CHARGE AND FIELD

Chapter 16

Equations

mqVv 2

2mcE c

Ep

qFE

VqW qEF xVE

qvBF

qBpr

204 r

qE

rqV

r 04:1

2

0

21

4 rqqF

Linear AcceleratorsChapter 16.1

between two flatcharged plates

+_

–

Shapes of electric fields

Acceleration: gravitational and electrical

A falling ball An accelerating charge

starts at highgravitationalpotential energy

gravitationalfield g

height h

gainskineticenergy

mass m

ends at lowgravitationalpotential energy

gravitationalpotential

ball falls down gravitationalpotential hill

gain of kinetic energy = loss of potential energy = mgh

electricpotential

charge ‘falls down’ electricalpotential hill

gain of kinetic energy = loss of potential energy = qV

starts at highelectricalpotentialenergy

potentialdifference V

gainskineticenergy

electricfield

ends at lowelectricalpotential energy

+

–

+

An electric field accelerates a charge as a gravitational field accelerates a mass

+charge q

forces oncharges fromcombinedattractions andrepulsions bycharges onplates

Forces act across empty space Electric field: forces act locally, field ‘fills space’

Two ways of describing electrical forces

Action at a distance Action via electric field

+

+

+

+

+

+

+

+

_

_

_

_

_

_

_

_

+ _

repelforces oncharges fromelectric field

+

+

+

+

+

+

+

+

_

_

_

_

_

_

_

_

+

–

+ _charges onplates produceelectric field

Defining electric field

+charge q force F

E = F/q

unit of E is N C–1

field E

attract

repel–attract

+

Field lines and equipotential surfaces

A uniform field

Field lines are always perpendicular to equipotential surfaces

1000V 0V750V 500V 250V

+

1000 V+ –

Equipotentials near a lightning conductor

field lines

equipotentialsnear conductor

equipotential followsconductor surface

+

+

no force in thesedirections so potentialis constant

force in thisdirection, sopotentialchanges

V

V +V V

electricfield field is at right

angles toequipotentialsurface

equipotentialsurface

Field strength and potential gradient

Field and potential gradient

slope = change in potential

= – V

distance x

electric field E = – Vx

or E = – dV if slope varies continuouslydx

negative slope is downhill, decreasing potential

1000 V 0 V750 V 500 V 250 V

1000 V+ –

1000

750

500

250

0 x

V

UNTAINAURANT

SKI CENTRE

Aonachan Nid

Ski Tows

Field strength and potential gradient

Contours and slopes

Slope is steep wherecontours are close.Direction of steepestslope is perpendicular tocontours

walk along contour to stay at same height

slope = change in height

= – h

distance x

negative slope is downhill, decreasing height

The accelerating field

field betweenelectrodes

zero fieldinside tube

negative electronaccelerated

+––

Principle of linear accelerator

Principle of linear accelerator

The alternating p.d. switches back and forth so that the electrons are accelerated as they pass between successive electrodes

Switching p.d.s to keep accelerating electrons

at one instant alternating highfrequency p.d.

+

–

bunches of electrons betweenelectrodes are accelerated

a little later

a little later still+

–

+ +– – –

bunches of electronsdrift through tube

bunches of electrons betweenelectrodes are further accelerated

+ +– – +

electrodes must be longer becauseelectrons are going faster

zero p.d.

–+

+–

When drop is held stationary:electric force F = gravitational force W

F = qE W = mg

Uniform electric fieldE = Vd

qE = mg

V = mgdq

V = mgdne

Vn = constant

If charges q arediscrete multiples n ofelectron charge e, thenq = ne

VdF = qE

W = mg

Charge on drop is changed by ionising air incell, using weak radioactive source

light source

Discreteness of charge: value of e

oil drop

5 mm

microscope

Millikan's ExperimentUsing this method Millikan was the first person to measure the charge on an electron (e)

e = 1.6 x 10-19 C

The ultimate speed: Bertozzi’s demonstration

The results:

9

6

3

00 2 4

accelerating p.d./MV

The difference made by relativityAs particles are accelerated speed v reaches a limit, c kinetic energy EK increases without limit momentum p increases without limit

