Transcript Diapositiva 1

Electricity. Electrostatic
The Electric Field
Electric charge.
Conductors and Insulators
Coulomb´s Law
The Electric field. Electric Field Lines
Calculating Electric Field for continous charge distribution
Gauss´s Law.
Charge and Field at Conductor Surface.
Motion of Point Charges in Electric Field.
Electric Dipoles in Electric Field
Electric Potential
Potential Difference
Calculating Electric Potential for a System of Point Charges and
for Continuous Charge distribution
Potential vs. Electric Field. Field Lines and Equipotential Surfaces
Electrostatic Energy and Capacitance
Electrostatic Potential Energy
Capacitance
The Storages of Electric Energy
Capacitors, Batteries and Circuits
Dielectrics. Molecular View of a Dielectric
The Electric Field
Electric charge
Conductors and Insulators
Coulomb´s Law
The Electric field. Electric Field Lines
Calculating Electric Field for continous
charge distribution
Gauss´s Law
Charge and Field at Conductor Surfaces
Motion of Point Charges in Electric Field.
Electric Dipoles in Electric Field
Electric Charge:
Objects carrying charges of opposite signs attract each other;
Objects carrying charges of the same sign repel each other
Positive, the charge acquired by a glass rod when is rubbed with a
piece of silk, (Franklin criteria), then electrons are transferred to silk.
The piece of silk acquires the same Negative charge
Law of Conservation of Charge
Charge Quantizacion
The SI unit of charge is the coulomb [C]
Fundamental Unit of Charge e = 1.602177 x 10-19 C
Conductors
Conductors
the
Insulators
Insulators
materials where the charges, usually electrons, are free to move about
the entire material, such as copper, iron,..
materials where the charges can not move freely, such as glass, wood
Charging by Induction
An object that has separated equal and opposite charges is said to be
What happens with
polarized
the charges
distribution on the
spheres once the
rod is removed?
Induction via grounding: using the earth as an infinitely large
conductor
Coulomb´s Law
k Coulomb constant:
8.99x109 N. m2/C2
Is the force exerted
by q1 on q2
Unit vector pointing from q1 to q2
In a hydrogen atom, the electron is separated from the proton by
an average distance of about 5.3x10-11m. Calculate the magnitude
of the electrostatic force of attraction. Compare the electrostatic
force with the gravity force between the proton and the electron.
Force exerted by a System of Charges:
Principle of superposition of forces
Coulomb´s torsion balance
Three charges are placed as shown in the
figure. Calculate the force exerted on the
particle at top of the isosceles triangle. Q= 10
μC; q= 500 nC; d = 10 cm
The Electric Field
F
E
q
F qE
The force exerted on the charge q
q is a small positive test charge
SI units [N/C]
[V/m]
Electric Field
for a system of
charges
To avoid the conceptual problem of the action at a distance –
instantaneous transmission- the concept of electric field is
introduced
[Suppose that a charged particle at some point is suddenly
moved. Does the force exerted on a particle some distance
away change instantaneously?]
Derive a general expression for the electric field on a point P due to a single
charge Q. P is placed at a distance r of the charge. Estimate the value of the
electric field for Q=10 nC and r= 15m.
Electric Dipole
A system of two equal and opposite charges separated by a
small distance is called a electric dipole .
Its strength and orientation is described by the electric
dipole moment
p  q L, L  2a
Exercise: Calculate the electric field of dipole in point P in
the dipole axis. Consider the situation when x»a.
E
2k p
x
3
The Electric Field Lines, or Lines of Force
At any given point, the field vector E is tangent to the
field line. They are also called lines of force because
they show the direction of the force exerted on the
positive test charge.
The density of the lines (the number of lines per unit
of area perpendicular to the lines) at any point is
proportional to the magnitude of the field at that point
The Electric Field Lines
The electric field lines for two conducting spheres are shown in
the figure. What is the relative sign and magnitudes of the
charges on the two spheres?
(A) Picture a uniform vertical electric field E = -2000 N/C.
(B)The same but the value of electric field is two times the
previous value.
Calculating Electric Field for Continous Charge Distribution
In the macroscopic world charge can usually be described as continuosly distributed.
When the charge is
distributed on a volume
When the charge is
distributed on a surface
When the charge is
distributed along a line
dQ
Q

;  average
Volume charge density
dV
V
dQ

Surface charge density
dS
dQ
Linear charge density

dl
Applying Coulomb´s
Law and the principle of
superposition
E
V
k dq
rˆ
2
r
Calculating Electric Field for Continous Charge Distribution.
E on the Axis of a finite
Line charge
E on the Axis of a finite
Line charge
E due to an infinite line charge
Gauss´s Law
Gauss´s Law is one of the so called Maxwell´s Equations that describe the
electromagnetics phenomena. For static charges, Coulomb´s Law and Gauss´s Law
are equivalent, but Gauss´s Law is more general.
Gauss´s Law can be
used to calculate the
electric field for charge
distribution with high
degree of symmetry
Gauss´s Law:The net number of lines out of any
surface enclosing the charges is proportional to
the net charge enclosed by the surface
Gauss´s Law
Electric Flux ϕ
E A
  E  nA
d  E  dS  E dSn
The number of field lines penetrating a surface is called the
electric flux. Units: N. m2/C
If we consider a surface A perpendicular to E,
In the case of surface that is not perpendicular to E, the
dot product enables us to obtain the value of area
perpendicular to the electric field.
Calculating E from Gauss´s Law.
The power of symmetry
Electric field for a single point charge
The electric field exhibits spherical symmetry around the charge.
Then we consider a spherical surface with center on the charge to
apply Gauss´s law. The value of E is constant in all points of this
sphere.
   E dA  E 4R 2  4 k Q
A
Ek
The flux is independent
from the selected sphere
Q
R2
Writing Gauss´s Law and Coulomb´s Law in terms
of permitivity of free space
2
1
o 
 8.85x1012 C
N  m2
4k
k
1
4  o
1
q
E
rˆ
2
4  o r
net 
Qinside
o
Calculating E from Gauss´s Law.
The power of symmetry
Electric field for a Thin Spherical Shell of Charge
The electric field exhibits spherical symmetry around the uniform
charge distribution . Then we consider a spherical surface with the
same center as the shell of charge to apply Gauss´s law. The value
of E is constant in all points of this sphere.
   E dA  E 4R 2  4 k Q
A
Ek
Q
1 Q

R 2 4o R 2
For a gaussian sphere
inside of the shell
charge
In the Earth´s atmosphere, the electric field
is 150 N/C downwards at an altitude of 250
m, and 170 N/C downwards at an altitude
of 400 m. Calculate the volume charge
density of the atmosphere assuming it to
be uniform between both altitudes.
   E dA  0
A
E 0
The flux is independent
from the selected sphere