ElectroStatics

Coloumb,s Law
Gauss’s Law
 Electromagnetism

Electromagnetism is one of the fundamental forces

in nature, and the the dominant force in a vast range
of natural and technological phenomena

 The electromagnetic force is solely responsible for the

structure of matter, organic, or inorganic substances.
 Physics, chemistry, biology, materials science, and Engineering

 The operation of most technological devices is based on

electromagnetic forces. From lights, motors, and batteries,
to communication and broadcasting systems, as well as
microelectronic devices.
 Electromagnetism
Electricity
Electromagnetism Magnetism
Optics

charge force field potential current

electric magnetic induction alternating waves
circuit field currents
reflection refraction image interference diffraction
 System of Units
We will use the SI system – SI  International System of Units

Fundamental Quantities
Length  meter [m]
Mass  kilogram [kg]
Time  second [s]

Other Units
Current  ampere [A]

Derived Quantities
Force  newton 1 N = 1 kg m / s2
Energy  joule 1J=1Nm
Charge  coulomb 1 C = 1 A s
Electric Potential  volt 1V=1J/C
Resistance  ohm 1=1V/A
Electrostatics
 Electric Charge

The Transfer of Charge

SILK

Glass Rod

Some materials attract electrons

more than others.
 Electric Charge

The Transfer of Charge

SILK
+ -

Glass Rod

As the glass rod is rubbed against silk,

electrons are pulled off the glass onto the silk.
 Electric Charge

The Transfer of Charge

SILK
+ -
+ -
Glass Rod

Usually matter is charge neutral, because the number of

electrons and protons are equal. But here the silk has an
excess of electrons and the rod a deficit.
 Electric Charge
The Transfer of Charge

+ SILK
+ - - -
+
+ - -
+
Glass Rod

Glass and silk are insulators:

charges stuck on them stay put.
 Electric Charge

+ +

Two positively charged rods

repel each other.
 Electric Charge
•
There is a property of matter called electric charge. (In the SI
system its units are Coulombs.)

– Charges can be negative (like electrons) or positive (like

protons).

– In matter, the positive charges are stuck in place in the nuclei.

Matter is negatively charged when extra electrons are added,
and positively charged when electrons are removed.

– Like charges repel, unlike charges attract.

– Charges travel in conductors, not in insulators

– Force of attraction or repulsion ~ 1 / r2

 Charge is Quantized
q = multiple of an elementary charge e:
e = 1.6 x 10-19 Coulombs

Charge Mass Diameter

electron -e 1 0
proton +e 1836 ~10-15m
neutron 0 1839 ~10-15m
 Electric Forces
• Electric charges exert forces of attraction or
repulsion on one another

• Charles Coulomb experimented to find the

factors that effect the magnitude of this force

– Force directly proportional to product of charges

– Force inversely proportional to square of distance

between charges
Coulomb’s Law

• Charles Coulomb measured

the magnitudes of electric
forces between two small
charged spheres
• He found the force
depended on the charges
and the distance between
them
 Coulomb’s Law
q1 q2 F12
r12

kq1 q 2
F 12  2
Force on 2 due to 1
r12
k = (4pe0)-1 = 9.0 x 109 Nm2/C2
e0 = permitivity of free space
= 8.86 x 10-12 C2/Nm2
Coulomb’s law describes the interaction between bodies due to their charges
Coulomb’s Law – Gives the electric force
between two point charges.
q1q2
F k 2 Inverse Square

r Law

k = Coulomb’s Constant = 9.0x109 Nm2/C2

q1 = charge on mass 1
q2 = charge on mass 2
r = the distance between the two charges

The electric force is much stronger than the

gravitational force.
Example 1
Two charges are separated by a distance r and have a force F on
each other. qq
F k 1 2
2
r
F q2 F
q1
r

If r is doubled then F is : ¼ of F

If q1 is doubled then F is : 2F

If q1 and q2 are doubled and r is halved then F is : 16F

Example 2
Two 40 gram masses each with a charge of 3μC are placed
50cm apart. Compare the gravitational force between the two
masses to the electric force between the two masses. (Ignore
the force of the earth on the two masses)