At low speeds

At all speeds

At high speedsp mvEK mv2

EK = qV

v c12 EK pc

actual speedof electrons

speed calculatedfrom mv2 = qV1

2

speed of light

oscilloscope

aluminium platedetects electronsarriving. Rise intemperaturechecks energybunch of

electrons

tube detectselectronspassing

acceleratedelectrons

8.4 m drift space

time

Powerful accelerators can’t increase the speed of particles above c, but they go onincreasing their energy and momentum

v

The Ultimate limitNewtonian physics predicts that it is very easy to get electrons up to and beyond the speed of light – but this is never observed

No matter how hard you try you cannot exceed the speed of light

Deflecting charged beams

Chapter 16.2

Deflections of electron beam by electric field

electron gun deflection plates

zero potential

hotcathode

spacing d

_

anode+ +V

acceleratingpotential difference

vertical force F = eE = e Vd

_

horizontalacceleration

_

constant velocity

_

vertical accelerationconstant horizontal velocity resultant velocity

_

force

F = e Vd

electric field E = Vd

electron gun

vertical deflectionplates (Y plates)

horizontal deflectionplates (X plates)

oscilloscope screen

horizontal deflection controlledby time-base

verticaldeflectioncontrolled bysignal

Deflection plates in CRO

Magnetic deflection

force

beam velocity

B-field

positiveions

force at right angles tofield and velocity ofcharge, F = qvB

force onpositive charge

Bv

Magnetic fields deflect moving charged particles in circular paths

+

–

negativecharges

e.g. electrons

current = charge flowtime

t

IL = q L = qvt

I = q

Force on current: force on moving charge

The force which drives electric motors is the same as the force which deflects moving charged particles

Electric motor Moving charge

Force on current I in length L

F = ILBForce on charge q at velocity v

B-field

electriccurrent

L

F F

qv

+I

F = qvB

Measuring the momentum of moving charged particles

force F

m v2 = qvBr

p = mv = qrB

Momentum of particle proportional to radius of curvature of path

at relativistic speedp = qrBis still true butp > mv

magnetic force

force F = qvB

motion in circle

m v2 = force Fr

velocity v

velocity v

circular path,radius r

charge q

magnetic field Binto screen Magnetic Fields

X is used to represent a field directed into the page

Dots are used to represent a field directed out of the page

Principle of synchrotron accelerator

radio frequencycavity toaccelerate beam

magnets to focus beam

electrostatic deflectorto extract beam anddirect it into targets

magnets to bend beam

electrostaticdeflector

inject beam at v c from smalleraccelerator

Electromagnetic waves from accelerating charges

radio ariel

e-mradiation

E-field

B-field

speed c

direction of travel

charges oscillatingin dipole aerial

N S N S

N NS S

intense e-mradiation used inmedical research

beam wiggler

alternating N-S and S-N poles deflect beam up and down

Synchrotron

a synchrotron radiatesbeacuse the particles areaccelerated towards the centre

synchrotronradiation

accelerationtowardscentre

When charges are accelerated, they emit electromagnetic waves

Synchrotron RadiationThis is a problem as it wastes energy and the radiation can be dangerous.

Particle accelerators are usually designed to minimise the amount of synchrotron radiation but there are exceptions e.g. Diamond Light

The Diamond Synchrotron – near Didcott

between church steeple andthundercloud

between conductorsin coaxial cable

Shapes of electric fields

+ +

+

+

Electric fields with cylindrical symmetry

From plane symmetry to cylindrical symmetry

bend incurve make

concentric

Field of a charged wire

radial fields

equipotentials

field very intense closeto wire. Equipotentialsclose together

L

r surface area2rL

Same number of field lines throughany cylinder surounding chargedwire

E 1r

++

+

++

+

+

Two ways of saying the same thing

Inverse square law and flux of lines through a surface

Experiment Inverse square law Gauss’ idea – flux of linesmeasure E-field of a charge atdifferent distances r

force F

test charge++q

r

F q

F 1r2

experiments done by Coulomb(1780s)

E qr2

Think of lines of E as continuous.Number of lines through area ofany sphere q

E density of lines

E q4r2

E kqr2

Example: field between parallel plates

field E

area A

charge density per unit area

E = density of lines

no. of lines through area A is EAno. of lines = charge enclosed/0charge on area A is q = A

