Transcript I-4

I-4 Simple Electrostatic Fields
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Main Topics
• Relation of the Potential and Intensity
• The Gradient
• Electric Field Lines and Equipotential
Surfaces.
• Motion of Charged Particles in Electrostatic
Fields.
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A Spherically Symmetric Field I
• A spherically symmetric field e.g. a field of a
point charge is another important field where the
relation between  and E can easily be calculated.
• Let’s have a single point charge Q in the origin.
We already know that the field lines are radial and
have a spherical symmetry:
 
kQ  0
E (r )  2 r
r
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A Spherically Symmetric Field II
• The magnitude of E depends only on r
kQ
E (r)  2
r
• Let’s move a “test” charge q equal to unity from
some point A to another point B. We study directly
the potential! Its change actually depends only on
changes of the radius. This is because during the
shifts at a constant radius work is not done.
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A Spherically Symmetric Field
III
• The conclusion: potential  of a spherically
symmetric field depends only on r and it
decreases as 1/r
kQ
 (r) 
r
• If we move a non-unity charge q we have
again to deal with its potential energy
kqQ
U (r ) 
r
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
The General Formula of E ( )
• The general formula is very simple
 

E (r )   grad  (r )

• Gradient of a scalar function f (r ) in some
point is a vector :
• It points to the direction of the fastest growth of
the function f.
• Its magnitude is equal to the change of the
function f, if we move a unit length into this
particular direction.
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
E ( ) in Uniform Fields
• In a uniform field the potential can change
only in the direction along the field lines. If
we identify this direction with the x-axis of
our coordinate system the general formula

simplifies to:

d (r )
E (r )  
dx


dU (r )
F (r )  
dx
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
E ( ) in Centrosymmetric Fields
• When the field has a spherical symmetry the
general formulas simplify to:



dU (r )

d (r )
and
F (r )  
E (r )  
dr
dr
• This can for instance be used to illustrate
the general shape of potential energy and its
impact to forces between particles in matter.
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The Equipotential Surfaces
• Equipotential surfaces are surfaces on
which the potential is constant.
• If a charged particle moves on a
equipotential surface the work done by the
field as well as by the external agent is zero.
This is possible only in the direction
perpendicular to the field lines.
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Equipotentials and the Field
Lines
• We can visualize every electric field by a set
of equipotential surfaces and field lines.
• In uniform fields equipotentials are planes
perpendicular to the field lines.
• In spherically symmetric fields equipotentials
are spherical surfaces centered on the center of
symmetry.
• Real and imaginary parts of an ordinary
complex function has the same relations.
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Motion of Charged Particles in
Electrostatic Fields I
• Free charged particles tend to move along
the field lines in the direction in which their
potential energy decreases.
• From the second Newton’s law:


dp
 qE
dt
• In non-relativistic case:



ma  qE  a 
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q
m

E
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Motion of Charged Particles in
Electrostatic Fields II
•
The ratio q/m, called the specific charge is an
important property of a particle.
1.
2.
3.
4.
•
electron, positron |q/m| = 1.76 1011 C/kg
proton, antiproton |q/m| = 9.58 107 C/kg
-particle (He core) |q/m| = 4.79 107 C/kg
other ions …
(1836 x)
(2 x)
Accelerations of elementary particles can be
enormous!
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Motion of Charged Particles in
Electrostatic Fields III
• Either the force or the energetic approach
is employed.
• Usually, the energetic approach is more
convenient. It uses the law of conservation
of energy and takes the advantage of the
existence of the potential energy.
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Motion IV – Energetic Approach
• If in the electrostatic field a free charged
particle is at a certain time in a point A and
after some time we find it in a point B and
work has not been done on it by an
external agent, then the total energy in
both points must be the same, regardless of
the time, path and complexity of the field :
EKA + UA = EKB + UB
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Motion V – Energetic Approach
• We can also say that changes in potential
energy must be compensated by changes in
kinetic energy and vice versa :
• ( EkB  EkA )  (U B  U A )  Ek  U  0
• Ek  q( B   A )  Ek  q  0
• Ek  q( B   A )  Ek  qVAB  0
• In high energy physics 1eV is used as a
unit of energy 1eV = 1.6 10-19 J.
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Motion of Charged Particles in
Electrostatic Fields II
• It is simple to calculate the gain in kinetic
energy of accelerated particles from :
Ek  qVAB
• When accelerating electrons by few tens of
volts we can neglect the original speed.
• But relativistic speeds can be reached at
easily reached voltages!
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Homework
• The homework from yesterday is due
Monday!
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Things to read
• This lecture covers :
Chapter 21-10, 23-5, 23-8
• Advance reading :
The rest of chapters 21, 22, 23
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Potential of the Spherically Symmetric Field
A->B
B 
rB

 ( B)   ( A)    E  dl    E (r )dr
A
rA
• We just substitute for E(r) and integrate:
rB
dr
1 1
 ( B)   ( A)  kQ  2  kQ(  )
r
rB rA
rA
• We see that  decreases with 1/r !
^
The Gradient I
f f f 

grad f (r )  [ , , ] (r )
x y z
It is a vector constructed from differentials of the
function f into the directions of each coordinate
axis.
It is used to estimate change ofthe function f if we
make an elementary shift dl.
The Gradient II





f (r  dl )  f (r )  dl  grad ( f (r ))
The change is the last term. It is a dot product.

It is the biggest if the elementary shift dl is
parallel to the grad.
In other words the grad has the direction of
the biggest change of the function f !
^
The Acceleration of an e and p I
What is the acceleration of an electron and a proton
in the electric field E = 2 104 V/m ?
ae = E q/m = 2 104 1.76 1011 = 3.5 1015 ms-2
ap = 2 104 9.58 107 = 1.92 1012 ms-2
[J/Cm C/kg = N/kg = m/s2]
^
The Acceleration of an Electron II
What would be the speed of an electron, if
accelerated from zero speed by a voltage (potential
difference) of 200 V?
ve 
mv2
Ek  Ek 
 (e)V
2
2Ve
11
6
1
 400 1.76 10  8.39 10 ms
m
Thermal motion speed ~ 103 m/s can be neglected
even in the case of protons (vp = 1.97 105 m/s)!
^
Relativistic Effects When Accelerating
an Electron
Relativistic effects start to be important when
the speed reaches about 10% of the speed of
light ~ c/10 = 3 107 ms-1.
What is the accelerating voltage to reach this
speed?
Conservation of energy: mv2/2 = q V
V = mv2/2e = 9 1014/4 1011 = 2.5 kV !
A proton would need V = 4.7 MV!
Relativistic Approach I
If we know the speeds will be relativistic we
have to use the famous Einstein’s formula:
E  mc  m0 c  EK  m0 c  qV
2
2
2
E is the total and EK is the kinetic energy,
m is the relativistic and m0 is the rest mass
m  m0  (  1)m0 c 2  qV 
1
qV




1
2
v2
m0 c
(1  c 2 )
^
Relativistic Approach II
The speed is usually expressed in multiples of the c
by means of  = v/c. Since  is very close to 1 a
trick has to be done not to overload the calculator.
 
1
(1   )
2

1

(1   )(1   )
1
2(1   )
So for  we have :
1
  1 2
2
^
Example of Relativistic Approach
Electrons in the X-ray ring of the NSLS have
kinetic energy Ek = 2.8 GeV. What is their
speed. What would be their delay in arriving
to -Centauri after light?
E0 = 0.51 MeV for electrons. So  = 5491 and
v = 0.999 999 983 c. The delay to make 4 ly
is dt = 2.1 s ! Not bad and the particle
would find the time even shorter!!
^