3μC 3μC
40g 40g

50cm
 m1m2
Fg  G 2
r
(.04)(.04) 13
 6.67  10 11
2
 4.27 10 N
(0.5)

q1q2
FE  k 2
r
6 6
(3  10 )(3  10 )
 9.0  10 9
2
 0.324 N
(0.5)
The electric force is much greater than the gravitational
force
 Gravitational and Electric Forces
in the Hydrogen Atom

M m = 9.1 10-31 kg
r12 -e
+e M = 1.7 10-27 kg
m r12 = 5.3 10-11 m

Gravitational force Electric Force

 Gravitational and Electric Forces
in the Hydrogen Atom

M m = 9.1 10-31 kg
r12 -e M = 1.7 10-27 kg
+e
m r12 = 5.3 10-11 m

Gravitational force Electric Force

 Mm
Fg  G 2 r
r12

Fg = 3.6 10-47 N
 Gravitational and Electric Forces
in the Hydrogen Atom

M m = 9.1 10-31 kg
r12 -e
+e M = 1.7 10-27 kg
m r12 = 5.3 10-11 m

Gravitational force Electric Force

 Mm   1  Qq
Fg  G 2 r Fe    2 r
r12  40  r12

Fg = 3.6 10-47 N Fe = 3.6 10-8N

 Superposition principle
• The resultant force on any one charge equals
the vector sum of the forces exerted by the
other individual charges that are present
– Remember to add the forces as vectors
• The resultant force on q1 is the vector sum of
all the forces exerted on it by other charges:

F1  F21  F31  F41
Superposition of forces from two charges

What is net force?

x
Superposition of forces from two charges

Consider effect of each charge separately:

y

x
Superposition of forces from two charges

Take each charge in turn:

y

x
Superposition of forces from two charges

Create vector sum:

y

x
Superposition of forces from two charges

Find resultant:
y

NET
FORCE
x
 Superposition Principle

F31 F F31
F31y
q1
F21
q2
F31x
F21
q3 F21y
Forces add vectorially
F21x
F = (F21x + F31x) x + (F21y + F31y) y
Example 3

Three charged objects are placed as shown. Find the net force
on the object with the charge of -4μC.
q1q2
F k
- 5μC r2
45º
(5 10 6 )(4 10 6 )
F1  9 10 9
 4.5 N
20 2  20 2  28cm (0.20) 2
20cm

(5 10 6 )(4 10 6 )

F2  9 10 9
2
 2.30 N
F1 45º (0.28)
5μC - 4μC
20cm F2

F1 and F2 must be added together as vectors.

 F1 2.3cos45≈1.6 - 2.9
45º θ
29º
2.3sin45≈1.6 - 1.6
F2 3.31

F1 = - 4.5 , 0.0 Fnet  2.9 2  1.6 2  3.31N

+ F2 = 1.6 , - 1.6   1.6 
  tan  1
  29 

Fnet = - 2.9 , - 1.6   2.9 

3.31N at 29º
• Two positive point charges ( Q1 = 50μC and Q2 =
1μC ) exert a repulsive force of 175 N. How far
apart are they?
• Find the position of q3 so that it has net force
equal to zero acting on it.
 Gauss’s Law

Flux ( or FLOW) is a general term associated with a FIELD that is

bound by a certain AREA. So ELECTRIC FLUX is any AREA that has
a ELECTRIC FIELD passing through it.
 Electric Flux

We define the electric flux ,

of the electric field E,
E through the surface A, as:

A
F=E.A

area A  = E A cos ()

Where:
A is a vector normal to the surface
(magnitude A, and direction normal to the surface).
 is the angle between E and A
 Electric Flux

You can think of the flux through some surface as a measure of

the number of field lines which pass through that surface.

Flux depends on the strength of E, on the surface area, and on

the relative orientation of the field and surface.

