ELEC TR O S TATIC S
Learning Outcomes for the Lesson:
state Coulombs Law
Use Coulombs Law to solve problems
Define Electric Field
-calculate net Electric field for two point charges
Describe electric field for simple charge distributions in
uniform and non-uniform situations
Solve problems for electric field
Use Electric Potential Energy equations to solve problems
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ELEC TR O S TATIC S
Vocabulary
Electric Force
Electric Charge
Electric Field
Electric Potential Energy
Electric Potential (Voltage)
electromotive force (emf),
potential difference,
Potential
Current
Coulomb
Ampere
Current
Conductor
Coulombs Constant
Coulombs Law
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Electric Force
Electric force is fundamental to our
very existence!
All atoms and molecules are held
together by this electric force.
Without the electric force there would
be no such things as molecules (NO
DNA & NO LIVING THINGS!)
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Electric Force
We can move because of this force:
(Muscle require chemical reactions to
operate.)
All forms of transportation require the
use of this force.
This is important Stuff!
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Electrostatic Force
Law of Electrostatic Charges
Like Charges repel.
Unlike Charges attract.
Charged objects attract neutral objects
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Electrostatic Force
+ + + + +
+ + +
Attraction
- - - - - - - - -
Charge is built up by rubbing
two objects together
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Electric Charge
Atoms are made of three sub-atomic
particles: protons, electrons, and
neutrons
Protons are found in the nucleus and
have a positive charge
Electrons are found outside the
nucleus and have a negative charge
Neutrons are found in the nucleus and
carry no charge
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Electric Charge
The simplest possible charge called the
elementary charge is the charge on an electron
(-) or proton (+)
It is given a value of e = 1.6 x 10-19 C
Like charges repel and unlike charges attract
(that is why electrons and protons are attracted
to each other)
1 Coulomb (C) of charge is equal to the number
of electrons that flow past a conductor in 1
second when 1 ampere of current flows (1 C =
6.25 x 1018 charges)
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Electrostatic Force
Electric Charge is always conserved.
When one object receives a positive
charge another object must receive an
equal but opposite negative charge.
The charges are due to the sub-atomic
particles within an atom
Proton (+)
Electron (-)
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Electrostatic Force
Coulombs Law:
Fe =
kQq
r2
Fe is the electric force (N)
Q is the charge of one object (C)
q is the charge of the other object in (C)
r = the distance between the two objects (m)
k is Coulombs Constant = 9.00 x 109 Nm2/C2
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Coulom bs Law
2.3 m
-15 C
24 C
What is the electrostatic force between the two charges?
Fe =
kQq
r2
9.00 x 109 x 24 x 10-6
=
Fe = -0.61 N
x -15 x 10-6
2.32
(Attractive)
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Coulom bs Law
22 C
45 C
-13 C
1.7 m
2.3 m
What is the net force on the -13 C due to the other two charges?
Solution:
Fe =
l
Find the force on each side of the center charge.
k Qq
r
Fe =
r
kQq
r2
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Coulom bs Law
Fe(left) =
Fe(right) =
9.00 x 109 x 22 x 10-6
x -13 x 10-6
= 0.891 N (Left)
1.72
9.00 x 109 x 45 x 10-6
x -13 x 10-6
2.32
= 0.995 N (right)
Therefore:
Fnet =
0.995 N (R) - 0.891 N (L)
0.104 N (R)
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Coulom bs Law vs G ravitationalLaw
Fe =
kxQxq
Fg =
GxMxm
r2
r2
Note the
similarity of
the two graphs
FE
Fg
r
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Coulom bs Law vs G ravitationalLaw
FE
Fg
m = KQq
1/r2
m = GMm
1/r2
Note the similarity between the two laws.
Scientists have been trying to link the
laws together for hundreds of years.
So far no connection exists!
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Electric Field
Remember when we talked about gravitational field to describe
how forces act over a distance well the same thing applies to
electric force.
Electric Field:
Shows the direction that a positive-point charged
would travel near another charge.
