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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:

02)( r

rkQ

rE

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A Spherically Symmetric Field II

• The magnitude of E depends only on 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.

2)(rkQ

rE

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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

• If we move a non-unity charge q we have again to deal with its potential energy

rkQ

r )(

r

kqQrU )(

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The General Formula of

• The general formula is very simple

• Gradient of a scalar function 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.

)()( rgradrE

)(E

)(rf

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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:

dx

rdrE

)()(

dx

rdUrF

)()(

)(E

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in Centrosymmetric Fields

• When the field has a spherical symmetry the general formulas simplify to:

and

• This can for instance be used to illustrate the general shape of potential energy and its impact to forces between particles in matter.

dr

rdrE

)()(

dr

rdUrF

)()(

)(E

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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:

• In non-relativistic case:

Eqdt

pd

EaEqam mq

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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. electron, positron |q/m| = 1.76 1011 C/kg

2. proton, antiproton |q/m| = 9.58 107 C/kg (1836 x)

3. -particle (He core) |q/m| = 4.79 107 C/kg (2 x)

4. other ions …

• 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 :

• • • • In high energy physics 1eV is used as a

unit of energy 1eV = 1.6 10-19 J.

0)()( UEUUEE kABkAkB

0)( qEqE kABk

0)( ABkABk qVEqE

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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 :

• When accelerating electrons by few tens of volts we can neglect the original speed.

• But relativistic speeds can be reached at easily reached voltages!

ABk qVE

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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

Potential of the Spherically Symmetric Field A->B

• We just substitute for E(r) and integrate:

• We see that decreases with 1/r !

rB

rA

B

A

drrEldEAB )()()(

rB

rA AB rrkQ

r

drkQAB )

11()()(

2

^

The Gradient I

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 .

)(],,[)( rz

f

y

f

x

frfgrad

ld

The Gradient II

The change is the last term. It is a dot product. It is the biggest if the elementary shift is parallel to the grad.

In other words the grad has the direction of the biggest change of the function f !

))(()()( rfgradldrfldrf

^

ld

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 IIWhat would be the speed of an electron, if accelerated from zero speed by

a voltage (potential difference) of 200 V?

Thermal motion speed ~ 103 m/s can be neglected even in the case of protons (vp = 1.97 105 m/s)!

^

Vemv

EE kk )(2

2

1611 1039.81076.14002 msm

Veve

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 VV = 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:

qVcmEcmmcE K 20

20

2

E is the total and EK is the kinetic energy,

m is the relativistic and m0 is the rest mass

1)1(

1

)1(

20

200

2

2

cm

qV

qVcmmm

cv

^

Relativistic Approach IIThe 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.

So for we have :

)1(2

1

)1)(1(

1

)1(

12

^

22

11

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!!

^