1. Electrostatics Learning Outcomes for the Lesson: state Coulomb’s Law
Use Coulomb’s 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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2. Electrostatics Vocabulary Electric Force Electric Charge
Electric Field
Electric Potential Energy
Electric Potential (Voltage)
electromotive force (emf),
potential difference,
Potential
Current
Coulomb
Ampere
Conductor
Coulomb’s Constant
Coulomb’s Law
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3. 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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4. 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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5. Electrostatic Force Law of Electrostatic Charges Like Charges repel.
Unlike Charges attract.
Charged objects attract neutral objects
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6. Electrostatic Force + + + + + + + + Attraction - - - - - - - - - -
Attraction
Charge is built up by rubbing two objects together
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7. 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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8. 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 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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9. 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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10. Electrostatic Force Coulomb’s Law: kQq Fe = 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 Coulomb’s Constant
= 9.00 x 109 Nm2/C2
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11. Coulomb’s Law
24 μC
-15 μC
2.3 m
What is the electrostatic force between the two charges?
9.00 x 109 x 24 x 10-6 x -15 x 10-6
2.32
Fe =
k Q q r2 =
Fe = N
(Attractive)
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12. Coulomb’s Law 22 μC -13 μC 2.3 m 45 μC 1.7 m
What is the net force on the -13 μC due to the other two charges?
Solution:
Find the force on each side of the center charge.
Fel =
k Qq r2
Fer =
kQq r2
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13. Coulomb’s Law Fe(left) = 9.00 x 109 x 22 x 10-6 x -13 x 10-6 1.72
= N (Left)
Fe(right) =
9.00 x 109 x 45 x 10-6 x -13 x 10-6
2.32
= N (right)
Therefore:
Fnet =
0.995 N (R) N (L)
= N (R)
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14. Coulomb’s Law vs Gravitational Law
Fe =
k x Q x q r2
Fg =
G x M x m r2
Note the similarity of the two graphs FE r Fg r
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15. Coulomb’s Law vs Gravitational LawFE
1/r2 Fg
1/r2
m = KQq
m = GMm
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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16. Electric Field 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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17. Electric Field Electric Field around a negative charge
Electric Field around two equal and opposite charges
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18. Electric Field Electric Field around two equal charges
Electric Field around various unequal charges
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19. Electric Field Electric Field between two charged plates+ -
Electric Field between two charged plates
The Electric field is constant between the plates (except at the ends).
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20. 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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21. Electric Field - Formula
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
F =
k x Q x q r2
and
Putting these two together results in:
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22. Electric Field - Formula
k x Q 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 Coulomb’s Constant
= 9.00 x 109 Nm2/C2
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23. Electric Field - CalculationsP
1.5 m
65 μC
What is the electric field intensity at a distance of 1.5 m away from a 65 μC charge?
E =
k x Q r2
New Equation
E =
9.00 x 109 x 65 x 10-6
1.52
E = N/C
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24. Electric Field - CalculationsEB P
45 μC
-15 μC
3.4 m
2.8 m ER
What is the net electric field at point P between the two oppositely charged particles?
Solution:
Find the E-fields due each charge separately
ER =
k x Q r2
EB =
k x Q r2
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25. Electric Field - Calculations
ER =
9.00 x 109 x 45 x 10-6
3.42
EB =
9.00 x 109 x -15 x 10-6
2.82
ER = N/C (Right)
EB = N/C (Right)
Enet = N/C (Right)
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26. Electric Potential Energy (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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27. Electric Potential Energy
Remember W = ΔEk
And W = F x d
Therefore:
ΔEk = ΔEp = F x d
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28. Electric Potential & Energy (Non-Uniform E Fields)
W = ΔEp = F x d
F =
KQ1Q2 R2
But:
Therefore:
KQ1Q2
Ep = R2
x d
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29. Electric Potential Energy (Non-Uniform E Fields)
Since d = R, one R will cancel, leaving:
KQ1Q2
Ep = R
Note:
This is very similar to Gravitational Potential Energy:
- GM1m2
Ep = R
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30. Electric Potential & Energy (Non-Uniform E Fields)
Example #1:
What is the potential energy of the –2.6 μC charge relative to the +37 μC Charge?
7.3 m
+37 μC
-2.6 μC
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31. Electric Potential & Energy (Non-Uniform E Fields)
Solution:
KQ1Q2
Ep = R
9.00 x 109 x 37 x 10-6 x –2.6 x 10-6
Ep =
7.3
Ep = J
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32. 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 μC
2.0 μC
What work is done in moving the 2.0 μC charge 0.75 m closer to the 4.0 μC charge?
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33. Electric Potential & Energy (Non-Uniform E Fields)
Solution:
Work = ΔEp
ΔEp = Ep Ep1
kQq r2
kQq r1
ΔEp =
9.00 x109 x 4.0 x 10-6 x 2.0 x 10-6
1.65
9.00 x109 x 4.0 x 10-6 x 2.0 x 10-6
2.4
ΔEp =
ΔEp = J J
= J
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34. 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 electron’s closest approach?
Fe atom
-26 elementary Charges
v = 6.0 x 105 m/s
d = ?
electron
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35. Electric Potential & Energy (Non-Uniform E Fields)
Solution:
Ek = Ep
KQ1Q2 R
½ m v2 =
½ x 9.11 x x (6.0 x 105)2 =
9.00 x 109 x 26 x x 10 –19 x x 10-19 R
R =
5.99 x 10-27
1.64 x 10-19
R =
3.7 x 10-8 m (about the size of an atom!!)
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36. 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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37. 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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38. Electric Potential Fg mg FA A B m Epi Epf Gravity Field FE qE FA B A q+
Applied Force +
Electric Field Fg mg FA A B m
Epi
Epf
Gravity Field
An object, with mass m, being raised in a uniform downward gravitational field by a force, F, experiences a downward gravitational force Fg = mg. FE qE FA B A q
Work must be done in moving a charge against an electric field.
Epi
Epf
ElectricField
An object, with charge q , being raised in a uniform downward electric field by a force, F, experiences a downward electric force FE = qE.

