Circuits (MTE 120) (Spring 2010)
Akrem El-ghazal
http://pami.uwaterloo.ca/~akrem/
University of Waterloo,
Electrical and Computer Engineering Dep.
Electric Charge
Objective:
1. To review and learn the properties of electric charge
2. To learn how to compute the force between two point charges (Coulomb‘s law).
3. To learn how to compute the force on a point charge due many point charges.
Electric Charge (Q):
Electric charge is one of the basic quantities in nature. An object is made up of fundamental
building blocks named atoms; each is composed of three main particles: electron, proton, and
neutron. A proton has a positive (+) charge, and an electron has a negative (-) charge. The
neutron has zero net electric charge (0).
+
Electron (-)
Proton (+)
Neutron (0)
+
+
-
A positive charge is a deficiency of electrons and a negative charge is an excess of electrons. It is
also important to note that only electrons are transferred from object to object, an example of
this transfer is friction. The object with equal amount of electrons and protons is electrically
neutral.
Quantization of Charge
The electric charge is said to be quantized: the charge of an object is an integer multiple of the
elementary charge, usually denoted e, which is the electric charge carried by a single proton, or
equivalently, the negative of the electric charge carried by a single electron. Any electric charge,
Q, occurs as integer multiples of the elementary charge e:
. N is the number of elementary charges, and e is the elementary charge.
Circuits (MTE 120) (Spring 2010)
The Unit of Electric Charge
A big amount of charge is used as a standard unit for charge. This unit is called the
Coulomb and is given the symbol C. One coulomb is equal to 6.24 × 1018 elementary charges (e).
Note:
• The smallest amount of free charge is the charge of the e=1.6×10-19 C.
•
In electrostatic the following sub-multiples are often used:
Mili-Coulomb
Micro-Coulomb
Nano-Coulomb
Pico-Coulomb
=10 -3 C
=10 -6 C
=10 -9 C
=10 -12 C
1 mC
1 C
1 nC
1 pC
Properties of Electric Charge
•
•
Electric charge can neither be created nor destroyed, but it can be transferred from one
object to another. The quantity of electric charge, the amount of positive charge minus
the amount of negative charge in the universe, is always conserved.
Two electrically charged objects exert a force on one another, called as electrostatic
force: like charges repel and unlike charges attract each other.
2
Circuits (MTE 120) (Spring 2010)
Coulomb’s Law
The magnitude of the electrostatic force of repulsion or attraction between two charges q1 and
q2 is given by:
is the electrostatic constant:
,- is the permittivity constant:
| | | |
1
8.99 x 10) *. / 4" #$
#$ 8.85 x 10/
*. r is the distance between the two charges q1 and q2
Remember: The force is a vector quantity and its direction is along the line between the two
charges.
Note:
From the figure below: The force (F12) acting on q1 due to q2 is equal in magnitude and
opposite in direction to the force (F21 )acting on q2 due to q1.
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Circuits (MTE 120) (Spring 2010)
Example #1:
What are the magnitude and the direction of the electrostatic force F12 acting on q1 due to q2.
Solution:
The magnitude of the force:
00000
| | | |
2 x 10/ x 1 x 10/2
)
8.99 x 10 x
5x 10/3 7.192 x 10/5 N
The direction of the force:
Since both particles are positively charged, particle 1 is repelled by particle 2.Thus the direction
of the electrostatic force F12 is in the negative direction of the x axis. Thus, the force F12 has the
following magnitude and direction of the x axis:
7.192 x 10/5 N ,
180$
We can also write the F12 in unit-vector notation as
00000
7.192 x 10/5 N 8̂
4
Circuits (MTE 120) (Spring 2010)
Exercise #1 (homework)
In the above example, what is the net force 0000000000
,:;< acting on q2 due to q1?
The Principle of Superposition
The electrostatic force obeys the principle of superposition. If we have n charged particles, the
net charge on any one of them, let us say particle 1, is given by the vector sum of the forces:
0000000000
00000
00000 00000
000000
,:;< 3 5 . . . :
Example #2
What is the net force 0000000000
,:;< acting on q1 due to q2 and q3?
Solution
Step #1: Identify the direction of each force acting on q1
Because q1 and q2 have charge of same signs, q1 is repelled along the line between q1 and q2.
Thus, force F12 is directed towards the negative x axis (1800) as shown in the figure below.
Because q1 and q3 have charge of opposite signs, q1 is attracted toward q3 along the line
between q1 and q3. Thus, force F13 is directed towards q3, at angle θ as shown in the figure
below.
