Electrostatics
The study of electric charges
Introduction
• Did you ever run a comb through your hair? What do
you notice.
• What causes the paper holes to jump onto the comb?
• There are electrical forces that are in place due to the
presence of charge on the comb.
The Atom
• An atom consists of
various charged and
uncharged particles.
• The central region is
called the nucleus.
• Protons (+) and
neutrons make up
the nucleus.
• Electrons (-) move
around the nucleus
in an orbital path.

 Nucleus

Neutrons
Protons (+)
Electrons (-)
The Significance of Charge
• As mentioned before, protons
are positive and electrons are
negative.
• An atom with balanced
charges is considered neutral.
• The overall charge can be
changed by adding or
removing electrons. This
makes the atom an ion.
Add e-
Take eOverall Charge: 1
01
(Negative
(Neutral
Ion)
(PositiveAtom)
Ion)
Actual Charge of Protons/Electrons
• Recall, charge is measured in Coulombs (C).
• Even though protons and electrons are very
small, they still have charge.
• Let us use q as a variable for charge.
Electron
qelectron  1.60 1019 C
Proton
qproton  1.60 1019 C
Sample Problem (Atomic Charge)
• A helium atom has a net electric charge
-19
of -8.0x10 C.
• Is it neutral or an ion?
• Are there extra electrons or a shortage of them?
• How many extra electrons are there?
– Charge Per e-: -1.60*10-19C
qNet
#e 
qe
19

8.0

10
C

#e 
1.6 1019 C


#e  5
Sample Problem
• How many excess electrons are on a ball that
has a charge of q = -4x10-17C?
qNet
#e 
qe

17

4.0

10
C

#e 
1.6 1019 C

# e  250
Sample Problem (Atomic Charge)
•
•
•
•
-19
An atom has a net electric charge of 4.8x10 C.
Are there extra electrons or a shortage of them?
How many electrons short is this atom?
Draw this atom given it is Boron.
Electric Forces
• Charges exert a force on other charges
Like Charges
Repel
Opposites
Attract
Electrostatic Demo’s
• Tape
• Electroscope
• Pith Balls
How do atoms get charged?
• Work can remove electrons from the atom.
– Results in a positively charged atom
• The free Electron can be transferred to
another atom.
– Results in negatively charged atoms
Coulomb’s Law
• The electrostatic force one charged object exerts
on an other
• The force is related to the amount of charge
– i.e more charge – more force
• The force is proportional to 1/d2
– i.e. the further apart the charges, the smaller
the force
Coulomb’s Law
F
Symbol
F
q1
q2
d
K
Kq1 q 2
Force
Charge
Charge
Distance
constant
d2
Unit
N
C
C
m
N m2 / C2
Ex. Coulomb’s Law
Object A has a charge of 6x10-6C. Object B has a
charge of 3x10-6C and is 0.03m away. Calculate
the force on A.
F
Kq1 q 2
d2
(9.0 x109 N m2 / C 2 )(6 x106 C )(3x106 C )
F
2
(.03m)
F  180 N
Separating Charge
• Charges are balanced in neutral objects.
• Work must be done to separate charge (free electrons).
• Once charge is separated, it can be used in
experiments.
Separation of Charge
• Bring a charge rod near a neutral conductor
• Un-like charges are attracted
• Like charges are repelled
Charge by conduction
• A charge rod touches a neutral conductor
• Like charges are repelled and uniformly
distribute
Charge by Induction
The
charges
oncharge
the
spheres
redistribute
to
A
Separation
charge
object
of
isthe
placed
takes
near
place
neutral
conductors
charge
object
is
removed
Contact
between
conducting
sphere
is maximum
broken
Result:
Two
spheres
charged
by
induction
separation
A
B
B
Charging by Polarization
• Certain substances, such as the one below, have polar
molecules. These molecules have opposite charges at
each end.
• Charging by polarization takes place when a charged
object is brought near, realigning the molecules in the
substance.
Magnification
Conductors and Insulators
• Electrical Conductors are similar to Heat
conductors.
• Electrical Conductors allow charge to
move easily.
• Electrical Insulators do not allow charge to
move easily
Conductors and Insulators
• Electrical Conductors all electrons to move
easily.
– Metals
– Graphite
• Electrical Insulators do not allow electrons
to move easily
– Glass
– Plastic
– Rubber
Lightning
• Lightning: An electrical discharge between
the clouds and oppositely charged ground.
• Charging by induction occurs during
thunderstorms
• The negatively charged cloud induces a
positive charge on the ground
Lightning
• Average Temperature 30,0000C
– (roughly 5x as hot as the sun)
• Typical charge (q) for lightning
10C to 25C
• How many electrons is this?
Lightning is actually a discharge of static electricity.
Charge differences are developed from the friction of
dust particles within the cloud.
When the concentration of charges becomes too great,
an electrical discharge results
Grounding
• The earth stores a seemingly infinite amount of charge,
both kinds.
• An object is grounded when it is connected to the earth or
another large object.
• Electrical devices often have a “ground,” which prevents
unwanted charge buildup.
• Grounding is also the principle behind lightning rods.
Outlet
Ground
The Electroscope
• An electroscope is a device that detects
electrical charge in objects brought near.
• Its metallic inner contents, which are usually
neutral, have to be separated from
surroundings by some type of insulator.
• There are two metal leaves that hang
inside.
• When a charged object is brought near, the
leaves separate.
• Charge can also be stored in the
electroscope by touching it with the rod.
Leaves