39. Electric Potential & Energy
charge
V = Ep Q
(Units = J/C = Volt)
Note:
Ep = QV
is a very useful equation to know!
And
Ep =
KQq R

40. Electric Potential & Energy (Non-Uniform E Fields)
Therefore:
Electric Potential or Voltage at a Point:
KQq
R q
V =
Therefore:
V = KQ R
Electric potential for a non-uniform E-field situation

41. Electric Potential (Non-Uniform E Fields)Q
Electric Field
Equipotentials for a small positive charge:
V3 = 100 V
V4 = 50 V
V1 = 200 V
V2 = 150 V
V = kQ R
V1 = 200 V
The same V exist anywhere along each of the equipotential concentric circles.
V2= 150V
V4 = 50 V
V3 = 100 V
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42. Electric Potential (Non-Uniform E Fields)
Example #1:
10.3 m P
What is the electric potential at point P due to the 32 μC charge?
32 μC

43. Electric Potential (Non-Uniform E Fields)
Solution:
V = KQ R
V =
9.00 x 109 x 32 x 10-6
10.3 m
V = V
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44. Electric Potential (Non-Uniform E Fields)
Example #2:
3.2 m
10 μC P
3 μC
What is the electric potential at point P midway between the two charges?
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45. Electric Potential (Non-Uniform E Fields)
Solution:
Calculate the Voltage (electric potential) due to each charge separately and then just add them.
VR = kQ r
VL = kQ r
VR =
9.00 x109 x 10 x 10-6
1.6
VL =
9.00 x109 x 3 x 10-6
1.6
VR = V
VL = V
V(at point P) = V
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46. Electric Potential (Non-Uniform E Fields)
For a real challenge try this:
What is the potential difference between points A and B shown below?
25 uC A B
3.0 m
1.7 m
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47. Electric Potential (Non-Uniform E Fields)
Solution:
Find the Voltage at each point due to each charge then find the voltage difference.
At Point A:
VA1 = kQ r
VA2 = kQ r
VA1 =
9.00 x109 x 25 x 10-6
3.0
VA2 =
9.00 x109 x 25 x 10-6
3.0
VA1 = V
VA2 = V
VA = V

48. Electric Potential (Non-Uniform E Fields)
At Point B:
VB1 = kQ r
VB2 = kQ r
VB1 =
9.00 x109 x 25 x 10-6
1.7
VB2 =
9.00 x109 x 25 x 10-6
1.7
VB1 = V
VB2 = V
VB = V
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49. Electric Potential (Non-Uniform E Fields)
Therefore the Voltage difference is
ΔV = VB - VA
ΔV = V V
ΔV = V
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50. Electric Potential & Energy for uniform Electric Field Situations+ -
P.S.
Uniform E Field
Found in between two charged plates or a capacitor.
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51. Electric Potential & Energy for uniform Electric Field Situations
V = Ep Q
(Units = J/C = Volt)
Re-arranging this equation results in a very useful form:
Ep = Q x V
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52. Electric Potential & Energy
Examine the following cathode-anode arrangements:
Which has the greatest electric potential and hence the greatest energy?
+200 V
0 V
-200 V
-400 V
-600 V a) b) c)
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53. Electric Potential & Energy
Example #1:
What is the impact speed of electrons on the red plate when accelerated by the uniform electric field shown?
350 V
v1 = 0
v2 = ? Ep Ek
Ep = QV
Ek= ½ mv2
But Ep is completely transferred to Ek!
Ep = Ek
QV = ½ mv2
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54. Electric Potential & Energy
Therefore:
350 V
v1 = 0
v2 = ?
2QV m
v2 =
v =
2 x 1.6 x x 350
9.11 x 10-31
v =
1.1 x 107 m/s
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55. Electric Potential & Energy
Example #2:
What is the impact speed of the electron in this case?
v2 = ?
125 V
v1 = 2.0 x 106 m/s
Ek1
Ep1 = QV
Ek2
Try:
Ek2 = Ek Ep1
(Ans: 6.9 x 106 m/s)
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56. 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)
E = F q
(Electric Field Equation)
And therefore:
F = q x E
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57. Electric Potential & Electric Field (Uniform E Field)
But: W = F x d
Therefore:
W = q x E x d
And: W = Ep = QV
Putting both of these equations together:
q x E x d = Q x V
Now: q = Q and cancels out on both sides
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58. 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)

59. 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 CRT’s and other devices that have charge plates or capacitors.
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60. Electric Potential & Electric Field (Uniform E Field)
For Example:
230 V
0.025 m
0.175 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?

61. KQq R2 KQq R V KQ R2 KQ R Summing up all the electrostatic Equations F
Non- Uniform Field Situation
x R F
(Force) Ep
(Energy)
KQq R2
KQq R
÷ q
(Electric Field) E V
(Electric Potential) KQ R2 KQ R

62. qE QV qV/d V V V d Summing up all the Electrostatic Equations F Ep E
Uniform Field Situation:
x d F
(Force) Ep
(Energy) qE QV or
qV/d
÷ q
(Electric Field) E V
(Electric Potential) V d V
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