 0. The equation becomes 𝐹
=𝑘
0 𝑜𝑛 1
𝑞 0*𝑞
(𝑟
1
2
)
01
𝑟
|𝑟
01
|
01
.
Elihu S. Conant-Haque
662020532
4. In the figure to the right, particle 1 has a negative charge q1 = - 9.00 nC and lies at the position P(0,
10.00, 0) with units in meters. Particle 0 has charge magnitude q0 = 4.00 nC, but the sign is unknown.
The direction of the force on particle 1 is to the left. a) What is the sign of charge q0? Explain.
Since q0 lies to the left of q1, the leftward force connotes attraction. Since only charges of
opposite sign can be attracted to one another, the sign of q0 must be opposite that of q1. q1’s sign is
negative, hence q0’s must be positive.
b) Rewrite equation 21a, putting numbers in the place of variables and including sign and vector direction
for the force on particle 0.
𝐹
9
2
2
= (9. 0 * 10 𝑁 * 𝑚 /𝐶 )
1 𝑜𝑛 0
−9
−9
(4*10 𝐶)*(−9*10 𝐶)
2
10
𝑟
|𝑟
10
|
10
=
−9
− 3. 24 * 10 𝑁
Answer indicates a force which goes along the vector from charge 1 to charge 0, but opposite in direction;
from charge 0 to charge 1. This makes sense; the charges should be attracted to each other because they
are opposite in sign.
5. Consider the same case, but with the addition of particle 2 at some point P(0, y’, 0) between particles 0
and 1. Particle 0 experiences an attraction from particle 1 and a repulsion to particle 2. Particle 2 has
charge q2 = 5.00 nC.
a) What is the sign of charge q2?
q2 is a positive charge.
b) For what value of y’ does particle 0 experience a net force of zero?
Equate magnitudes of attractive force from charge q1 and repulsive force from charge q2.
Simplify equation by canceling k and q0. Cross-multiply and solve for y’. y’=7.45 m
c) For this position, find the x and y components of the total force on particle 2.
The x-component is clearly 0. The y-component is found by adding the magnitudes of the force
by 0 on 2, and by 1 on 2, because both forces tend to push 2 in the same direction (it is attracted to 1 and
repelled from 0). k((q0q2/7.454^2) + (q1q2/2.564^2)) = 6.485*10^-8 N
Consider the case when the charged particle 2 from question 5 is now located at point P(0, 0, 1.80) on the
z-axis.
a) Draw two vectors representing the forces on particle 2 due to each of the two charges 0 and 1.
Elihu S. Conant-Haque
b) What is the magnitude of F0on2?
|F0on2| = k*|q0|*|q2|/1.8^2 = 5.55*10^-8
c) What is the magnitude of F1on2?
First find hypotenuse = 10.1607 m
|F1on2| = k*|q1|*|q2/10.1607^2 = 3.923*10^-9
d) What angle does 𝐹⃑ 1𝑜𝑛2 make with the +y-axis?
Theta = arctan(10/1.8) = 79.8 degrees
662020532