Ex. Coulomb’s Law
The distance that separates electrons in a typical
atom is 1.6E-12m. What is the electrostatic force
between them?
Kq1q2
F
d2
(9.0 x109 N m2 / C 2 )(1.6 x1019 C )(1.6 x1019 C )
F
(1.6 x1012 m)2
5
F  9.0 x10 N
Electro-static Applications
Electrostatic filter
Ex. Coulomb’s Law
Three charges are aligned as shown.
Calculate the force on q2 due to q1.
q1  6.25 x10 4 C
q2  3.5 x10 4 C
4
q3  5.25 x10 C
q1
Kq1q2
F
d2
q2
Electric Charge Positions
q3
Scale: 1 Square = 0.05m
(9.0 x109 N m2 / C 2 )(6.25x104 C )(3.5x104 C )
F12 
(.15m)2
F12  87500 N
Ex. Coulomb’s Law
Three charges are aligned as shown.
Calculate the force on q2 due to q3.
q1  6.25 x10 4 C
q2  3.5 x10 4 C
q3  5.25 x10 4 C
q1
q2
Electric Charge Positions
q3
Scale: 1 Square = 0.05m
Kq1q2
F
d2
(9.0 x109 N m2 / C 2 )(3.5 x104 C )(5.25 x104 C )
F23 
(.3m)2
F23  18375N
Ex. Coulomb’s Law
Three charges are aligned as
shown. Calculate the force on q2
due to q3.
q1  6.25 x10 4 C
q2  3.5 x10 4 C
4
q3  5.25 x10 C
q1
Electric Charge Positions
Fnet  87500Nright 18375right
Fnet  105875right
q2
q3
Scale: 1 Square = 0.05m
Coulomb’s Law in 2-D
•
•
•
•
To find Fnet with 3 or more charges
Calculate each Force vector.
It helps to have a grid system on which to work.
Use vector addition to find the resultant Fnet
q2
q3
F13
q1
F14
F12
FNet
q4
Coulomb’s Law in 2-D (cont.)
• Find the net force acting on q1.
• First find the distance between q1 and the others.
• Use the Pythagorean Theorem to find these distances.
A B C
2
2
1st Triangle
2
C  A B
2
2nd Triangle
2
C  22  32
C  13
A
q2
C
C
q3
A
B
q1
B
C  A2  B2
C  22  42
C  20
The variable C from each triangle’s
hypotenuse is the variable d used in
the Coulomb’s Law equation.
d12  13
d13  20
Coulomb’s Law in 2-D (cont.)
kq1q2
F12 
d12 2
Charge (C)
q1
3.0 X 10-4
q2
-2.6 X 10-5
q3
7.2 X 10-6
F12