Shown as vector arrows.
Electric field lines point
away from a positive
charge.
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Electric Field
Electric Field around a negative charge
Electric Field around two equal
and opposite charges
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Electric Field
Electric Field around two equal
charges
Electric Field around
various unequal charges
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Electric Field
+
Electric Field between
two charged plates
The Electric field is
constant between the plates
(except at the ends).
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Electric Field
If two metal plates (called a capacitor) are
oppositely charged by a power supply a uniform
electric field is created between the plates
+
P.S.
-
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Electric Field -Form ula
Electric Field strength is analogous to gravitational field strength.
(Remember: g = F/m and g = GM/r2)
By definition, electric field equals force per charge:
E=
F
q
and
kxQxq
F =
r2
Putting these two together results in:
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Electric Field -Form ula
kxQ
E =
r2
E is the electric field intensity (N/C)
Q is the charge of one object (C)
r = the distance between the two objects (m)
k is Coulombs Constant = 9.00 x 109 Nm2/C2
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Electric Field -Calculations
P
1.5 m
65 C
What is the electric field
intensity at a distance of 1.5 m
away from a 65 C charge?
E =
kxQ
New Equation
r2
E =
9.00 x 109 x 65 x 10-6
1.52
E = 260000 N/C
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Electric Field -Calculations
EB
ER
45 C
3.4 m
-15 C
P
2.8 m
What is the net electric field at point P between the two
oppositely charged particles?
Solution:
ER =
Find the E-fields due each charge separately
kxQ
r2
EB =
kxQ
r2
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Electric Field -Calculations
9.00 x 109 x 45 x 10-6
ER =
3.42
ER = 35034 N/C (Right)
9.00 x 109 x -15 x 10-6
EB =
2.82
EB = 17219 N/C (Right)
Enet = 52000 N/C (Right)
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Electric PotentialEnergy (Non-Uniform
E Fields)
We are now going to look at how electric
potential energy and electric force is related in
non-uniform e-fields (outside of charged
plates)
We have to generate some new equations
because of this new situation.
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ELEC TR IC P O TEN TIA L EN ER G Y
Remember W =
Ek
And W = F x d
Therefore:
Ek = Ep = F x d
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Electric Potential& Energy (Non-Uniform
E Fields)
W = Ep = F x d
But:
F=
KQ1Q2
R2
Therefore:
Ep =
KQ1Q2
xd
R2
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Electric PotentialEnergy (Non-Uniform
E Fields)
Since d = R, one R will cancel, leaving:
Ep =
KQ1Q2
R
Note:
This is very similar to Gravitational
Potential Energy:
Ep =
- GM1m2
R
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Electric Potential& Energy (Non-Uniform
E Fields)
Example #1:
What is the potential energy of the 2.6 C charge relative to
the +37 C Charge?
7.3 m
-2.6 C
+37 C
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Electric Potential& Energy (Non-Uniform
E Fields)
Solution:
Ep =
Ep =
KQ1Q2
R
9.00 x 109 x 37 x 10-6 x 2.6 x 10-6
7.3
Ep = - 0.119 J
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Electric Potential& Energy (Non-Uniform
E Fields)
Example #2
(A more realistic problem is when one charge is forced to move or is
allowed to move on its own.)
2.4 m
0.75 m
4.0 C
2.0 C
What work is done in moving the 2.0 C charge 0.75 m closer to
the 4.0 C charge?