EA = q =

A

0

0

E =

0

experimentally k = 8.99 109 V C–1 m

k 14

0

0 = 8.85 10–12 C V–1 m–1

+q

+q

0 = constant

– – –––

– –

E q4

0r2

+ + ++

+++

area 4r2

number of lines = charge enclosed

0

Gravity and electricity – an analogy

Vgrav

r

rad ius r

00

r

fie ld g = –slope= –Vgrav /r

V grav

Velec

r

rad ius r

00

r

fie ld E = –slope= –V elec/r

V elec

Gravitational potential and radius Electric potential and radius

Relationships between force, field, energy and potential

ElectricityInteraction oftwo charges divide by

charge

Behaviour ofisolated charge

Forceunit N

F = q1 q24

0r2

Fieldunit N C–1

E = q

40r2

Potential energyunit J

potential =

q1 q24

0r

Potentialunit V = J C–1

Velectric = q

40r

slope ofgraph ofpotentialenergyorpotentialagainst r

areaundergraph offorce orfieldagainst r

multiply by charge

r

radius r

00

g

Gravitational field and radius

r

gVgrav = area gr

r

radius r

00

E

r

EVelec= area Er

Gravity and electricity – an analogy

Electric field and radius

HintsQuestion sheets

Hints for 10W

5. Remember to convert millimetres to metres.

7. Use the magnitude of the electronic charge. The negative sign is best ignored in this calculation.

8. Remember to convert electron volts into joules.

Hints for 80W

3. Suppose a length XY of the beam will pass Y in time t. What is the distance XY?

Hints for 60S

10. 1 mg = 10–3 g and 1 g = 10–3 kg. The sideways force on the particle is constant.

Hints for 90S

1. Remember that the left-hand motor rule works for conventional current.

2. Remember that the left-hand motor rule works for conventional current.

Hints for 150S

5. Think about the charge and mass of the proton.

10. What happens to the speed of the proton as it loses energy?

Hints for 160S4. The charge and mass of the electron are given in the ‘Instructions and information’.

6. The electric and magnetic forces are equal in magnitude.

7. You will need to change the units before calculating the electric field strength.

8. You will need to consider the equation for the electron gun used to accelerate the electrons.

9. You are used to calculating the speed of electrons when accelerated. The mass of a deuteron is not the same as the mass of an electron.

10. The question gives the diameter of the orbit and not the radius..

Hints for 180S4. The value of 1 /(4 pi e0) is given under ‘Instructions and information’.

5. Think about how the electric field strength varies with distance.

6. Think about how the electrical potential varies with distance.

7. The value of 1 /(4 pi e0) and the electronic charge are both given under ‘Instructions and information’.

8. The value of 1 /(4 pi e0) and the electronic charge are both given under ‘Instructions and information’. Remember that the proton and electron do not have the same sign of charge.

Hints for 190S1. 1 mg = 10–3 g and 1 g = 10–3 kg.

4. Consider the forces between identical charges.

6. Remember that potential varies inversely with distance.

7. Remember that potential varies inversely with distance.

9. There will be forces between the positive charges on the spheres.

Hints for 200S

1. 1 g = 10–3 kg

3. The potential will be the value at the surface of the sphere when it has the charge calculated. Use the value of charge you obtained before rounding to one significant figure.

Hints for 250S2. What would be inside the tube if it was not evacuated?

4. You will need to use the electronic charge given in the instructions and information.

5. Remember that force equals rate of change of momentum.

11. Start by eliminating velocity v from the equations given.

Hints for 20M3. Remember that the electric field is uniform between parallel plates.

4. What factors determine the force on a charge in a uniform electric field?

5. What determines the direction of an electric field?

6. What factors determine the acceleration of a charge in a uniform electric field?

7. The mass of the ion depends on the total number of protons and neutrons in the nucleus.

8. It may help to think of C V as charge when considering the responses D and E. This is not the same charge as q of course.

10. Use V = W / q to check the potential difference.

Hints for 110M1. Consider what would happen if the electrons travelled at different speeds.