9.0 10
9 Nm 2
C2
  3.0 10 C  2.6 10 C 
4

13m

5
2
F12  5.4 N
q2 F12
20
q3
13
F13
q1
kq1q3
F13 
d132
F13


9.0 109
Nm 2
C2
  3.0 10 C  7.2 10 C 
4

20m

6
2
F13  0.972 N
Coulomb’s Law in 2-D Sample
• Determine the direction of each of the forces prior to
vector addition.
opp
tan  
adj
F12
 2 
  tan  
 3 
2
  tan  
4
  33.7
  26.6
Quad II Adjust
Quad III Adjust
  33.7  180
  26.6  180
  146.3
  206.6
1
opp
q2
5.4N
hyp
adj
0.972N
hyp
q3
F13
1
opp
q1
adj
Coulomb’s Law in 2-D Sample
• The remaining
task is to use
analytical
vector addition.
Mag
Ang
X
Y
Q
F12
5.4N
146.3°
-4.49
3.00
II
F13
0.972N
206.6°
-0.87 -0.44
III
FNet
5.94N
154.5°
-5.36
II
F12 x  F12 cos
F12 x  5.4N  cos 146.3
F12 x  4.49 N
F12 y  F12 sin 
F12 y  5.4N  sin 146.3
F12 y  3.00N
F13x  F13 cos
F13x   0.972N  cos  206.6
FNet
F13x  0.87 N
F13 y  F13 sin 
F13 y   0.972N  sin  206.6FNet
F13 y  0.44N
FNet
 YTot 
tan   

X
 Tot 
Y 
  tan 1  Tot 
 X Tot 
 2.56 N 
  tan 1 

 5.36 N 
  25.5
XTot  F12 x  F13x
XTot  4.49N  0.87 N
XTot  5.36N
YTot  F12 y  F13 y
YTot  2.56 N
 X Tot 2  YTot 2

 5.36N    2.56N 
 5.94N
2
2.56
Quad II Adjust
2
  25.5  180
  154.5
FNet  5.94N @154.5
Conclusion
• Electrostatics, the study of the forces between charges
at rest.
And in it
there’s no
lightning
I had a
dream…
Coulomb’s Law in 2-D Sample
q1  4.5 x104 C
Three charges are aligned
as shown. Find the net
force on q1.
q2  3.2 x104 C
q3  7.3x104 C
q2
Mag
Ang
X
Y
Q
q1
Electric Charge Positions
F12
5.4N
146.3°
-4.49
3.00
II
F13
0.972N
20.6°
-0.87
-0.44
III
FNet
5.94N
154.5°
-5.36
2.56
II
q3
Scale: 1 Square = 0.5
cm
End Ch 20
Example: Charge Distribution
• What is the total charge of three conducting spheres with
charges of 6q,-1q, and 0q
qtotal  6q  (1q)  0q
qtotal  5q
A
6q
B
-1q
C
0q
Example: Charge Distribution cont.
• What is the final charge distribution if sphere A and B
touch?
A
6q
2.5q
B
-1q
2.5q
C
0q
Example: Charge Distribution cont.
• What is the final charge distribution if sphere B and C
touch?
A
2.5q
B
1.25q
2.5q
C
0q
1.25q
Example: Charge Distribution cont.
• What is the total charge of three conducting spheres
qtotal  2.5q  1.25q  1.25q
qtotal  5q
A
2.5q
B
1.25q
C
1.25q
Example: Charge Distribution cont.
• Sphere B is twice as large as sphere A, what will be the
charge distribution after they touch?
A
5q
B
5q
Charge by Conduction
• Bring a charge rod near a neutral conductor
• Un-like charges are attracted
• Like charges are repelled
A
5q
B
-1q
C
0q
q2
q1
q3
Scale: 1 square =0.1cm
q1
q2
q3