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Electric Potential& Energy (Non-Uniform
Solution:
E Fields)
Work = Ep
Ep = Ep2 - Ep1
Ep =
kQq
kQq
r2
r1
9.00 x109 x 4.0 x 10-6 x 2.0 x 10-6
Ep =
1.65
Ep = 0.0436 J - 0.030 J
9.00 x109 x 4.0 x 10-6 x 2.0 x 10-6
2.4
= 0.0136 J
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Electric Potential& Energy (Non-Uniform
E Fields)
Now Try this:
An electron travelling at 6.0 x 105 m/s when it is heading
straight for this iron atom. What is the distance for the
electrons closest approach?
v = 6.0 x 105 m/s
d=?
electron
Fe atom
-26 elementary Charges
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Electric Potential& Energy (Non-Uniform
Solution:
E Fields)
Ek = E p
m v2 =
KQ1Q2
R
x 9.11 x 10
-31
x (6.0 x 10 ) =
5 2
R=
R=
9.00 x 109 x 26 x - 1.6 x 10 19 x - 1.6 x 10-19
5.99 x 10
-27
1.64 x 10-19
3.7 x 10-8 m
(about the size of an atom!!)
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Electric Potential
(Voltage)
Not to be Confused
with electric
potential energy!
Other words for the same term:
Voltage,
electromotive force (emf),
potential difference,
Potential
The driving force that causes
electrons (and other charges) to
move from one place to another
electrical pressure
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Electric Potential
Relates the potential energy (or work) done on an
object and the amount of charge on that object
Book Definition:
Electric potential equals potential energy per
unit charge.
For example:
a 1.5 V dry cell can give 1.5 Joule of energy
per coulomb of charge moved.
120V circuit can give 120 J of energy per
coulomb of charge
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Electric Potential
m
Gravity
Field
FA
B
A
Fg
Applied Force
mg
Ep
Ep
+
f
An object, with mass m, being raised in a uniform
downward gravitational field by a force, F, experiences
a downward gravitational force Fg = mg.
Electric
Field
A
FE
qE
Electric Field
Work must be done in moving a
charge against an electric field.
FA
+
+ + +
+
+ +
+
+
+
+ +
+
+
E pf
E pi
An object, with charge q , being raised in a uniform
downward electric field by a force, F, experiences a
downward electric force FE = qE.
38
Electric Potential& Energy
Electric Potential Energy
Electric Potential =
charge
Ep
(Units = J/C = Volt)
V =
Note:
Ep = QV
is a very useful equation to
know!
And
Ep =
KQq
R
39
Electric Potential& Energy
(Non-Uniform E Fields)
Therefore:
Electric Potential or Voltage at a Point:
V =
KQq
Rq
Therefore:
V=
KQ
R
Electric potential
for a non-uniform
E-field situation
40
Electric Potential(Non-Uniform
E Fields)
Equipotentials for a small
positive charge:
V3 = 100 V
V4 = 50 V
V1 = 200 V
Electric Field
Q
V2 = 150 V
kQ
V1 = 200 V
V4 = 50 V
V=
The same V exist
anywhere along
each of the
equipotential
concentric circles.
V2= 150V
V3 = 100 V
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Electric Potential(Non-Uniform
E Fields)
Example #1:
10.3 m
32 C
What is the electric
potential at point P due
to the 32 C charge?
42
Electric Potential(Non-Uniform
Solution:
V=
E Fields)
KQ
R
9.00 x 109 x 32 x 10-6
V=
10.3 m
V = +28000 V
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Electric Potential(Non-Uniform
E Fields)
Example #2:
3.2 m
P
10 C
3 C
What is the electric potential at point P midway between
the two charges?
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Electric Potential(Non-Uniform
E Fields)
Solution: Calculate the Voltage (electric potential) due to each
charge separately and then just add them.
VR =
kQ
VL =
9.00 x109 x 10 x 10-6
VR =
VR = 56250 V
1.6
VL =
kQ
r
9.00 x109 x 3 x 10-6
1.6
VL = 16875 V
V(at point P) = 73125 V
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Electric Potential(Non-Uniform
E Fields)
For a real challenge try this:
What is the potential difference between points
A and B shown below?