2. What is necessary to change the direction of the beam?

3. Remember that work done is the product of force and the direction moved in the direction of the force.

4. How much deflection can an electron have in a uniform electric field?

5. Start with q v B = m v2 / r.

6. Start with q v B = m v2 / r.

7. What is the resultant force on a charge carrier when the potential difference, V, is steady?

8. What determines the radius of the track?

Hints for 240M1. Which factor will have no effect on the force acting on the electron or the time for which that force acts?

2. How is v related to V?

3. Try to visualise the equipotential lines.

4. Start by using B e v = m v2 / r.

7. Remember that E is proportional to D V / D r and E is constant in this case.

9. Think about ways of making the force due to the magnetic field both very small and very large.

Hints for 260M1. You will need to assume that the dome is a sphere. An estimate of the radius of the sphere will be necessary and you will need to estimate the electric field strength at the surface.

2. Start by calculating the speed of the electrons as they leave the electron gun. And then consider both the magnitude and the direction of the forces due to the gravitational and magnetic fields. You will need to assume that the electrons experience a constant acceleration due to both forces, and calculate the displacements resulting from each.

3. Think about the charge on the nucleus once the alpha particle has been released. You will also need to estimate the radius of a uranium nucleus.

4. You will need to estimate the size of a nucleon.

5. You will need to estimate the size of an atom and assume the maximum photon energy would be the same as the potential energy of an atomic electron.

Hints for 40C

Estimate the number of lightning flashes from a Thundercloud over a 30 minute period using data from the section entitled ‘The thundercloud’.

9. Start by calculating the force on a single ion.

Hints for 50C

9. The electric field strength is given in the paragraph.

10. See the second figure. ‘Noise’ is rapid changes (fluctuations) in the current, which make the graph of current against p.d. look spiky.

12. Consider each microparticle as a cube of side 1 mm

Hints for 70C

4. What can you say about the forces involved?

6. Remember that d is given in mm.

Hints for 130D4. Remember that the current is the total charge per second. The total charge is N e in a time t, where N is the number of protons passing in a time t.

6. Think about the direction of the field across the gap at the accelerating point.

7. How many accelerating points does the proton pass in each revolution?

8. Compare the energy at the start with the energy at the end.

9. How long would the linear accelerator be?

12. What happens to the time for the protons to move from one accelerating electrode to the next one?

13. You should have established that B is directly proportional to momentum in this case.

Hints for 170D

1. When entering a number such as 2.5 ´ 10–9 in a spreadsheet cell you should write 2.5E–9.

4. Remember that electric field is a vector. The value of the resultant field at d = 2 ´ 10–10 m should be 1.28 ´ 1011 N C–1.

Hints for 210D

1. The numbers on the ruler represent centimetres. The relationship only requires distances and forces to be measured in arbitrary units but it may help you with question 3 if the distances are measured in metres at this stage.

4. You must use SI units for this part. A good approach would be to work from the gradient of your graph drawn in question 1 but you must make sure that r is measured in metres (or a conversion factor is applied afterwards). The force F is W tan q which becomes W d since tan q = d / 1. By converting d on the graph into F in newtons, the gradient will yield the value of [(force constant) ´ (charge on one ball) ´ (charge on other ball)] and hence the force constant.

Hints for 30X

1. You could start by calculating the voltage needed to cause a spark in air at this electrode spacing. You should think about the molecular nature of a gas. Think about the energy needed to ionise an air molecule.

2. Start by considering the processes involved in accelerating ions.

Hints for 120X

1. You might start by calculating the velocity of the electrons as they leave the electron gun.

2. You might like to start by explaining why a charged particle can move in a circular path in a magnetic field.

Hints for 230X

1. You could start by calculating the sizes of the forces for two protons within the nucleus.

2. You could start by considering the energy changes involved. Treat the protons as if they were particles in an ideal gas.

Hints for 120X

1. You might start by calculating the velocity of the electrons as they leave the electron gun.

2. You might like to start by explaining why a charged particle can move in a circular path in a magnetic field.

Hints for 120X

1. You might start by calculating the velocity of the electrons as they leave the electron gun.

2. You might like to start by explaining why a charged particle can move in a circular path in a magnetic field.