E E

Normal to surface,
magnitude A
area A
Here the flux is A 
F=E·A
 Electric Flux

The flux also depends on orientation

F = E . A = E A cos q

area A
E E area A

q
q
A cos q A A cos q A

The number of field lines through the tilted surface equals the
number through its projection . Hence, the flux through the tilted
surface is simply given by the flux through its projection: E (A cosq).
What if the surface is curved, or the field varies with position ??

=E.A 1. We divide the surface into small

regions with area dA

2. The flux through dA is

dF = E dA cos q
dA dF = E . dA
q
A 3. To obtain the total flux we need
E
to integrate over the surface A

F =  d =  E . dA
 In the case of a closed surface
q
  d   E  dA  inside
0
The loop means the integral is over a closed surface.

dA
q

E
 Electric Flux
Visually we can try to understand that the flux is simply the
number of electric field lines passing through any given
area.
In the left figure, the flux is zero.

In the right figure, the flux is 2.

• When E lines pass outward through a closed surface, the

FLUX is positive
• When E lines go into a closed surface, the FLUX is negative
 Gauss’ Law

In the case of electric fields the source of the field is the CHARGE!
So we can now say that the SUM OF THE SOURCES WITHIN A CLOSED
SURFACE IS EQUAL TO THE TOTAL FLUX THROUGH THE SURFACE.
This has become known as Gauss' Law
Gauss’ Law
The electric flux (flow) is in direct
proportion to the charge that is
enclosed within some type of
surface, which we call Gaussian.

qenc
 E  dA  o
 Gauss’ Law – How does it work?
Consider a POSITIVE POINT CHARGE, Q.
Step 1 – Is there a source of
symmetry?
Yes, it is spherical symmetry!
You then draw a shape in such a way as
to obey the symmetry and ENCLOSE the
charge. In this case, we enclose the
charge within a sphere. This surface is
called a GAUSSIAN SURFACE.
Step 2 – What do you know about the electric
field at all points on this surface? qenc
It is constant. E  da 
o
The “E” is then brought out of the integral.
 Gauss’ Law – How does it work?
Step 3 – Identify the area of the Gaussian
surface?
In this case, summing each and every dA
gives us the surface area of a sphere.

qenc
E (4r )  2

o
Step 4 – Identify the charge enclosed?
The charge enclosed is Q!

Q Q
E (4r )  2
 E
o 4r 2 o
 Gauss’ Law and cylindrical symmetry
Consider a line( or rod) of charge that is very long (infinite)

We can ENCLOSE it within a CYLINDER. Thus our Gaussian

surface is a cylinder.
+
+
+ qenc qenc
+ E  da  E (2rL) 
+ o o
+
+ L Q
+ E (2rL)  RECALL : Macro   
+ o L
+
+  Q  L  qenc
+ E
2r o Acylinder  2rL

Where λ is linear charge density which is quantity of charge per unit length.
 Gauss’ Law for insulating sheets and disks
A charge is distributed with a uniform charge density over an infinite plane
INSULATING thin sheet. Determine E outside the sheet.

For an insulating sheet the charge resides INSIDE the sheet. Thus
there is an electric field on BOTH sides of the plane.
qenc
+
 E  dA  o
Q Q
EA  EA   2 EA 
o o
Q A
  , 2 EA 
A o

E
2 o
 Gauss’ Law for conducting sheets and disks
A charge is distributed with a uniform charge density over an infinite thick
conducting sheet. Determine E outside the sheet.

For a thick conducting sheet, the charge exists on the surface only

+ qenc
+
 E  dA  o
E =0 Q
+
EA 
+ o
+
Q A
+  , EA 
+ A o

E
o
 In summary
Whether you use electric charge distributions or Gauss’ Law you get the SAME
electric field functions for symmetrical situations.
Q dq
E  dE 
4 o r 2 4 o r 2
qenc
 E  dA  o
Function Point, hoop, Disk or Sheet Line, rod, or
or Sphere (AREA) cylinder
(Volume) “insulating (LINEAR)
and thin”

Equation Q  
E E E
4 o r 2 2 o 2 o r