A
3.0 m
3.0 m
B
1.7 m
25 uC
1.7 m
25 uC
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Electric Potential(Non-Uniform
E Fields)
Solution:
At Point A:
VA1 =
Find the Voltage at each point due to each charge then
find the voltage difference.
kQ
VA2 =
9.00 x109 x 25 x 10-6
VA1 =
3.0
kQ
r
9.00 x109 x 25 x 10-6
VA2 =
VA1 = 75000 V
3.0
VA2 = 75000 V
VA = 150000 V
47
Electric Potential(Non-Uniform
E Fields)
At Point B:
VB1 =
kQ
VB2 =
9.00 x109 x 25 x 10-6
VB1 =
1.7
kQ
r
9.00 x109 x 25 x 10-6
VB2 =
VB1 = 132352 V
1.7
VB = 132352 V
2
VB = 264704 V
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Electric Potential(Non-Uniform
E Fields)
Therefore the Voltage difference is
V = VB - VA
V = 264704 V - 150000 V
V = 114704 V
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Electric Potential& Energy for uniform
Electric Field Situations
Found in between
two charged plates
or a capacitor.
Uniform E Field
+
P.
S.
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Electric Potential& Energy
foruniform Electric Field
Situations
Ep
(Units = J/C = Volt)
V =
Q
Re-arranging this equation results in a very useful
form:
Ep = Q x V
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Electric Potential& Energy
Examine the following cathode-anode
arrangements:
Which has the greatest electric potential
and hence the greatest energy?
+200 V
a)
0V
+200 V
b)
-200 V
-400 V -600 V
c)
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Electric Potential& Energy
Example
#1:
What is the impact speed of electrons on the
red plate when accelerated by the uniform
electric field shown?
Ep
Ek
v2 = ?
v1 = 0
Ep = QV
Ek= mv2
But Ep is completely transferred to Ek!
Ep = Ek
350 V
QV = mv2
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Electric Potential& Energy
Therefore:
v2 = ?
v1 = 0
2QV
v2 =
v =
350 V
v =
2 x 1.6 x 10-19 x 350
9.11 x 10-31
1.1 x 107 m/s
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Electric Potential& Energy
Example #2: What is the impact speed of the electron in
this case?
Ek1
v1 = 2.0 x 106 m/s
v2 = ?
Ep1 = QV
Ek2
Try:
Ek2 = Ek1 + Ep1
(Ans: 6.9 x 106 m/s)
125 V
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Electric Potential& Electric Field
(Uniform E
Field)
Let us take a look how these two ideas
can be related:
W = Ep = QV
(Potential electric Energy)
F
E =
q
And therefore:
(Electric Field Equation)
F = qxE
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Electric Potential& Electric Field
But:
Therefore:
(Uniform E Field)
W = F xd
W = qxExd
And: W = Ep = QV
Putting both of these equations together:
qxExd = QxV
Now: q = Q and cancels out on both sides
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Electric Potential& Electric Field
(Uniform E
Field)
This leaves you with a very useful
equation that relates electric potential
and electric field.
V = E x d
V = voltage
across plates
(V)
E = electric field
in the plates
(N/C) or (V/m)
d = distance
between the
plates (m)
58
Electric Potential& Electric Field
(Uniform E Field)
By a little re-arranging:
Voltage across the plates
E=
V
d
Distance bewteen the
plates
This again a useful equation in solving
problems inside CRTs and other devices
that have charge plates or capacitors.
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Electric Potential& Electric Field
For
Example:
(Uniform E Field)
0.175 m
230 V
0.025 m
a) What is the E-field (Magnitude & Direction in between the plates?
b) What is the electric force on the electron while it is in the plates?
c) What is the acceleration of these electrons?
d) Assuming they start from rest what would the speed of these
electrons just before they hit the positive plate?
60
Summing up all the electrostatic Equations
Non- Uniform Field Situation
xR
(Force)
Ep
KQq
R2
(Electric Field)
KQ
R2
(Energy)
KQq
R
V
(Electric Potential)
KQ
R
61
Summing up all the Electrostatic Equations
Uniform Field Situation:
xd
(Force)
qE
Ep
QV
or
qV/d
q
(Electric Field)
V
d
(Energy)
(Electric Potential)
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