Electrostatics
Electric circuits
Ohm’s Law
Türmer Kata
2012.
Electric Charge

Ancient Greeks
amber: elecktron [gr]
◦ Found that amber, when rubbed, became electrified
and attracted pieces of straw or feathers.
◦ (Magnetic forces were discovered by observing
magnetite attracting iron).




Repulsion occurs between 2
charged glass rods.
All glass rods rubbed with
silk are charged similarly.
The charges in glass rods are
thus identical.
Same charges repel each
other.



!
Attraction occurs between a
charged ebonite rod and a
charged glass rod.
Charges in the ebonite rod
and glass rod are different.
Different charges attract
each other.
Law of charges:
Same charges repel, and different charges attract.

An atom is made up of 3 different subatomic particles.
-
-
+ +
+
+ +
+ +
-
-



Benjamin Franklin (1706-1790): the charge
left on a glass rod after rubbing with rabbit
fur was given the name “positive”, while that
left on amber was called “negative”.
Positive: electron deficiency.
Negative: electron excess.

Coulomb (C)
q=n·e
 q: electric charge
 e: electronic charge (charge of 1 single electron)
 n: integer
◦ q of an electron: -1.6 · 10-19 C
◦ q of a proton: +1.6 · 10-19 C
◦ q of a neutron: 0

In an isolated (closed) system:
!
◦ Charge is not created, only exchanged.
◦ Objects become charged because negative charge is
transferred from one object to another.
!
The net charge of an isolated system remains
constant.
Electrostatic Charging

Conductors: electrons have relatively high mobility.
◦ Metals
◦ When a conductor is charged in a small region, the charge
distributes itself over the entire surface.

Insulators: electrons are more tightly bound to the
atom.
◦ Glass, rubber
◦ When insulators are charged by rubbing, only the rubbed area
becomes charged.
 There is no tendency for the charge to move to other regions of the
material.



A charged object (the
rod) is placed in contact
with another object (the
sphere).
Some electrons on the
rod can move to the
sphere.
When the rod is
removed, the sphere is
left with a charge (the
same as the object
doing the charging).
An object is connected to a conducting wire or pipe buried
in the earth  grounded.






A negatively charged rubber rod is brought near
an uncharged sphere.
The charges in the sphere are redistributed.
A grounded conducting wire is connected to the
sphere  some electrons move from the sphere to
the ground.
Grounding is removed, the sphere is left with an
excess of induced positive charge.
The positive charge on the sphere is evenly
distributed due to the repulsion between the
positive charges.
Charging by induction requires no contact with the
object inducing the charge.



Charging by induction
does not have to involve
a removal of charge
from an object.
Charge can be moved
within an object to give
different
regions
of
charge..
In this case induction
brings
about
polarization (separation
of charge).
Electric Force and Electric Field

The magnitude of the force between q1 and
q2 is described by Coulomb’s Law:
!




k  q1q2
F
2
r
F= electric force (N) vector!
q= charge (C)
k= Coulomb constant (9,0 × 109 N∙m2 / C2)
r= distance between the charges (m)

Two point charges of -1.0 µC and +2.0 µC
are separated by a distance of 0.30 m. What
is the electrostatic force on each particle?
(1µC = 10-6 C)
◦
◦
◦
◦
q1 = -1.0 µC = -1.0 × 10-6 C
q2 = +2.0 µC = +2.0 × 10-6 C
r = 0.30 m
k = 9.0 × 109 Nm2/C2
k  q1q2
F
r2
◦ Find: F
2

9 Nm 
 9 10
  - 1.0  10-6 C  2.0  10-6 C
2 
C 
 18 10 3

Solution: F 

 0,2 N
0,30m 2
0,09

An electric field exists in space around a charged
object.
◦ When another charged object enters this electric field, the
field exerts a force on the second charged object.
F
E
q0



kq
E 2
r
Unit:
E: magnitude of the electric field
q0: positive test charge
k: Coulomb constant (9,0 × 109 N∙m2 / C2)
N
C





Electric fields are represented by electric field lines
(imaginary).
Electric field direction is from a positive charge to a
negative charge.
Electric field lines do not touch each other.
Electric field vector represents the force direction
of a small positive charge in the electric field.
An electric field is the region where an electric
force is exerted on any electric charge placed
within the influence of the electric field.