It is a straight forward to calculate the angle θ :
3
θ tan/ A C 36.87$
4
5
Circuits (MTE 120) (Spring 2010)
Step # 2: Calculate the magnitude (|F12| and |F13|) of each force acting on q1:
00000
00000
3 | | | |
1 x 10/ x 2 x 10/
)
8.99
x
10
x
4x 10/3 | | |3 |
3 1.1238 x 10/) N
It is a straight forward to calculate the distance (r13) between q1 and q3:
3 E3 4 5 F
)
00000
3 8.99 x 10 x
1 x 10/ x 3 x 10/
5x 10/3 1.079 x 10/) N
Step # 3: Calculate the net force by adding the forces as vectors:
0000000000
00000 00000
,:;< 3
Since the forces 00000
GH 00000
3 are not directed along the same axis we CANNOT sum them up by
combining their magnitude. Instead, we MUST add them as vectors:
00000
00000
Each force (
GH 3 has x and y components that can be calculated as follows:
3,I |3 | cosM 1.079 x 10/) cos36.87$ 8.63 x 10/$ N
3,N |3 | sinM 1.079 x 10/) N sin36.87$ 6.47 x 10/$ N
,I | | cos180 1.1238 x 10/) N
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Circuits (MTE 120) (Spring 2010)
,N |3 | sin180 0 N
Now, sum up the components axis by axis:
The sum of the x components:
,:;<,P 3,P ,P 8.63 x 10/$ 1.1238 x 10/) 2.61x 10/$ N
The sum of the y components:
,:;<,Q 3,Q ,Q 6.47 x 10/$ 0 6.47 x 10/$ N
The net force ,:;< has a magnitude:
,:;< R,:;<,P ,:;<,Q E2.61x 10/$ 6.47 x 10/$ 6.98x 10/$ N
F
F
The net force ,:;< has a direction:
,:;<,Q
6.47 x 10/$
S tan/ T
U tan/ T
U
,:;<,P
2.61x 10/$
According to the above equation
S 68.03 $ or 68.03 180 111.97$
0000000000
However, the S 68.03 is an unreasonable result because the net force ,:;< must have a
direction between the direction of 00000
GH 00000
3 . Therefore, the correctS is 68.03 180 $
111.97 .
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Circuits (MTE 120) (Spring 2010)
Exercise #1 (homework)
In the above example:
What is the net force 0000000000
3,:;< acting on q3 due to q1 and q2?
What is the net force 0000000000
,:;< acting on q2 due to q1 and q3?
Example:
In the following, a particles A is fixed at the origin and a particle B is fixed at a distance L from
the origin. At what point (other than infinitely far away) can a proton of charge qp be placed so
that the net force on it is zero (Equilibrium).
Solution:
WY , then
If the force on the proton due to qA is 000000
WX and the force on the proton due to qB is 000000
the point that we seek must satisfy the following condition:
000000
WX 000000
WY 0
This condition requires:
000000
WX 000000
WY
This tell us that at the point we seek, the forces acting on the proton due to the other two
particles(A and B) must be of equal magnitudes Z0000000
[\ Z Z0000000
[] Z) and must have opposite
directions.
Three locations should be studied:
y
y
L
qp
qA = 8q
+
L
FPB
qp FPB
-
+
FPA
qB = -2 q
+
x
FPA
(1)
+
qA = 8q
(2)
8
qB = -2 q
x
Circuits (MTE 120) (Spring 2010)
000000
In Figure (1): the two forces 000000
WX and WY are in the same direction and not in opposite
direction as required. The configuration of Figure (1) fails to satisfy the direction condition.
000000
In Figure (2): the two forces 000000
WX and WY are in opposite directions as required. However,
000000
coulomb’s law equation tell us that the 000000
WX and WY cannot have equal magnitudes. From
000000
000000
000000
coulomb’s law equation WX must be greater than
WY , because WX is produced by a closer
charge (with less r) and with greater magnitude (8q versus 2q). The configuration of Figure (2)
fails to satisfy the magnitude condition.
000000
000000
In Figure (3): the two forces WX and WY are in opposite directions as required. However,
because now the charge of greater amount qA is farther away from the proton (qp) than the
000000
000000
charge of lesser amount, there is a point at which WX is equal to
WY .
000000
From the coulomb’s law equation, the magnitudes of the two forces 000000
WX and WY are equal if:
From the above equation:
Rearrange the above equation:
|8| _ 2|| _ `
` a
8
2
` a
`
` a 1
`
4
` a 1
`
2
` 2a
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Circuits (MTE 120) (Spring 2010)
Conductors and Insulators
Not only can electric charge exist on an object, but it can also move through an object.
Therefore, it is convenient to classify substances in terms of their ability to conduct electric
charge:
Conductors: are materials through which electric charges can move rather freely, examples
include metals (such as copper), the human body and tap water.
Insulators: are materials in which electric charges cannot move freely; examples include
rubber, plastic and glass.
Semiconductors: are materials that are intermediate between conductors and insulators;
examples include silicon and germanium.
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