Electric field lines (imaginary):
◦ Direction: from a positive charge to a negative charge.
◦ Field lines do not touch each other.
◦ An electric field is the region where an electric force is
exerted on any electric charge placed within the influence
of the electric field.
◦ Electric field vector represents the force direction of a small
positive charge in the electric field.
+
-
-
+

The ‘density’ of field lines
determines the strength of the
field.

The ‘density’ of field lines determines the strength of the field.




Any excess charge on an isolated conductor
resides entirely on the surface of the conductor
The electric field is zero everywhere inside a
charged conductor
The electric field at the outer surface of a charged
conductor is perpendicular to the surface
Charge tends to accumulate at sharp points, or
locations of greatest curvature, on asymmetric
charged conductors (lightning rod!)
Electrical Energy and Electric Potential


When 2 or more charges are brought closer
together or further apart, work is done, and
energy is expended or stored
Electrostatic potential energy:
W
V
Q

Change in potential energy=Electrical
potential = Voltage
W
V 


AB
Q
W: work done in bringing the test charge in from
infinity
Unit: volt (J/C)  V
Capacitance and Dielectrics

Capacitors store electrical energy
!



Q  CV
Q: charge
V: voltage
C: capacitance
(constant of proportionality)
unit: F (farad)
Battery
!



C
0 A
d
A: surface area
d: distance between plates
ε0: permittivity of vacuum (8.85 × 10-12 C2/Nm2)

What would be the area of the plates of a 1.0
F parallel-plate capacior with a plate
separaion of 1.0 mm?
◦ C = 1.0 F
◦ d = 1.0 mm = 10-3 m
◦ ε0 =8.85 × 10-12 C2/Nm2
C
0 A
d
A
◦ Find: A
Solution:
1F  10 3 m
9
2
A

1
,
13

10
m
2
C
8,85  10 12
Nm 2
(We use µF in practice!)
C d
0
Current and electrical circuits

Closed electrical network

Elements (devices) of the electrical circuit:




(electrical network which has closed
loop giving a return path for the current).
Source of the voltage (battery)
Transmission lines (wires)
Resistors
(Capacitors)




Converts chemical energy into
electric energy.
Anode:
positively
charged
terminal of the battery.
Cathode:
negatively
charged
terminal of the battery.
Electromotive
force
(electric
potential or voltage): electric
potential difference between the
terminals of the battery.
V
Vac 
qo
V: electrostatic potential energy
qo: charge



Source of the voltage (electric potential) that
supplies
the
electric
energy
through
conversation of other forms of energy.
Conductor which must possess mobile
charged particles e.g. ions, electrons.
Closed electrical circuit: gives a return path
for the current.




Consider charges moving in a conductor – such as a wire. If an electric
field is applied there is a net flow of charges in the conductor.
Electric current (I): net charge flowing through the cross-sectional area (A)
in time:
q
I
t
!
1A  1
Unit of current: A (ampere)
The current flowing through the cross-sectional area (A) is called Current
Density:
I
J
A

C
sec
Unit of current density:
A
m2

A current of 0.50 A flows in a circuit for 2.0
min. How much charge passes through a
cross-sectional area of one of the connecting
wires during this time?
◦ I = 0.50 A
◦ t = 2.0 min = 120 s
Q
I
t
◦ Find: Q
Solution:
Q  0,50 A 120s  60C
Q  I t
Ohm’s Law and Resistance

Current is directly proportional to the voltage
I ~V

The slope of the straight line gives the
resistance (R) of the system


Resistance (R) is inverse proportional to the current
unit: Ohm (Ω)
1
I~
R
R

V
I
Ohm’s law: shows the connection between current,
voltage and resistance
V
I
R
V  RI
V
R
I
(Ohmic conductor: obeys to Ohm’s Law)
!
Electric Circuits

Series circuit
◦ Resistors are connected end to
end.
◦ The current going through
each resistor is equal to the
current of the source.
◦ The voltage of the source
decreases at each resistor.
◦ The equivalent resistance (Rs)
of the circuit.
Rs  R1  Rn  R3
I  I1  I 2  I 3
V  V1  Vn  V3  IR1  IR2  IR3  I R1  R2  R3 

Parallel circuit
◦ Resistors are connected with one end to another.
◦ The voltage drop on each resistor is equal to the voltage
of the source.
◦ The current divides among the resistors proportionally.
◦ The equivalent resistance (Rp) of the circuit.
1
1
1
1



R p R1 R2 R3
V  V1  V2  V3
I  I1  I n  I 3 
 1
V V
V
1
1  V


 V  
  
R1 R2 R3
 R1 R2 R3  R p
Thank you for the attention!